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eHC^ 1
WENTWORTH'S SERIES OF MATHEMATICS First
Steps
in
Number.
Primary Arithmetic.
Grammar School Arithmetic High School Arithmetic. Exercises
in
Arithmetic.
Shorter Course
in
Algebra.
Elements of Algebra.
Complete Algebra.
College Algebra.
Exercises
in
Algebra.
Plane Geometry.
Plane and Solid Geometry. Exercises in Geometry. PI.
and
Sol.
Geometry and
PI,
Trigonometry.
Plane Trigonometry and Tables. Plane and Spherical Trigonometry, Surveying. PI.
and Sph. Trigonometry, Surveying, and Tables
Trigonometry, Surveying, and Navigation.
Trigonometry Formulas. Logarithmic and Trigonometric Tables (Seven). Log. and Trig. Tables (Complete Edition). Analytic Geometry, Special
Terms and Circular on
Application,
PLANE AND SPHEEICAL
TRIGONOMETRY, AND
SURVEYING BY G. A.
WENTWORTH,
A.M.,
PBOFEBSOR OF JE^^MATICS IN PHILLIPS EXETER
^
ACADEJttT.
2Ceacl)ers* lEtiitton*
BOSTON COLLEGE PHYSICS DEPT. BOSTON,
U.S.A.:
GINN & COMPANY, PTTBLISHEES. 1891.
D/^ Jj
-y
t
X
Entered, according to the Act of Congress, in the year 1886, by
G, A.
h4
WENTWORTH,
in the Office of the Librarian of Congress, at Washington.
Typography by
J. S.
Gushing &
Prksswork. by Ginn
JUN19
&
Co.,
Boston, U.S.A.
Co., Boston, U.S.A.
/
PREFACE.
THIS
edition
is
intended for teachers,
publishers will under no
and for them
circumstances
teachers of Wentworth's Trigonometry
;
sell
only.
The
the book except to
and every teacher must con-
sider himself in honor bound not to leave his copy where pupils can
have
access to
Messrs. Ginn It
is
it,
and not
to sell his
copy except to the publishers,
& Company,
hoped that young teachers will derive great advantage from
studying the systematic arrangement of the work, and that ers
who
obliged
are to
pressed for time will
work out every problem
find in
great relief
the
all
teach-
by not being
Trigonometry and Sur-
veying. G. A.
^\
WENTWORTH.
TEIGONOMETET. Exercise 1.
What
are the functions of the
other acute angle
ABC{Yig. sin
2)?
B = -1 c
B
of the triangle
I.
Page
5.
1)
TRIGONOMETRY. ,
(
.,
—
A = 120
A, = 119
.
VI.) Sin
^
cos A
tan
—
A=
A = 120
,
cot
-
,
A = 169
CSC
>
169
A
119"
120 4. filled
What b,
make them angle? in
condition must be ful-
by the lengths
lines a,
Is
Example
c (Fig.
5.
of the
three
order to
2) in
the sides of a right this condition
3 a^
tions
.
119
120 sec A
.
169
169
^
tri-
fulfilled
?
+
b^
=
c^.
Find the values of the funcof A, if a, h, c respectively
have the following values (i.)
(ii.)
2mn, m^ — n'^,
_^,
x-y
x-vy,
mn?
:
-\-
n^.
TEACHERS EDITION. 6.
in
c,
Prove that the values of a, h, and (ii.), Example 5, satisfy
them the _^ 52
—= 24
cos B,
— =— =
cot B,
—=
tan B,
„ A = 143 = sm B,
cos J
sides of a right triangle. (i-)
a2
=
145
make
necessary to
condition
the
sin J.
(i.)
tan J.
^ g2^
.
145 94 143
mnf + (m^ — n^Y = (m'^ + n^)^, 4 rn^n^ + m* — 2 m?v? + n* = m* + 2 m^v? + n*, (2
cot J.
=
24
—
o A = 145 = CSC B,
sec A
•
143 A CSC -4
(li.)
2xy\
3^ x-yj •2
x^
o — Axy
+
+(aj
x^
+
3/^
-+
+
x2
2a;2/+2/2
B when a = 0.264, c = 0.265. ^2 = g2 _ 0^2
_ a;*+2a;V+.V^ — 2 + 2/2 a;?/
a;''
4a;y + a;*-2a;y + 2/* = a;* + 2cey + 2/*, a* + 2a;y + 2/* = a;* + 2a;y + 2/*. 7.
What
must be
.-.
= 0.070225-0.069696 = 0.000529. 6 = 0.023.
a, 5, c, in (iii.)
cos
^=- =
tanJl
=?=
sent the sides of a right triangle.
cot .4
or
Jp^
_|_
+
g2j,2g2
2^8^
_
a
^=
csc^ =
^
= ?^ = sec.g.
a
(iv.)
csc5,
23
6
=pH'^.
cot5,
264
sec^ = ^ =
7'2g2p2^
^=
— =tan5,
=- =
(iii.)
sin B,
23
6
repre-
—= 265
c
by the values of and (iv.), Example 5,
may
265
c
equations of condition
in order that the values
?^ = cos5,
sin^ = ^ =
satisfied
^2g2^2
264
2/1 12
t2/>-)2 1
10.
m^n's^
or 8.
and
A
Compute the functions of
9.
and
y^
\-
r>
B.
24
=
2/)'
—
= 145 = sec
+ m^p^v^ = n^(fr^.
Compute the
fun<;tions of J.
B when a = 24, & = 143. c
= =
\/(24)2
+
\/21025
= 145.
(143)2
and
Compute the functions oi
B when 6 = 9.5, c = 19.3. a^ = c^ — 52 = 372.49 - 90.25 = 282.24. .-.
a
= 16.8.
sin.4=^=l^ = c
193
cos5,
A
TRIGONOMETEY. 95 A= 5- = — .
cos
tan
cot
sec
b
95
^=-
95
a
168
CSC
2
"
Vq^ + pq
«
\/pi
tan 5,
b
95
c
193
a
168
—=
^=- =
.
P+
cot B,
—=
^=-=
+pq y/f i i-i = sm B, A = -b = — .
cos
c
—=
A=- =
.
.J, = sm ij,
193
c
CSC
is,
sec
D B.
-j-pq
sec
AA = ^ = fP + Q^ - = CSC B, " Vq^ +
CSC
A= _ p+ q = ^ Vp^ + pq
T>
j-
pq
c
Compute the functions
11.
and
of
A
B when = Vp^ + g^,
a
b
= V2pq
13.
and
+ 2pg' + p + q = c. p^
sin
cos
A=
=
^-^
A=- = c
Compute the functions
a2
= 2->/pq, + 52 = c\
a?
+ ipq =p'^ + 2pq +
a
=p — q'
b
c'^.
q
\/2pq
sin B,
p+q
sm
tI
cos J.
= - = ^2pq = tan B, ^
y/p"^
o
y/2pq
+
c
^ 12.
and
_ p
-\-
y/p^
tan -4
B,
cot J.
_ = sec
of
c=p + q,
A
CSC
14.
A = a-
2V^q p P+
tan 5,
.
pq P+9 =
p-q
CSC
B,
= 2&, + 62 = c2, + 62_c2,
462
= c2, C-6V5.
562
(
sec B.
2 6.
a
q
.
Compute the functions
when a =
a2
Vp2 +pq ^ —^ ^ i-2 = cos B, AA = -=
P+
= cos B,
9.
(^
= c^ — Q? = (f -{-pq. b = Vq^ + pq. c
^
9.
sec J. == ^
5,
b^
sm
=-= a
B when
.'.
— —
^y/pq
Compute the functions
a=Vp'^ +pq,
q^,
p+q „ i-i = sm B, = 2Vpq p+q pj-q cot B,
q^
q
+
= -b c
= esc
^
=p + q.
c
=- =^ c
.
CSC
A
cos 5,
V2pq cot J.
of
B when
a
p+
sec B.
of
A
TEACHERS EDITION. sm CO?
-d.
= -a = _2b_ = f V5 = 0.89443,
16.
Compute the functions
when a +
b
A=
= ^c. _ g2^ + 2a5 = ffc2, +
J2
a2-2a6 +
62
a2
bV5 a2
+
62
2.
b
cotA
b
=- =h a
2
b
a b
cscA = '- = ^'=iV5. 2b
a
15.
^
Compute the functions
when a =
I
of J.
c.
a
= fc,
^ ^2 _ 6 = Vc^ — a^
52
0^2^
2
sm
-, ^ = - = —- = 3' 3
p
&
cosul
=- =
tan J.
=-= ^
a
2
2^ — = ^\/5, -T
= fV5,
^V5
cot^ = ^
= :^.
^=-
fa ^"'
sec
2
csc^ = ^ = ?. a
2
=fV5,
= _7_c2,
—b=
of
A
'
TRIGONOMETRY.
6 17.
Compute the functions
of J.
CSC
when
^=-= «
a
V3I +
4 18.
a2-2a5 +
Z.2
+
52
G
_
2ab
+ + 2a6 +
a
6
=
c
g
16
20.5
5a =
I.
19.
I'
4
2a = -V31+-4
4 ..
a
26
= -(\/31 +
= - VSI - -. 4
4 .-.
6
l).
= |(V3l-l).
^(V3l +
1)
5'
61.5,
a =12.3.
= 31 c^
— = -,
5
^3
15c^
- Vsl, 4
sin J.
if
sin^ = ^ = ^.
16
a +
a
c2
&2
62
Find
= 20.5.
16
a2
1
—6=—
=
|
and
.
; ;
TEACHERS 23.
a
=
Find
CSC J.
.
c
6.45
esc J.
if
and
35.6
EDITIOISr.
28.
= - = -^ = a
35.6
c =
229.62.
pute the
6.45.
sm •
.-.
24.
given
Construct a right triangle c
= Q,
tan^ = a
.*.
Draw
=3
;
=
-.
3 and &
AB -= 2,
join
=f
tan -4
= 2.
and .5C
± to ^5
Cand A.
Prolong vie to D, making
AD = 6.
DE 1. to AB produced. A ADE will be similar to
Draw Rt.
rt.
^ACB. .'.
ADE
is
the
rt.
A required.
25. Construct a right triangle given a = 3.5, cos A = ^.
;
A A^B'C so that y=l, cos A = Construct A ABC similar to Construct
c^=
2.
Then
>}.
= 3.5.
A^B'C^, and having a
26. Construct a right triangle given 5 = 2, sin J. = 0.6.
Construct
rt.
A
A^B^C^, m'aking
a^= 6, and c = 10. Then sin A^= j%. Construct
A ABC
A^B'C^, and having
6
=
similar
to
2.
27. Construct a right triangle; given & = 4, cscA = 4.
A
Construct rt. 4 and a^= 1.
A^B^C'', having
c''=
Then construct A J.5(7 similar and having 6 = 4.
A A^B'C,
to
c
= 2.5
cos A = 0.8;
com-
In a right triangle,
miles, sin J.
= 0.6,
legs.
^ A — A= e
TRIGONOMETRY.
Exercise
II.
Page
Eepresent by lines the func-
1.
8.
Construct the angle x
x
if esc
tions of a larger angle than that
shown
in Fig.
Let O
3.
ABM
be a unit circle, with 0; construct -ST tangent to
centre
B = 2 OA
the circle at
then J. OT
OT;
0M= I radius
Take at P.
that sinx
is
less
than
tana;. 1,
but .'.
3.
OM: PM::OA:
AS,
0M< OA. PM< AS.
Show that sec x is
In
rt.
.'.
sec
4.
>
Show
greater than
OaS'> side
AS
greater than
cot>r.
OT=csc, BT=cot In A BOT, Hyp. 0T> .'.
CSC
>
Hence, construct an
Z
Construct the angle
Construct
side
rt.
PM= 2 OM.
x
x.
45°.
a;
if
sin x
Z PMO, making
Draw OP.
POM\b
Construct
10.
4 sin X is
if sin
a;
PM= sin x and 0M= cos
Then
tan.
that esc x
angle required.
by hypothesis, PM--= OM. by Geometry, x = 45°.
9.
AS=id.n.
A OAS, Hyp.
POif the
= 2 cos X.
tana;.
0/S'=sec,
is
But, .•.
In Fig.
M
cos X.
Let
Show
2.
At
OA.
meet the circumference Draw OP.
Construct the angle
8. 1.
cos x
if
± to
Then
Fig.
a;
2^-
erect a
=
connect
;
required angle.
Construct the angle
7.
—
is
the angle required. the
angle
if
= tan x.
Take J J/ erect a
of radius
OA
to
M.
At
meet the circumferDraw OP. ence at P. Then POil/ is the required angle. _L to
BT. 11.
cot.
angle
Show is
that the sine
of
an
equal to one-half the chord
tan x of twice the angle. Have given Z POA. Construct POB = 2 POA. Draw unit circle, he a with Let Then it is ± to OA tan- chord PB. centre 0; then construct and PM, its half, is the sine of POA. gent to the circle at J. = 3 OA .•. sin a; = 2 chord 2a;. then J. or is required angle. 5.
Construct the angle x
if
O BAM
AT
•
;
;
TEACHEES EDITION.
9
AB
Let be the sine of the Z a; in 12. Find x if since is equal to one-half the side of a regular in- a circle whose centre is 0. Draw perpendicular to the scribed decagon.
AC
Let AC\)Q a side of a decagon.
Then
360°
=
vertical diameter.
Then CO = AB. Take CF on vertical diameter
or^Oa
36°
10
=
Draw
CO.
FD
perpendicular to
Draw OB bisecting AO. Then Z ^0(7 will be bisected, and Z ^05
vertical diameter,
and meeting
cir-
=
cumference at D. Draw perpendicular to
OB
18°.
But
and draw OD. 0F= 2 CO by construction.
AOB = 18°.
X or
.'.
DE
oiAOB=^\AQ.
the sine
ED=FO;
FO being the projecGiven x and y {x + y being tion of the radius OD. construct the value less than 90°) .-. = 2 AB, and DOB = &nglQ — sin of sin 13.
DE
;
{x
Let
+ y)
x.
required.
AB = sin
whose centre
(a;
0,
is
+ y) and
in a circle
CD = sin
x.
16. Given an angle x construct Then, with a radius equal to CI), an angle y such that cos 3/ = J cos a;. describe an arc from B, as centre, Let OB = cos AOB. cutting &t E. Erect a ± CD at C, the middle Then BJA will be the constructed point of OB, and meeting the cir— value of sin (x + y) sin x. cumference at D. Draw DO. Then DOB is the angle required. 14. Given x and (x + y being ;
AB
y
less
than 90°)
of tan (x
+
construct the value
;
y)
—
sin {x
+
y)
+
tan x
— sin X. Let
CI)
also
and
AB
Let be the tangent of x. Prolong to C, making AB, and draw OC from 0, the
AB
a;
3
3/),
0F=i2,xxx.
AC=
centre of the circle.
F with
a radius
= AB
take
From ^with
a radius
=
GF
add
From
;
AB = sin {x + y), = sin EF = tan {x +
and
17. Given an angle x construct an angle y such that tan 3/ = 3 tan a;.
COA
is
the required angle.
18. Given an angle x construct an angle y such that sec y = esc a;. ;
HI.
From / with
a radius
= CD
take
Since sec
Then ^^will be the constructed
£
— sin {x + y) +
b
value of tan {x tan X
+ y)
— sin x.
.'.
a
=
esc,
—£ a
=
b.
Hence, construct an isosceles right 15. Given an angle x construct triangle. an angle y such that siny = 2 sin a?. The required angle will be 45°. ;
:
;
:
TRIGONOMETRY.
10 19.
Show by
2 sin J.
construction
that
sin ( J.
.-.
+ P)
> sin 2 A.
< sin ^ + sin B.
Construct Z BOC and Z COA 21. Given sin a; in a unit circle each equal to the given Z A. find the length of a line corre^^^'^ = 2 sin A, and AD, the ing in position to sin a; 11 .„uio Then to OB, = sin 2 ^. whose radius is r. J_ let fall from ^
AB
A
But AB > AD. Hence 2 sin J.
> sin
1
r sin x required line. length of line required = r sin
:
2 A. .•.
:
:
:
x.
Given two angles A and B 22. In a right triangle, given the being less than 90°), show hypotenuse c, and also sin A = 7?i, that sin {A + B) sin A + sin B. = Z ^, and COH cos A = n find the legs. Construct 20.
{A^ B
<
HOK
;
= ZB. Then
(A + B)
sin
=
CP, sin
A=
UK, sin ^ = CD. Now CP
(7P<
c .'.
cos
EK>DE.
and
sm ^ = - = m. a
=
cm.
A — - = n. c
CD+HK. Exercise
.'.
III.
Page
b
=
en.
11.
2. Express the following funcExpress the following funcas functions of the comple- tions as functions of an angle less than 45° mentary angle 1.
tions
sin 30°.
CSC 18°
cos 45°.
cos37°24^
tan 89°.
cot82° 19^
cot 15°.
CSC 54°
10^
46^
= cos (90° - 30°) = cos 60°. cos 45° = sin (90° -45°) = sin 45°. tan 89° = cot (90° -89°) = cot 1°. cot 15° = tan (90° -15°) = tan 75°. esc 18° 10^= sec (90° - 18° 10^ = sec 71° 50^. cos 37° 24^= sin (90° - 37° 24^ = sin 52° 36^. cot 82° 19^= tan (90° - 82° 19^ = tan 7° 41^ esc 54° 46^= sec (90° - 54° 46^ = sec 35° 14^
sin 30°
sin 60°.
;
TEACHEES EDITION. 3.
tan 30°
Given
= ^ V3
;
Given cot ^ J. = tan A find A.
8.
find
11
;
cot 60°.
tan
= cot(90°-30°) = cot 60°. .•.cot60° = J\/3,
J^ = 90°-^, ^ = 180° -2^, 3^ = 180°.
tan30°
.-.
4.
Given tan tan
90°
;
find
A.
A = cot (90° - A),
9.
find
- J. = A, 2^ = 90°. .-.
5.
A = cot A
A = 60°.
Given tan (45° + ^)
=
cot
A
^
A = tan (90° - A), -A) = tan (45° + A), 90° - J. = 45° + A, 2 J. = 45°. A = 22° 30^ cot
tan (90°
A = 45°.
^ = sin 2 J. find J.. cos A = sin (90° - A),
Given cos
A = cot (90° - A),
.'.
;
A if sin ^ = cos 4 A. sin A = cos (90° - A),
10. Find
dO°-A = 2A, 3^ = 90°. A = 30°.
dO°-A = 4:A, 5 A = 90°. A = 18°.
.-.
.'.
6.
Given sin ^
= cos 2 .4 find A. sin A = cos (90° - A\ 90°-A = 2A, 3 A = 90°. /.
;
11.
Find
^ if cot ^ = tan 8^. A = tan (90° - A), 8A = 90°-A, 9 A = 90°. A = 10°.
cot
A = 30°.
.'.
7.
find
Given cos
J.
= sin (-15° — J ^
12. ;
J..
A = sin (90° - A), 90°- J. = 45° -J J., cos
180°
-2^ = 90° -A .-.
^ if cot J. = tan nA. cot A = tan (90° - A), 90°- A = nA, 90° = A{n+ 1).
Find
.:A = ^^.
A = 90°.
Exercise IV. 1. Prove Formulas [1] - [3], using for the functions the line values in
unit circle given in §
3,
n+
Page
1
12.
+
[1].
sin^^
[2].
tan^=^-HL4.
cos^yl
cos J.
=
1.
^ '
.
TEIGONOMETRY.
12 [3].
AxcscA = l, AxsQc A = l, tan J.X cot J. = l. sin
BH^BC
.
(?5
*
cos
but
(?5
CX)
=
1,
BC^l.
BH=—. CB BHxCB =
l,
=
1.
CSC J.
X
sin J.
cos
A = BB,
sec J.
In similar
A
=
BF.
5i^^ and BCB,
BF: BE:: BC BB. :
BF _BC BE BB' .
Fig.
2.
DC= sin J., BD = cos J.,
[1].
DC^ + BD^ = but
BC^l.
(7^2
,'.
= 1. BC^ + BD"" = 1. sin^A + cosM = 1. CB^
BF=^
BFxBB =
.-.
.'.
BE=1,
but
sec
~
BB l,
A X cos J. = 1. A = EF, cotA=GR.
tan
BC= sin A, BB = cos A,
[2]
In similar
A GSB and FEB,
OH: OB:: BE:
EF=iQ.nA.
GH_BE.
A FBE and ^Ci) are similar. FE BE:: CB BB. .-.
:
FE_CB_. BE BB'
but
BE==1.
but
BE==1.
GH-
2.
CB = sin J., 55"= CSC A.
[3].
In similar :
/i^
HGB and
QB
::
BC:
Prove that
tan^ = 7,
1 +tan2J.
sec^
= sec'A = -. c
b a-'
(75i),
CB.
1
FE GBxFE=l. cot A X tan A = l.
FE = CB BB sin A tan J.= cos A
^S"
GB FE' GB = 1.
:
Or.
FE,
Dividing
all
+ b^ = c\ the terms
by
5^,
TEACHERS EDITION. 52
a^
Substituting for
^ g2
—
Substitutins ^ for
and
—
values cot^J. and
1
= sec^J..
Prove that
1
+
cot^ CSC 0,2
Dividing
all
_j.
cot^^
4.
- csc^^.
=-',
^
c^.
the terms
O^
52
^ g2
a?
a?
0?
—
their
a?
csc^J.,
we have
cot2^
= csc^J..
Prove that cot
cot^ sin
a
A=— a 52
+
and
we have
+
tan^^
3.
sec^J.,
— o?
their 1
values idiU^A and
13
by
o?,
A
=
A = cos A sin A
.
TRIGONOMETKY.
14 cot
^ = 0.75.
sec
A=
— 0.6
3.
A = 1.6667.
CSC
J_ A = 0.8'
CSC
A = 1.25,
Find the values of the other
functions
when
cos
+
cos^
sin^
sm
sec
1
A = ff =
1,
TEACHERS EDITION.
15
-^l^h4 COS
.-.
J.=
sin
A = iVS.
tan
A = cos
tan J. sec
1
A^
COS CSC
=
A
I V2 1
^= sin
A
V2.
= ^—— =
.
A
Vs.
1
2
=
V2.
1
cot^ =
1
^
tan^
\y/2
iV3.
CSC 7.
Find the values of the other
functions
tan -4
when
sin-4
= 0.5.
cot J.
=
2.
A sin A tan^ = cot
9.
0.5
cos -4
2 cos
A = sin
4 cos^ J.
— sin^^ =
(squaring)
A=
esc
— = V2
\/2.
sin
J.=
cos
^ = Vl - (^ V2)2 =
1V2,
vT^
1
=
=1
5 cos^J.
when
A.
cosM + sin^^ =
cos J.
Find the values of the other
functions
= a/- = 0.45.
v|=jA
tan^=i^ = l, ^V2
=4 4cos2^— sin2^ = 5sin2J. = 4 4cos2J.4- 4sin2J.
cot
1 sec
sin
= 0.90 A = -^'5
sec
A=
1
cos CSC
A=
A
1 sin
A
^ = 1 = 1, A = r-^ =
= 2.22.
10.
Find the values of the other
functions
= 1.11.
cos
cos
when
A= sec
sin
A
sec -4
=
A = Vl — cos^^
sin
A = 7n.
= sin ^
m
cos
Vl — TO^
A
w Vl — m'^
2.
\
= -, 2
when
A = Vl — sinM = Vl — m^,
Find the values of the other
functions
V2.
iV2
tan -4 8.
'
cot^
— w?
1
tan yl
1
—7n?
mVl — m'^
16
TRIGONOMETRY.
sec
1
A=
cos
CSC
,
^i _ ^2
^
1
A=
sin
11.
1
A
A
1
m
Find the values of the other ^
A
2m
TEACHERS EDITION.
cos 30°
17
TRIGONOMETRY.
18
.*.
4 + 2\/2 =v V4-2V2
=
COS
20, Given tan 90° = other functions of 90°.
tan 90°=
= aV2+V2. cot
=
find the
;
00.
—=—= tan
^1_2±V2 sm =
00
0.
00
sm cos
\
'4-2-V2
=.^'2z:^=iV2:vI. Given siB(F
18.
= CO cos^ sin^ + cos^ = 1 — COS^ = CO CO cos^ = 1. sin^
= 0;
find
the
other fanctions of 0°.
= Vl — sm^ = Vl -0. cos = 1. cos
.*.
•
=
sin
= 00.
sin
=
sec
= -1 = 00.
CSC
=
tan=?^=? = 0. cos cot
=
1
= - = 00. tan
sec
CSC
=
=
—= 1
1 -
cos
1
1
= 11.
—=
COS
-xi
\ CO
cos^
—
1
0.
1.
1.
21. Express the values of all the other fanctions in terms of sin A.
= _1 = 00.
By
sin
formnlae on pages 11 and 12,
A = sin A, cos A = Vl — sin^^, sin ^ tan^ =
sin
19. Given sin 90°= 1; other functions of 90°.
find the
,
sin 90°
= 1.
.
Vl — sin'^^l
= Vl — sin^ = 0. 1 sin _ tan = —- = - = CO. cos
cot
J.=
Vl - sin^^ sin
.
cos cot
=
— =—= tan
sec
=
1
A=
0.
CO
— =-= 1
sec
CSC CO.
A=
A
1
Vl - sinM 1
sin^
COS 1 CSC= -r-
sm
1 1 = 7= i. 1
22. Express the values of all the other fanctions in terms of cos A.
TEACHEE-S
By formula on sin
A=
pages 11 and 12,
EDITION.
19
TRIGONOMETRY.
20
= sin^ J. + 4 sin^ J. = 5 siu^A = 1, sin^J.
+
cos^J.
1.
1.
sinM = i5 sin
A
iV5. '
".
26. sin
A
cos J.
n
= |V5.
Given 4 sin ^ and tan J.. J,
But
sin
= tan ^
^
;
find
TEACHERS EDITION. tan
A
sin
_
cos cos
cot J.=
sin
A A A A
sin^^ cos
tan^ + cot^=^HL^ + ^-^^
A
cos
sin
A
EXEECISE VI. 1,
In Case
of finding
c,
cos
II.
give another
after h has
way
been found.
A = -, c
h = c c
cos A, h
=
COS
2.
A
In Case III. give another way c, after a has been found.
of finding
«
A
c
c&va.
A = a, c
a
=
sin
3.
A
In Case IV. give another
of finding
6,
after the angles
way have
been found. cos
A = -y c 6
4.
=c
cos A.
In Case V. give another
of finding
c,
after the angles
been found. ^ A = —, G
cos A
c cos J.
=
c
=
h,
cos J.
21
way have
=
^va?A
+
tan J.
+ cot A-
cos^J.
1.
+
cos^^
A sin A
22
TRIGONOMETEY. sin
c &'m
A A = a, =
c
sin J.
8.
Given
b
and
cos J.
=
c
-.
;
find A,
b, a.
;;
TEACHEES log h
= 0.30103,
log c
h
=2.
log cos
A is an isosceles ^ = ^ = 45°.
the
.*.
.-.
required b
=
A.
log b
16.
c
=2011.6
B = (90° - ^) = 66""
Given
required b
575.0.
= c sm A. a = log c + log
log
log log
= sin A = = a
=
a
sin
log
A.
log c log sin log
2.39797
a
a
250.02.
log
log
log
cos A.
- 2.79727 cos A = 9.96240
log c
=2.75967
log b
log cos
b
= 575.
h
17.
required
c
2011.6.
B = 61°
55^
= c sin A. « = log c + log a
log log log
log
a
A,
= 1.85824 A = 9.80412 = 1.66236 = 45.958. = c cos A. 6 = log c + log
=
cos A.
1.85824
sin
^.
= 3.35793 sin A = 9.67280 = 3.03073 a = 1073.3. c
5
log 5
= c cos A. = log c + log
cos A.
A = 9.88699 = 1.74523 = 55.620.
Given g=-1, A = 36°; required a = 0.58779, 5 = 0.80902.
= 2280, A = 28° 5^ 5 = 54°, B = 61° 55^, a = 1073.3,
Given
15.
=
sin
c
log 6
6
= c sin J., a = log c + log
6
6
log
26^
a
9.60070
= c cos A. = 6 log c + log
= 72. 15, A = 39° 34^; 26^ a = 45.958,
^ = 50°
30^.
2.79727
c
c
B = 50°
= 55.620.
a
log
= 3.35793 A = 9.94560 = 3.30353
b
= 627, A = 23^ 30^ B = 66° 30^ a = 250.02,
Given
14.
rt.
23
EDITION".
;
TRIGONOMETRY.
24
= c cos A. 6 = log c + log
logc
b
log
log cos
A = 9.90796 = 9.90796 = 0.80902.
log b b
Given
=
=
c
200,
^ = 68°
required b
A.
= 0.00000
log c
18.
cos
10
B = 21°
13^
=
a
47^
185.72,
74.219.
13^
sm
-d
=— c
a log a
= c sin ^. = log c + log
sin
A.
COS
A.
= 2.30103 sin A = 9.96783 = 2.26886 a = 185.73.
log c log log
a
cos .4
= -• c
&
log 6 log G
log cos log b
6
19.
=
= 2.30103 A = 9.56949 = 1.87052 = 74.22.
Given
required 6
= c COS .4. = log c + log
c
=
93.4,
A = 13°
35^
B = 76° 25^ a
=
90.788.
A = 13°
35^
= c sin A. a = log c + log
a log
sin
;
21.936,
A.
TEACHERS EDITION.
25
26
TRIGONOMETRY.
teachers' edition.
B = 52° h cos A = c =c c =
h
¥.
cos
^.
h
COS J.
log
log 5
colog COS log c c
= log h + Colog = 0.60206 A = 0.10307 = 0.70513 = 5.0714. G
a
=
86.53,
3^
log a
= 1.85600 = 8.06283 A = 9.91883 = 56° 3^ = 33° 57^
log sin
A .5
= log a + log a = 1.85600 cot J. = 9.82817 = 1.68417 b = 48.324. log 6
log log &
= 0.60206 log tan A = 9.89177 = 0.49383 log a = 3.1176. a
=
A = log a + colog
COS A.
= b tan ^. = log & + log tan A.
a
71.78
B = 33°
log sin
colog c
A = --
log a
c
A = 56°
;
57^
= 48.324.
log
tan
Given
30.
required b
27
c.
cot A.
log h
Given
31.
required
6
Given
29.
6
=
^
c
=
31°
8590,
24^
a
=
4476
B = 58°
colog c log a
A=—
log sin
colog c log sin
A B
a
=
8.49
;
46^
A
J.
=— <^
= 90° - .4. = 9.02919 = 0.92891 A = 9.95810 = 65° 14^ = 24° 46^ .5
36^
= log a + colog c. = 3.65089 = 6.06601 A = 9.71690 = 31° 24^. = 58° 36^
log sin J. log a
9.35,
c
7332.8. Sin
=
= 3.917. sm
required
c
A = 65° 14^ B = 24°
A B cos
A=— c
= c COS A. = 0.97081 cos A = 9.62214 = 0.59295 b = 3.917. 6
log c cot
A=—
= log a + log = 3.65089 a cot A = 10.21438 b = 3.86527 log h
log log
log b
log
a
= 7332.8.
log cot
A. b
32.
Given c=2194, 6
required
a
=
1758.
^=53°
15',
=
1312.7;
B = 36°
45^
TRIGONOMETRY.
28
COS
log 6
A=
TEACHERS EDITION. Given a = 415.38, & = 62.080; required ^ = 81° 30^ .S = 8° 30^ 36.
c
= 420. tan log a
A==
2.61845
= 8.20705 - 10 log tan A = 10.82550 = 81° 30^ A = 8°30^ B
colog b
•
Bin
^
AA = --
29
TRIGONOMETRY.
sin
A = c— sin
log a
colog sin log c c
A
= log a + colog sin A. = 0.69897 A = 0.19375 = 0.89272 = 7.8112.
log
c
axh
F
30
15.
Compute the unknown parts
40.
the and a = 0.615, c = also
I
.
.
area,
having given
70.
a
sm A^-G
log sin J.
= log
log a
=9.78888-10
colog c
log sin
A £ tan
log a
log b
b
c.
= 8.15490-10 A = 7.94378 = 30^ 12^^. = 89° 29^ 48'^ A^--
log &
colog tan
a + colog
= log a + colog tan A. = 9.78888 - 10
A = 2.05626 = 1.84514 = 70.007.
TEACHERS EDITION.
= log a + colog sin A. = 0.84510 A = 0.50461 = 1.34971
log log a
colog sin log c
c
sm
tan
J.
=
^ = c-. = c sin^. a = log c + log sin A. a
log
= 22.372.
c
31
log
c
log sin -•
log a
a
= 1.13792 A = 9.68739 = 0.82531 = 6.6882.
tan J. log h
- 0.84510
log a
colog tan
A = 0.48224 - 10
b
logi^
logF
F
the
also 12,
area,
c
parts
also
sm
a log a
log b
colog COS log c
c
c
log sin
h
A
= log b + colog COS A.
= 1.07918 A = 0.05874 = 1.13792
= 13.738.
= c sin A. = log c + log
=
log c
COS
log
A = -' c
c
=
given
having given
A=— c
the area, having
= 68, ^ = 69°54^ A = 69° 54^ B = 20° 6^
^=29°8^ A = 29° 8^. B = 60° 52^ COS
Compute the unknown parts
44.
and
"
Compute the unknown
43.
=
colog 2
= 1.32734 = 9.69897-10 = 1.87141 = 74.371.
log 6
colog 2
6
log 5
F=^ab. = 0.84510
log a
= 0.82531 = 1.07918 = 9.69897-10 = 1.60346 = 40.129.
log a
= 1.32734 = 21.249.
log b
and
F= I ab. log F= log a + log 5 + colog 2.
= log a + colog tan ^.
A.
1.83251
A = 9.97271
log a
=1.80522
a
= 63.859. cos
sin
A = -. c
6
log 6
=c
cos
= log
c
^.
+
log cos A.
TRIGONOMETKY.
32
=
log c
log cos
46.
1.83251
and
A = 9.53613
a
= 1.36864 = 23.369.
log b b
Compute the unknown also
the area, having 48° 49^.
= 47, ^ =
A = 41° IK = a cot -4, b = log a + log cot A.
b
F= J ab.
log i^
F
=746.15.
log b colog 2
log a log cot
log& b
Compute the unknown parts
45.
and c
log
= 1.80522 = 1.36864 = 9.69897-10 = 2.87283
log a
the area, having given
also
= 27, B = M°4:'. A = 45° 56^ a
=
c sin
A.
= log c + log = 1.43136 A = 9.85645 = 1.28781 = 19.40.
log a
f
log c log sin
log a
a
sin
log c log cos log b
&
= 1.43136 A = 9.84229 = 1.27365 = 18.778.
F= J a6. log a log b
colog 2
log -F
F
A.
= c cos A. 5 = log c + log cos A.
6
log
= 1.28781 = 1.27365 =9.69897-10
= 2.26043 = 182.15.
parts
given
= A= =
1.67210 10.05803
1.73013
= 53.719.
TEACHERS EDITION. c
log c
=
mi A
= log
a
colog sin A.
4-
=1.11332
log a
colog sin ^"==0.08523
= 1.19855 = 15.7960
log c c
i^-
^ ah.
= 1.11332 = 0.95424 = 9.69897-10 = 1.76653
log a log 6
colog 2
logi^
= 58.416. Compute the unknown
48.
and c
=
also
8.462,
the area,
B = m°¥. ^ = 3° 56^ a
log a log c log sin
parts
having given
= c sin A. = log c + log
sin
A.
= 0.92747 A = 8.83630
log a
=9.76377-10
a
= 0.58046.
= c cos A. = log c +log = 0.92747 A = 9.99898 = 0.92645 = 8.442. h
log h log c
log cos log b 6
cos ^.
33
TRIGONOMETRY.
34 53.
Given i^=
58,
a
=
10
;
solve
tan
A = -'
the triangle.
F= I ah.
log tan log a
a log 6
log2P
colog 6
= log
2 i^ + colog a.
=2.06446
= 9.00000 - 10 = 1.06446 = 11.6.
colog a log h
h
tan log tan
A=A = loga + colog b.
= 1.00000 = 8.93554 - 10 colog b log tan^ = 9.93554 = 40° 45^ 48^^ A log a
B
=49°14M2^>'.
sin^ log € log a
colog sin
.4
log c c
54.
= log
a
+ colog sin A.
= 1.00000 = 0.18513 = 1.18513 = 15.315.
Given i^=
18,
6
= 5;
solve
the triangle.
F= I ah. 2F log a
log2i?' colog b
log a
a
= log
2 i^ + colog
=1.55630
= 9.30103 - 10 = 0.85733 = 7.2.
"
6.
log
A = log a + colog b. = =
0.85733 9.30103
-10
;
35
TEACHERS EDITION. tan 29°
= & tan 29°. log a = log b 4- log tan 29°. a
log 6
= 0.81823
log tan 29°
=
log a
= 0.56198 = 3.6474.
a
9.74375
but .-.
cos
«
A
•
= 2 )4.26867 = 2.13434 2.13434 = log 136.25. 62 - 242 = 136.25. 62 = 378.25. log 6 = Hlog 378.25). = 1.28889. 6 = 19.449.
log\/l8564
= ^.
c
log COS c
= sin 29°
log c
log a colog sin
= log a + colog sin 29°.
c
Given i^= 100,
56.
c
= 22;
the triangle.
F= ab a
=
1
ab
=
100.
200.
= 2_00
'
b
40000 62
o3
+
52
= c2 = 484.
Substitute,
40000
+
62
= 484.
62
= 484 62. 6* -484 62 = -40000.
40000
¥-{
)
+
6*
+(242)2= 18564.
solve
colog
= 1.28889
colog c
= 8.65758 A = 9.94647 = 27° 52^
A B
= 0.87641 = 7.5233.
A = log 6 +
log 6
log cos
= 0.56198 29° = 0.31443
log c
A = -'
=62° 8^
c.
TRIGONOMETRY. log
a
'
TEACHERS EDITION. 61.
120
At a
log a
horizontal distance of
from the foot of a
feet
steeple,
log cot J.
the angle of elevation of the top to be 60° 30^ find the
log b
height of the steeple.
b
was found
;
A=—
tan
a = b tan A. a = log b + log tan A.
log
a
breadth measured along the bank, the point A being directly opposite a tree C on The angle the other side. = 96 was also measured. If 21° 14^ find the feet, and to find the
of a river a distance
ABO=
breadth of the river. If
ABC= 45°, what would be
breadth of the river 62.
From
326
out of the
feet
water, the angle of depression of a boat was found to be 24° find the
log log
distance of the boat from the foot of the rock.
A=— a = a + cot A. b = log a + log cot a.
b
log a log cot
log 6 6
B = AC ^AB. AC=ABxt&nB. AC = log AB + log tan B.
AB = 1.98227 log tan B = 9.58944
;
log
the
?
tsin
the top of a rock that
rises vertically
cot
AB was
ABC AB
= 2.07918 tan A = 10.24736 = 2.32654 a = 212.1.
log 5
log
= 2.30103 = 1.21351 = 3.51454 = 3270.
In order
64.
b
log
37
log^C
=1.57171
AC
=
37.3 feet.
AC = log AB + log tan B. = 1.98227 log AB log tan B = 10.00000 log
= 2.51322 A = 10.35142 = 2.86464 = 732.22.
log^C
=
AC
=96
1.98227 feet.
65. Find the angle of elevation a monument, in of the sun when a tower a feet high a level plain, from the eye, if the casts a horizontal shadow b feet height of the monument is 200 feet long. Find the angle when a = 120, 63.
How
far
is
and the angle of elevation of the top 3° 30^
?
cot J. b
log 6
5
=
70.
tan^ =
=—
b
a
=a
-.
cot A.
= log
a
+
log cot A.
tan log tan
A
120 70
^ = log
120
+
colog 70.
TRIGONOMETRY.
38
= 2.07918 = 8.15490 - 10 colog 70 log tan J. = 10.23408 = 59° 44^ 35^^ A log 120
vessel,
=
and equal one hour's progress
10 miles.
A C= distance due east passed over in one hour.
As the
direction of the ship
is
north-east,
How high
A = 45°.
a tree that casts a horizontal shadow b feet in length when the angle of elevation of the sun is A° ? Find the height of the 66.
tree
when tan
b
is
= 80°, A = 50°.
log
log
a= log
= log b + log tan A. = 1.90309
a
log 6 log tan
h tan A.
cos A.
= 1.00000 cos A = 9.84949 = 0.84949 b = 7.0712 miles due east,
log 10 log
A = -•
= c cos A. b = log c + log
b
b
and
due north, since
also
AP= AO.
A = 10.07619 = 1.97928 = 95.34.
log a
a
In front of a window 20
69.
high
is
feet
a flower-bed 6 feet wide.
How long must a ladder be to reach from the edge of the bed to the 67. What is the angle of elevawindow ? if it inclined plane rises tion of an 1 foot in a horizontal distance of 40 tan J. = -• feet?
tan
log tan
A = -'
A = log 20 + colog 6. = =
log 20
log tan
A = loga + colog b.
= 0.00000 = 8.39794-10 colog b log tan A = 8.39794 log a
A
=1°
colog 6 log tan
1.30103
9.22185-10
A = 10.52288 = 73°
A
25' 56'^
c
=
18^
a sin -4
A
due northeast with a velocity of 10 miles an log hour. Find the rate at which she colog is moving due north and also due 68.
ship
is
AB
be the direction of the
c
c
sin
= log 20
-f-
= 1.30103 A = 0.01871 = 1.31974 = 20.88. a
log c
east.
Let
log
sailing
colog sin A.
TEACHERS EDITION. 70.
A ladder 40
so placed that
dow
it
feet
long
may be
will reach a win-
33 feet high on one side of the and by turning it over with-
street,
out
moving
window
its
Find
side.
foot
it
will reach a
21 feet high on the other the
breadth
of
street.
cos
33 Bp =^ —
40 log 33
colog 40 log cos
B tan
= 1.51851 = 8.39794 - 10 B = 9.91645 = 34° 2V 45^^
5 = —.
the
39
40
TRIGONOMETRY.
and at a point 100
feet nearer the
How high is
fort it is 45°.
the fort
A ABC, ZBAC= 30°,
In
rt.
?
Z ABC=
and
.-.
60°.
BC=^AB = ^x212>.20 = 136.60 feet.
A
73. From a certain point on the ground the angles of elevation of the belfry of a church and of the top of a steeple were found to be 40° and 51° respectively. From a point 300 feet farther off, on a horizontal
100
C
A'
ft.
B represent
Let
horizontal plane,
the
fort,
AC
the
£C a ± from
line,
the angle of elevation of the
top
of the
33°
45''.
fort
steeple is found to be Find the distance from
the belfry to the top of the steeple.
to plane,
BAC= angle
made by
= 30°. BA^Q= 45° = angle
line from
eye of observer 100
of elevation
feet nearer.
From A^ draw A^Nl. In rt. A AA^N, Z.
and
^
NAA^=^
AB.
30°,
= 60°. NA^= 50 feet.
iV^.4^^ .-.
AN= \/(100)2- (50)2
/.
= In
to
rt.
\/7500
= 50\/3
= 86.602. A BNA'
^^ = cot NBA'= cot
DE ± to AB from A BED
Draw In
= ^ BD 15°,
^Z)
NA' and log
B]Sr=
NA'
log cot 15
log^iV
BN AN AB
NA' cot 15°.
= 1.69897 = 0.57195 = 2.27092
= 186.60 = 86.60 = 273.20
log 300 log sin 33° log
sin 33° 45^.
= 300xsin33°45^ = 2.47712
45^= 9.74474
ED
=
2.22186
Z EAD = 180°- 33° 51°) = 17° 15^ In
D.
45^
-
A ADE ED = sin 17° 15^.
AD
(ISO"
TEACHERS EDITION.
AD = log
ED
ED sin 17° 15^
= 2.22186
colog sin 17° 15^= 0.52791
log
AD
= 2.74977
A ADC DC = cos 51°. AD DC= AD cos 51°. = 2.74977 log AD
In
= 9.79887 = 2.54864
log cos 51° log
In
DO A ADO
^= DC
tan 51°.
AC=DOiQ.uf>l°. log
DC
41
42
TRIGONOMETRY.
Exercise VII.
Page
25.
TEACHERS EDITION.
= 2(90°-^). 2a
a c
8. c
In an
and h
;
= 2a
cos
A.
isosceles triangle,
find A, C, a.
given
43
TRIGONOMETKY.
44 log Jc
=
TEACHERS EDITION. 14. In an isosceles triangle, given c
= 9,
^
= 20
tan
;
colog h log
A,
c,
a.
^C=i^.
log tan ^ c
log^c
find
= log J c +
colog
=0.65321
= 8.69897 - 10
A.
45
TRIGONOMETRY.
46
F= i cA.
^c
A=
COS
Jc log a
= log I c + colog cos A.
logic
=0.41497
colog cos
A = 0.81547 =
log a
tan J
A
21.
1.23044
C.
J'=A(/itaniC)
= h^ barn
tan J
is
40
of the roof
pitch
=17.
a
=A
C.
X
is
80 feet, the 45°; find the
length of the rafters and the area
of both sides of the roof. 17.
In an
isosceles triangle, find
F in terms F= J ch.
the value of
of
a and
40-2 = 20 = ^c. cos^ = fc^a = 20 a. 20 = a cos A.
c.
-f-
4 a2
=v
-
a
c2
= log 20 + = 1.30103 A = 0.15051 = 1.45154
log a log 20
Z^=2c(jV4a2-c2)
colog cos
= |c\/4a2-c2.
log a
isosceles triangle, find
F in terms of a and F= I ch.
the value of
28.284 C.
= a sin f C. h = a cos J 0. F= a sin ^ Cx a cos J C.
Jc
19.
In an
22. In a unit circle what is the length of the chord corresponding to the angle 45° at the centre ? sin J
of a and A.
F=lch.
F= a cos Ax a sin A sin J. cos
log a
= log a log = 0.00000 -j-
sin J C.
logsin ^(7= 9.58284
ic= a cos A. h = a sin A.
= a^
(7= i^. a
log J c
isosceles triangle, find
F in terms
X 80 = 2262.72. X 2 = 4525.44.
2262.72
= a^ sin J C cos J C. the value of
colog cos A.
= 28.284.
a In an
20 cos -4
= *V4a2-c2.
18.
=
logjc
=9.58284-10
Jc
=0.382683.
G
= 0.76537.
A.
= 30, = 44, find
23. If the radius of a circle
In an isosceles triangle, find and the length of a chord the value of in terms of h and C. the angle at the centre. 20.
F
•
TEACHERS EDITION.
= 3,
then
sin h
C= —
Let a
sinK'=— a
logsin JC= log
2^
c
+ colog
^
a.
log sin i
=1.34242
logic
=
colog a
8.52288
- 10
log 1
colog 3
log sin ^(7= 9.86530
47 c
= 2,
1.
3
C= log 1 + colog 3. = 0.00000 = 9.52288 - 10
IC
=47° 10^
log sin J (7= 9.52288
C
= 94°
i
C C
20^
and J c =
=19° 19°
28^ 16|^^.
= 38°
56^ 33^^^
24. Find the radius of a circle if 26. Find the area of a circular a chord whose length is 5 subtends sector if the radius of the circle = 12 at the centre an angle of 133°. and the angle of the sector = 30°. sin i (7= i^-
Area
O = irE^.
Area
sector
a
= log
log a
logic
Jc
+ colog
sin J C.
=0.39794
=0.43554
a
=
25.
What
is
360
is
2.7261.
the angle at the cen-
equal to | of the radius
+
log
TT
log 30
tre of a circle if the corresponding
chord
360 log area sector
colog sin -^(7= 0.03760 log a
=
?
+
=
log 30
2 log R.
-f
colog
48
TE-IGONOMETRY.
TEACHERS EDITION.
49
TRIGONOMETRY.
50 tan J (7= t^.
h
14 tanJ(7=^-^-L^=^. 14 p 196 logp = I (log 196 + log tan ^ C).
=
log 196 log tan
^C
2.29226
=9.68271
51
TEACHERS EDITION. having n sides, BA the side of the Then polygon having 2n sides. OA is ± to 5 C at its middle point
log
= 0.38278 = 0.90309 = 1.28587 = 19.3139.
log h
D.
ZBOA^ 360° 2n
ZOBC=dO°-
logi^
180°
n 180°
Ic^=-^
n
A'= 180^
n
logic^
^90o_ 180 n
^90° n
= cos 90°"
^c = 1
n
b
90° J.
cos
n
Whence, h
he
=
90°
o cos 2
cos
n
12.
Compute the
—n 90°
difference
be-
tween the areas of a regular octagon and a regular nonagon if the perimeter of each
^
is
16.
2n 1
16
QAO
^ = i^=22°30^ n
log A
,
=
logjc log cot log A
= log I c +
^
0.00000
=10.38278
=
0.38278
20°.
log A^= log ^ c'-f log cot A\
A ABC= Z OB A- Z OBC
ic i—
180°
18
n'
Q0°
/9Qo_90^\
•
= — = 0.8889.
2n^
The A BOA is isosceles. .\/.OBA=-(l^Q)'
F= log h + log Ijp.
log cot A.
TKIGONOMETRY.
52 |j2
=
log 240 log tan
= 240
^C
tan J
2.38021
=9.86126
C.
•
TEACHERS EDITION.
tani)^0=—
c
he
= log J c +logtan 54°. = 0.56916 = 10.13874 = 0.70790
log h
logjc log tan 54° log A
p = ic X log
2 n.
F= log jc+log n+log h. =0.56916
logjc log n
= 0.69897 = 0.70790
log h logi?'
=1.97603
F
= 94.63.
16.
The area of an inscribed
ular pentagon
331.8;
is
reg-
find the
area of a regular polygon of 11 sides
same
inscribed in the
Let
AB be
circle.
a side of a regular in-
scribed pentagon,
and
AD the side
of a regular inscribed polygon of 11 sides.
Let
a be
the radius of the circle
whose centre is 0, and h and h^ the apothems of the 2 polygons, respectively.
Given
=
F
the area
Find
331.8.
F\
of pentagon
the area of 11-
sided polygon.
Let p and p^ and
c
and
c' repre-
sent the perimeters and sides of the
pentagon and 11-sided polygon, spectively.
F== Iph. 331.8
= hph,
ph = h
663.6.
663.6
P
re-
=
54
TEIGONOMETRY.
;
TEACHERS EDITION. log r log cos J C^
log
= 0.75703 - 9.99040 = 0.74743
ic^ — = sm -
r
2
J c^ = r X sin J
=0.75703
log r
log sin
ia = 9.31788
log ^c^
=0.07491
log
C\
F= log J c^+ log n + log h^.
log
= 0.07491 = 1.17609 = 0.74743
F
= 99.640.
log log 15
1.99843
19. A regular dodecagon is circumscribed about a circle, the circumference of which is equal to 1 find the perimeter of the dodecagon.
Given circumference of inscribed
= 1, n=12; 2 Trr r
find p.
= circumference. = circ.
55
TRIGONOMETRY.
56
Page
Exercise IX.
36.
1. Construct the functions of an 239° is in Quadrant III. angle in Quadrant II. What are sin = — tan = +
their signs
cos = —
?
cot
=
-f
sec
=—
esc
=—
Sines and tangents extending up-
wards from horizontal diameter are 145° is in Quadrant II. downwards, negative. Cosin = tan = — sines and cotangents extending from cos = — cot = — vertical diameter towards the right positive
=— CSC =
;
sec
-t-
are positive
are
;
towards the
left,
nega-
Signs of secant and cosecant
tive.
made
to agree
sine, respectively.
sin
cos
and and
400
= 360°
= signs
40°
-f
-f-
of func-
tions of 40°.
with cosine and Hence, 40°
sec are
in
Quadrant
— —
700°
I.
tan = -f cot = +
=+ cos = +
esc are -f
tan and cot are
is
sin
= 360° -f
340°
=+ CSC = sec
-t-
= signs
of the
functions of 340°.
2. Construct the functions of an angle in Quadrant III. What are 340°
their signs
?
sin
— are —
sin
and
esc are
cos
and
sec
tan and
cot are
cos
is
=— =+
1200°
-h
Quadrant IV.
in
tan
=
cot
3
X
=— =—
360°
sec esc
+
120°
=
=+ =— signs
of the functions of 120°. 3.
Construct the functions of an
angle in
What
Quadrant IV.
their signs
are
?
—
sin
and
esc are
cos
and
sec are
+
tan and
cot are
—
120^
is
in
Quadrant
=
-h
tan
=—
sec
=—
cos
--=
—
cot
=—
CSC
=+
3800°
=
10
X
360°
is
in
Quadrant
the
sin
functions of the following angles 340°, 239°, 145°, 400°, 700°, 1200°,
cos
=— =—
5.
How many
What
are
the signs
of
:
3800°
?
340° sin
cos
is
in
=— =+
Quadrant IV. tan cot
=— =—
+
200°
signs
of the functions of 200°.
200° 4.
II.
sin
=+ — = esc sec
tan cot
III.
= -f =+
sec CSC
=— =—
angles less than
360° have the value of the sine equal to -I- f, and in what quadrants 'do they lie?
.
TEACHERS EDITION. Since the sine angles can
lie in
57
.•. two values, one in Quadrant is +, by § 24, the but two quadrants, and one in Quadrant III.
I.
first and second. (v.) Sign being — the angle can In the first quadrant, by | 3, the be in Quadrant II. or IV. sine increases from to 1, and in .•. two values, one in Quadrant II. the second, decreases from 1 to 0. and one in Quadrant IV. This is a continually increasing and
the
,
what
limits must the cosx = — |? if cot a; angle whose sine is equal to + f in = 4? if sec a; = 80? ifcsca; = -3? each quadrant, the first and second. {x to be less than 360°.)
decreasing quantity.
8.
Therefore there can be but one
"Within
angle x
lie
if
If cos X = — f X must lie in the values less than second or third quadrant, or between 720° can the angle x have if cos x 90° and 270°. ,
6.
=
How many
+f, and
they
what quadrants do
in
720°
is
If cot
a;
= 4,
90°, or 180°
lie ?
twice
moving radius
360°;
will
hence the
make
If sec
a;
X is between 0° and and 270°.
= 80,
a;
is
between 0° and
exactly 2 90° or 270° and 360°.
complete revolutions. If CSC a; = — 3, a; is between 180° The cosine has the + sign in the and 360°. first and fourth quadrants, hence it 9. In what quadrant does an two in Quadwill have four values angle lie if sine and cosine are both rant I. and two in Quadrant IV. negative ? if cosine and tangent are 7. If we take into account only both negative ? if the cotangent is angles less than 180°, how many positive and the sine negative? :
values can x have cos
=
a;
= ^?
if
f? ifcotaj
cos x
if
sin
a;
=—|?
= if
(i.) Sine is negative in Quadrants f ? if tan x II, and III. cosine is negative in ;
= -7?
Quadrants III. and IV. .•. angles having both sine and angle being the can (i.) Sign +, cosine negative are in Quadrant be in Quadrant I. or II. III. .-. two values, one in Quadrant I. (ii.) Cosine is negative in Quadand one in Quadrant II. rants II. and III. tangent is nega(ii.) Sign being -f, the angle is in tive in quadrants II. and IV. Quadrant I. or IV. .•. angles having both cosine and .•. two values, one in Quadrant I. tangent negative are in Quadrant and one in Quadrant IV. ;
(iii.)
Sign being
—
II. ,
the angle can
(iii.) Cotangent is positive in Quadrants I. and III. sine is neg.•. two values, one in Quadrant II. ative in Quadrants III. and IV. and one in Quadrant III. .'. angles having cotangent posithe tive and sine negative are in Quadbeing angle can (iv.) Sign +, rant III. be in Quadrant I. or III.
be in Quadrant
II. or III.
;
TRIGONOMETRY.
58
Between 0° and 3600° how many angles are there whose sines have the absolute value f ? Of 10.
cot X
=
= tan X
—=—
—1
1.
how many are positive 13. Given tana;= V3; find the and how many negative ? other functions when x is an angle Between 0° and 3600° there are in Quadrant III. 10 revolutions, and in each there tana3= VS. are 4 angles whose sines have the these sines
absolute value
The
angles.
.*.
f.
sine
are
40
positive
in
there
is
Quadrants I. and II., and negative ing Quadrants III. and IV. .•. there are 20 angles with the sine positive, and 20 with the sine negative. 11.
cot X
tancc
tan X
Vs
In finding cos x by means of cos x = ± Vl — sin^ x,
3
when must we choose the positive sign and when the negative sign ? Since cosines only of angles in I. or IV. are positive, we use the sign + only when angle x
within these limits.
4
Also, since cosines of angles in II. and III. are nega-
we known to
use the sign
tive,
12.
lie in either
Given cos a;
in
Quadrant sin^x sin X
esc
X
sec x
— when ,
x
is
;
find the
1.
= Vl — cos^o; = Vl - {-VlY =
11/9 =— =
sm
r:
x
VJ.
"v^-
V^
= -V2.
= cos 03
tan 05 =
sin^^_Vj_ ^ _ cos
a;
is
an angle
II.
+ cos^a; =
=
x
of these.
= - V|
when
other functions
cos
—VI
1
X a;
sin
X
cos
a;
cos x
= sin x.
= sin x.
— sin^a; = cos^x + sin^a; = = cos^a;
cos%
1
1
±i
cos X
The angle being the cosine
Quadrants
^y/S.
cos'^a;
Quadrants
lies
—=
-
V3
equation
the
=
.".
cos X
is
in
Quadrant
negative.
III.
TEACHEES EDITION. cos X
= sec
sma;
a;
^
= - f V3, a;
=
tana;
=
CSC
cot
a;
49
>'49
59
:
TRIGONOMETRY.
60 negative value
When
its cotangent. 0° or 360°, co-
of
the angle
is
tangent is parallel to the horizontal diameter and cannot meet it. But cotangent 360° may be regarded as extending either in the positive
or in the negative
hence either 19.
+
direction
— oo
or
oo
;
and
.
Obtain by means of Formu-
the angles given (i.)
90°
tan
cos 180°
(iii.)
cot 270°
(iv.)
esc 360°
= oo. = - 1. = 0. = - 00.
cot 90°
tan 270°
= - = oo.
cos 270° 0.
sm
270°
=
= 0. sin2 270°-hcos2 270° = l. = l. sin2 270° + sin2 270° = l. sin 270° = - 1. sin 270°
CSC 360°
= - oo.
sin 360°
=
—
sin2 360°
+
cos2
= oo=l. =~ CO
cos2
cos
sin 90°
cos 90° cos 90°
= 0.
(IV.)
(i-)
tan90°
cot 270°
cos 270°
las [l]-[3] the other functions of
(ii.)
(ill-)
— 00
= - 0. 360°
= l.
= l. 360° = 1. 360°
^=-
tan 360°
=
cot 360°
= -!- = - 00.
0.
-0 sin 90° = 0.
cos2 90°
+
sin^
90°
=
1.
= 1. 90° = 1.
-0
sin2 90°
sin
20. Find the values of sin 450°, tan 540°, cos 630°, cot 720°, sin 810°,
(ii.)
cos 180° sin2
180°
sin2
180°
sin 180°
tan 180°
CSC 900°.
= -1. + +
cos2 180° 1
= 1.
= 0. sin 180°
cos 180°
-1
= -0. cot 180°
= cos 180° _ - 1 sin 180°
= sin (360° + 90°) = sin 90° = 1. tan 540° = tan (360° + 180°) = tan 180° = 0. 630° = cos (360° + 270°) cos = cos 270° sin 450°
= 1.
=
0.
TEACHERS EDITION. cot 720°
sin 810°
esc
900°
= cot (360° + = cot 360° = 00.
360°)
Substituting,
axO + 6xO-cxO=0.
= sin (2 x 360° + = sin 90° = 1.
Compute the value of a cos 90° - h tan 180° + c cot 90°.
23.
90°)
= 0. tan 180° = 0. cos 90°
= esc (2 x 360° +180°) = CSC 180°
cot 90°
ax0-6x0 + cx0 = 0.
21. For what angle in each quadrant are the absolute values of
the sine and cosine alike sine
24.
45°),
?
and cosine of 45° are
+
(270°
45°),
sin
+
45°)
cos
Hence the
sines
and
cosines
of 45°, 135°, 225°, 315°,
etc.,
are all
cos
Compute the value
a sin 0° +
5 cos 90°
25.
180°.
Compute the value
(a2-52)cos 360°- 4 ah
of
sin 270°.
= 1. 270° = - 1.
cos 360°
of
- c tan 180°.
sin
= 0. cos 90° = 0. tan 180° = 0.
sin 0°
Substituting,
(a2-52)xl-4a&X-l
= a2-62 + 4a5. Exercise X.
2.
of
axl-6xl+(a-5)x-l = 0.
equal in absolute value. 22.
- 5 cos 360° + (a - 6) cos 90° = 1. 360° = 1. 180° = - 1.
Substituting,
in
the second, third, and fourth quadrants.
Compute the value
a sin 90°
Correequal in absolute value. sponding to the angle of 45° in the first quadrant are the angles (90° +
(180°
=0.
Substituting,
= 00.
The
61
Page
41.
cos 100°
Express sin 172° in terms of
the functions of angles less than
= cos (90° + 10°) = - sin 10°.
45°.
sin 172°
= sin
(180°
= sin 8°.
- 8°) 4.
Express tan 125° in terms of of angles less than
the functions
Express cos 100° in terms of 45°. the functions of angles less than 3.
45°.
tan 125°
= tan (90° - 35°) = - cot 35°.
TRIGONOMETRY.
62
Express cot 91° in terms of
5.
Express
13.
the functions of angles less than the functions
271° in terms of less than
45°.
45°.
cot 91°
6.
esc
of angles
= cot (90° + = - tan 1°.
CSC 271°
1°)
= CSC (270° - 1°) = -secl°.
Express sec 110° in terms of 14. Express sin 163° 49'' in terms of angles less than of the functions of angles less than
the functions 45°.
sec 110°
7.
Express
the functions
= sec (90° + = - CSC 20°.
45°.
20°)
sin 163°
less
than
45°. CSC 157°
8.
= esc (180° -23°) = CSC 23°.
Express sin 204° in terms of of angles less than
the functions 45°.
sin 204°
sin
= sin
esc 157° in terms of
of angles
49^=
15.
(180°- 16° 110 16° IV.
Express cos 195° 33^ in terms
of the functions of angles less than 45°.
cos 195°
33^=
cos (180°+ 15°
= - cos
16. Express tan 269° 15^ in terms
than
of the functions of angles less
= sin (180° + 24°) = - sin 24°.
33^
15° 33^
45°.
tan 269° 15^= tan (270°- 450
= cot45^ Express cos 359° in terms of the functions of angles less than 9.
17.
Express cot 139° 17^
in
terms
of the functions of angles less than
45°.
cos 359°
= cos (360° - 1°) = cos 1°.
45°.
cot 139° 17^= cot (180°- 40°
430
= - cot 40° 43^ 10.
Express tan 300° in terms of of angles less than
the functions 45°.
tan 300°
= tan (270° + 30°) = - cot 30°.
18.
Express sec 299°
45''
in terms
of the functions of angles less than 45°.
sec 299°
45^=
sec (270°+ 29°
450
= CSC 29° 45^. 11. Express cot 264° in terms of the functions of angles less than 45°.
cot 264°
= cot (270° -6°) = tan 6°.
19.
Express esc 92°
25''
in terms
of the functions of angles less than 45°. esc 92°
25^=
CSC
(90°+ 2° 250
= sec 2° 25^
Express sec 244° in terms of the functions of angles less than 12.
20. Express all the functions of
45°. sec 244°
= sec (270° -26°) = -csc26°.
— 75°
in terms of those of positive angles less than 45°.
TEACHERS EDITION,
= sin (270° + = - cos 15°. cos (- 75°) = cos (270° + = sin 15°. sin
(-
tan(-
75°)
75°)
=
tan (270°
+
15°)
Express all the functions of in terms of those of positive angles less than 45°. 24.
— 52° 37''
15°)
sin
15°)
= - cot 15°. cot (- 75°) = cot (270° + 15°) = - tan 15°. Express all the functions of in terms of those of positive angles less than 45°. 21.
— 127°
= sin (270° - 37°) = - cos 37°. cos (- 127°) = cos (270° - 37°) = - sin 37°. tan (- 127°) = tan (270° - 37°) = cot 37°. cot (- 127°) = cot (270° - 37°) = tan 37°. sin
(- 52° 370
tan (- 52°
370 = tan (270°+ 37° 230
= - cot 37° 23^. cot (- 52° 370 = cot (270°+ 37° 230 = -tan37°23^
(- 127°)
Express all the functions of in terms of those of positive angles less than 45°. 25.
— 196° 54-^
= sin (180°-16°540 = sin 16° 54^ cos (- 196° 540 = cos (180°-16° 540 = - cos 16° 54^, tan (- 196° 540 = tan (180°-16° 540 sin
(- 196° 540
Express all the functions of — 200° in terms of those of positive cot (- 196° 540 angles less than 45°.
= sin (180° - 20°) = sin 20°. cos (- 200°) = cos (180° - 20°) = - cos 20°.
= -tanl6°54^ = cot (180°-16° 540 = -cotl6°54^
(- 200°)
= tan (180° - 20°) = -tan20°. cot (- 200°) = cot (180° - 20°) = - cot 20°. tan (- 200°)
Express all the functions of 345° in terms of those of positive angles less than 45°. 23.
—
sin
= sin (270°- 37° 23^
= - cos 37° 23^ cos (- 52° 370 = cos (270°+ 37° 23^ = sin 37° 23^
22.
sin
63
(- 345°)
= sin
15°, etc.
26. Find the functions of 120°. sin 120°
= sin (90° + = |V3.
30°)
= cos 30°
= cos (90° + 30°) = - sin 30° = - J. tan 120° = tan (90° + 30°) = - tan 30° = -Vs. cot 120° = cot (90° + 30°) = -cot30° = -^V3. sec 120° = - 2. CSC 120° = f Vs. cos 120°
TRIGONOMETRY.
64 27.
Find the functions of 135°.
sin 135°
= sin
cos 45°
cos 135°
tan 135°
+
(90°
=
45°)
I V2.
= cos (90° + 45°) = -sin45° = - jVi
= sin
135°
cos 135°
cotl35°
= ^"^^^^° = -l. sin 135°
sec 135°
^— = - \/2. V2.
Find the functions of 150°.
= sin = sin cos 150° = COS sin 150°
(180°
30°
- 30°)
= J.
- 30°) = -cos30° = -jV3. tanl50° = tan(180°-30°) (180°
= -tan30° =
- cos 30°
cot
sin 30°
= -V3. = sec
(180°
= - sec
30°
- 30°) =
= sin (180° + 45°) = -sin45° = -|\/2. 225° = cos (180° + 45°) cos = -cos45° = - J\/2. tan 225° = tan (180° + 45°) = tan 45° = 1. cot 225° = cot (180° + 45°) = cot45° = l. 31. sin
(180°
= CSC
30°
Find the functions of 240°.
240°= sin
(270°
- 30°)
= -cos30° = -jV3. cos 240° = cos (270° - 30°) = - sin 30° = -h tan 240° = tan (270° -30°) = cot 30° = Vs. cot 240° = cot (270° - 30°) = tan 30° = ^ Vs. 32.
= CSC
- 30°)
= sin 30°
= 2.
Find the functions of 225°.
- cos 30°
= -fV3. CSC 150°
30.
sin 30°
= -iV3. cot 150° = cot (180° - 30°) - cos 30° 30° = =-
sec 150°
= sin (180° + 30°) = - sin 30° = - |. cos 210° = cos (180° + 30°) = -cos30° = -A\/3. 210° = tan (180° + 30°) tan = tan30° = J\/3. cot 210° = cot (180° + 30°) = cot 30° = V3. sin 210°
sin 225°
1
=
sin 135°
28.
Find the functions of 210°.
1
=
cos 135° CSC 135°
29.
Find the functions of 300°.
= sin (270° + 30°) = -cos30° = -jV3. cos 300° = cos (270° + 30°)
sin 300°
1
2-
TEACHERS EDITION. tan
300°= tan (270° +
cot 300°
sin 225^
30°)
= - cot 30° = - V3. = cot (270° + 30°) = -tan30°=-jV3.
Find the functions of - 30°. 30° = - sin 30° = - i. sin cos - 30° = cos 30° = J Vs. = 30° tan -tan30° = -^V'3. 33.
cot sec esc
- 30° = - cot 30° = - Va. - 30° = sec 30° = |V3. - 30° = - CSC 30° = - 2. Find the functions of -225°.
34,
- 225° = 90° + 45°. - 225° = sin (90° + 45°) = cos 45° = \V2. 225° = cos (90° + 45°) cos = - sin 45° = ~ I V2. tan - 225° = tan (90° + 45°) = - cot 45° = - 1. 225° = cot cot (90° + 45°) = - tan 45° = - 1. sin
sec
1
- 225° =
+
cos (90°
CSC
- 225°
sin (90°
Given
35.
negative of
X,
;
45°)
= -V2. 1 =— sin
a;
+ 45°)
= —Vl,
=V2.
and cos a?
find the other functions
and the value of
x.
Since sin 45° == V|, and the signs of both the sine and cosine are negative, the angle III.,
and must
180°
+ 45° =
must be
in
Quadrant
be, therefore,
225°.
Then cos45°=\/J. Hence cos (180° + 45°)
65
= - VJ.
TKIGONOMETBY.
66
Find the functions of 3540°.
37.
in
smg00^ ^
-|V3
sign in
i
1
1
tan 300°
_v^
cot 300°
= V3
and the
negative, the angle must be
Quadrant
II. or IV.,
and must
=
What
40.
values of » between O*' satisfy the equation
and 720° will sin X = 4r 2?
1
= - = 2L €Os3(X>'^
Since cot 30°
is
be therefore 180° - 30° 150°, or 360° - 30° = 330°, or - 30°. Hence the angles are 150° or — 30°.
-|V3. secSCKT
are 135°, 225°, or
-225°. (ii.)
cos SOO'*
and must
II. or III.,
Hence the angles
2 tan 300^ =
Quadrant
be therefore 180° - 45° = 135°, or 180° + 45° = 225°. Also, the functions of — 225° are the same as the functions of 360° - 225° = 135°.
= 9 X 360° + 300°. sin 300° = sin {S&)P - m^} = sin 60° = - 1 VS. cos aOO^' = cos (360° - 60°) cos 60^ = i. 3540«
i
= -| and the sign is the angle must be in Quad-
Since sin 30°
esemF = sin 300*^
positive,,
^V3
rant
I.
or
II.,
and must be therefore
180°- 30°
30°, or
=
150°, the first
In the second revoluthese angles must be increased
revolution.
Sa What equal to (i.)
— V3
^aee
^an
angles less
have a sine equal to
—J?
360°'
and 510°.
150°, 390°.
?
30^= J and the
sin
tion,
a tangent by 360°. Hence the angles are 30°, sign
angle nrast be in 41. In each of the following eases Qnadraat III. or IV., and mnst be find the other angle between 0° and therefore 180° + 30° = 210°, or 360° 360° for which the corresponding is
negative,
tlie
-30° = 330°.
function (sign included)
Since tan 60°
=V3
the
has
and the same value sin 12°, cos 26°, tan 45°, sign is negative, the angle must be cot 72° sin 191°, cos 120°, tan 244°, in Quadrant II. or IV., and must cot 357°. (ii.)
:
;
be therefore 180°-60° 360° -60° = 300°.
Which
= 12CP,
or
sin 12°
mentioned in Examples 27-34 have a 39.
of the angles
cosine equal to
equal to (i.)
sign
— V3 ?
Since
is
In order that the sign shall be the same,
— VJ?
must be in Quadrant
= sin(lSO^-
12°)
II.
= sin
168°.
26° miist be in Quadrant IV. a cotangent cos 26°) cos 334°. cos (360°
cos45°=V^ and
=
-
=
the tan 45° must be in Quadrant III. = tan (180° + 45°) = taa 225^
negative, the angle must be
TEACHERS EDITION. cot 72°
must be
Quadrant
in
= cot (180° + sin 191°
must be
= sin cos 120°
(360°
in
= cot
(180°
in
+
III.
252°.
Quadrant IV.
- 11°) = sin
must be
= cos
72°)
349°.
Quadrant
60°)
=
III.
cos 240°.
tan 244° must be in Quadrant
I.
= tan (244° - 180°) = tan 64°. cot 357°
must be
= cot 42.
(357°
in
Quadrant
II.
- 180°) = cot
Given tan 238°
=
177°.
find
1.6;
sin 122°.
tan 238°
= tan (180° +
58°)
-tan 58°.
= sin (180° - 58°) = sin 58°. 238° = 1.6. tan tan 58° = 1.6.
sin 122° ,
But .-.
tan 58°
= sin
58°
cos 58° sin
58°
1
Vi-sin258° 2.56
- 2.56 sin2 58° = sin2 58°.
3.56 sin2 58°
sm
58°
= 2.56. = -\/ ^'3.56
= 43.
0.848.
Given cos 333°
= 0.89
tan 117°.
= 0.89. = cos (270° + 63°) = sin 63° = tan (180° -63°) = - tan 63°. Bin2 63° + cos2 63° = 1. (0.89)2 + cos2 63° = 1. cos 333°
;
find
cos2 63°
67
=
0.2079.
TRIGONOMETRY.
68
example
sin (180°
+ x) = — cos X. —y) = — sin y. + x)^ — sin x.
sin (270°
— y) = — cos y.
sine increases in value,
COS (180° cos (270°
Hence
As X
+
tan (—
3/.
is
-tan
2/)
=—
tan
(180° -3/)
1/.
=
tan
Hence the expression
— cos x.
increases from 0° to 45°, the
and cosine
to this
point sin
= cosine. .r
— cos
it
For the remainder of sine is greater than co-
negative.
Quadrant I. and consequently the expression sin X — cos X is positive. In Quadrant II. sine is positive and sine,
— tan (180° — y).
tan (— y)
the question of last
for sin x
Hence up
Simplify the expression
tan x
Answer
decreases, until at 45° sine
the expression
= cos X sin y — sin x cos 50.
52.
cosine negative, so the expression 2/.
=
tan
x.
— cos X
uniformly positive. is negative and cosine negative hence, so long the as sine is less than cosine, the exsin X
is
In Quadrant III. sine ;
For what values of x is expression sin x + cos x positive, pression is positive, viz., to 225°; and for what values negative ? Rep- after that point, sine is greater than resent the result by a drawing in cosine, and sin x — cos x is negative. which the sectors corresponding to In Quadrant IV. sine is negative the negative values are shaded. and cosine positive therefore sin x The If X be any angle in Quadrant I., — cos a; is uniformly negative. sin X + cos X must be positive since expression is, then, negative beboth the sine and cosine are posi- tween 0° and 45°, and 225° and In Quadrant II. the sine is 360°; positive between 45° and tive. positive and cosine negative hence, 225°. 51.
:
;
so long as the sine
is
greater than,
53.
or equal to, the cosine, the expres-
sion
sin
a?
-f
cos
re
after passing the
rant
Zx
is
II., viz.,
is
positive;
but (x
middle of Quad- ^^^-
135°, the cosine of
negative.
III. both sine
pression
sin X
+ cos X
is
X- 90° = 360° - (90° - x)
sin {x
In Quadrant and cosine are negative, and hence their sum must be In Quadrant IV. the negative. sine is negative and cosine positive. The sine and cosine are equal at 315°, after which the cosine is greater than sine. Hence the exis
in terms of the functions
= 270° -f X. - 90°) - sin (270° + x) = — cos X. cos {x - 90°) - cos (270° x)
greater than sine, and the ex-
pression
Find the functions of
— 90°)
-1-
= sin X. tan {x - 90°) - tan (270° -f x) = — cot X. cot {x - 90°) = cot (270° + x) = — tan X.
negative
Find the functions of
from 135° to 315°, and positive between 0° and 135°, and 315° and
{x
— 180°)
360°.
of
cc.
54.
in terms of the functions
69
TEACHERS EDITION.
= — cos X. - 180°) = tan (180° - x) tan = tan X. cot {x - 180°) = cot (180° + x) = cot X.
- 180° = 360° - (180° - X) = 180° + X. sin {x - 180°) = sin (180° + x) = — sin X. cos {x - 180°) = cos (180° + x) X
(a;
Page
48.
sin (90°
+ y)
Exercise XI. Find the value of
1.
and
cos {x
+ y) when
=
sin
-tV. cosy
f,
sin (x
2/
sin
sin {x
= |,
a?
+ y) cos x
= sin 90° cos y + cos 90° siny
12
= (1 Xcosy) + (Oxsiny) = cos y.
13-
= sin x cos y + cos x sin y
+ y)
cos (90°
'5^13 + 5^13
cos
(a?
+ 3/) =
36
20
_ 56^
65
65
65*
cos X cos
:«
(tan 90°
—
sin x sin
2.
Find
15
33
65
65
65"
sin (90°
— y)
-y) by making Formulas [8] and [9].
a;
= and
cos
= 90°
— cos 90° cos 90° = 0.
cos
y
in
4.
the
+ y)
—
-,
sin y.
= (0 X cos y) + (1 X sin y) = sin y.
sin
V ^ cosy
3.
Find, by Formulas [4H11], four functions of 90° + y.
first
= — tan y. ,
by Formulas
[4]-[ll],
four functions of 180°
— y.
- y) = sin 180° cosy — cos 180° siny = (0 X cos y) — (— 1 X sin y) = sin y, cos (180° - y) = cos 180° cos y + sin 180° sin y = (— 1 X cos y) + (0 X sin y)
= — cosy. tan (180° - y) =
^ cosy
cot (180°
the
,
+ y)
Find, first
= — cot y.
sin (180°
= 1. sin (90° - y) = (1 X cosy) — (0 X sin y) = cos y. cos (90° - y) = cos 90° cos y + sin 90° sin y
sin 90°
y)
cos y ^ sin y
cot (90°
- y)
= sin 90° .-.
(5^13
48
(90°
sin (90°
1/
=
"'5^13)
+
= cos 90° cos y — sin 90° sin y = (0 X cos y) — (1 X sin y) = — sin y.
=
= — tan y.
- y)
—V^ = — cot y.
cos ;
sin
,
y
TRIGONOMETEY.
70 Find, by Formulas
5.
the
first
[4]-[ll],
four functions of 180°
sin (180°
= cos 270° cos y — sin 270° sin y = (0 X cos 2/) — (— 1 X sin y) = sin y.
+ y)
= sin 180° cos y + = (0 X cos y) + = — sin y.
(-
cos 180° sin 1
X
y
sin y)
+ y)
cos (270°
+ y.
+ y)
tan (270°
— cosy^ = — =— :
cos (180°
sm y
+ y)
= cos 180° cos y — sin 180° sin y = (— 1 X cos — (0 X sin y) = — cos y. tan (180°
+ y)
= — sin ^v = — cosy
=
cot (180°
tan y.
V "^
:
Find, by Formulas [4]-[ll], the first four functions of 270° — y.
cos (360°
6.
- y) = sin 270° cos y — cos 270° sin y = (— 1 X cos — (0 X sin y) = — cos y. cos (270° - y) = cos 270° cos y + sin 270° sin y = (0 X cos + (- 1 X sin y) = — sin y.
- y) — sin v = =— -^
cosy
=
— —
= — cot y.
r-"^
sin
the
by Formulas
Find,
first
y.
sin (360°
+
+
y.
y)
2/.
[4]-[ll],
four functions of 270°
sin (270°
+
= sin 360° cos y + cos 360° sin y = (0 X cos 2/) + (1 X sin y) = sin cos (360°
by Formulas
[4]-[ll],
four functions of 360°
2/
Find,
,
2/
.
= :i^iM = tan2/. — cos
first
,
tan y.
cot (360° -2/) cos y
9.
tan (270° -2/)
- y)
2/.
2/)
;
— y.
- y)
tan (360°
2/)
— sin y cot (270° - y)
[4]-[ll],
= cos 360° cos y + sin 360° sin y = (1 X cos 2/) + (0 X sin y) = cos
sin (270°
the
by Formulas
= sin 360° cos y - cos 360° sin y = (0 X cos 2/) — (1 X sin y) = — sin y.
+ y) ,
7.
i.
four functions of 360°
sin (360°
—v^ = cot = — cos
= — tan y.
— cos y
first
—^y = cot y. = — cos — sin y
—
sin
Find,
8.
the
,
+ y)
cot (270°
2/)
,
cot y.
+ y.
y)
= sin 270° cos y + cos 270° sin y = (- 1 X cos 2/) + (0 X sin y) = — cos y.
+ y)
= cos 360° cos y — sin 360° sin y = (1 X cos y) — {Ox sin y) = cos y. tan (360° +2/)
_ sm_y ^ ^^^ ^^ cos
2/
TEACHERS EDITION. cot (360°
=
+
sin
y)
—^ = cot
cos y —
,
siny
first
-
270°)
= sin X cos 270° — cos x sin 270° = sin X — cos X (— 1) = cos X. - 270°) cos = cos X cos 270° + sin x sin 270° = cos X + sin x (— 1) = — sin X. a;
a;
Find, by Formulas [4]-[ll],
10.
the
y.
(a;
1
four functions of x
— 90°.
- 90°) = sin X cos 90° — cos x sin 90° = (0 X sin — (1 X cos x) = — cos X.
sin {x
(a;
a;
a;)
cos (x
tan
-270°)
(a;
cos X
- 90°)
= + = (0 X cos x) +{lx sin x) = sin X. - 90°) tan — cos = —
sin x sin 90°
cos X cos 90°
= — cot X. — sm X - 270°) cot — _ sin X _ — (a;
tan
a;.
cos X
(a;
a;
cot X.
13.
sin X
cot {x
_
- 90°) sin X
— cosx 11.
the
the
sin {x
four functions of
by Formulas
four functions of
[4]-[ll],
— y.
- y) = sin 0° cos y — cos 0° sin y = (0 X cos y) — (1 X sin y) = — sin y. cos (0° - y) = cos 0° cos y + sin 0° sin y = (1 X cos 2/) + (0 X sin y) = cos y. sin (0°
= — tan X.
Find, by Formulas [4]-[ll],
first
Find,
first
a;
— 180°.
- 180°)
= sin x cos 180° — cos x sin 180° = sin x (— 1) — cos X = — sin X. - 180°) cos = cos X cos 180° + sin x sin 180° = cos X (— 1) + sin X a;
tan(0°-3/)
(a;
= — sin ^V = — tan y. .
cosy
a;
= — cos X. - 180°) tan — sin X = — cos X - 180°) cot — cos X
cot(0°-y)
=
(a;
tan
14.
the
12.
Find,
first
y -"^
by Formulas
Find,
first
= — cot v. ,
[4]-[ll],
four functions of a;— 270°.
by Formulas
[4]-[ll],
four functions of 45°
— y.
— y) = sin 45° cos y — cos 45° sin y = 2^V2 cos y — ^\/2 siny = ^ V2 (cos y — sin y).
sin (45°
a;.
sin X
the
cos
X.
(a;
cot
—
— siny
72
TRIGONOMETEY.
— y) = cos 45° cos y + sin 45° siny = |V2 cos + ^ y/2 sin y = \yf2 (cos + sin y). tan (45° - y) _ cos y — sin y _ 1 — tan y COS (45°
2/
2/
cos
cot (45°
y +
sin
1
3/
+
+
tan (30°
^
^ V3
_
1
tan y
tan y
+ y)
_ V3
_ cos y + sin y _ cot y + 1 cos y — sin y cot y — 1
+
— jV3 tany
cot (30°
- y)
y)
_ cos y + \/3 sin y Vo cos y — sin y divide each term by VS cos y,
cos
y
— sin y by
divide each term 15.
the
Find,
first
by Formulas
sin (45°
_
+ y)
cos
+ y) y + sin y _
1
cos
y — sin y
1 — tan y
cot (45°
17.
the
the
Find,
first
sin
4-
coty +
y
by Formulas
+
cos y
+
[4] -[11],
+ y. y
= \ (cos y + \/3 sin y). cos (30°
+ y)
= cos 30° cos y — sin 30° sin y
= J ( V3 cos y — sin y).
— y.
cos
y
y — sin y
+ V3
sin
y
_ V3 — tan y 1 + VS tan y
1
cos 30° sin
[4]-[ll].
- y)
_ Vs
y)
= sin 30°
by Formulas
- y) = sin 60° cos y — cos 60° sin y — K"^ cos y — sin y). cos (60° - y) = cos 60° cos y + sin 60° sin y = 2 (cos y + \/3 sin y).
cos
four functions of 30°
sin (30°
VS
four functions of 60°
cot (60°
16.
—1
sin (60°
+ y) y
Find,
first
tan (60°
+ tan y
_ cos y — sin y _ cot y — 1 cos
sin y,
+ y)
= cos 45° cos y — sin 45° sin y 5= J V2 cos y — ^\/2 sin y = J V2 (cos y — sin y). tan (45°
"V^ cot y
coty +
+ y.
= sin 45° cos y + cos 45° sin y = iV2 cosy + jVlsiny = J V2 (cos y + sin y). cos (45°
=
[4]-[ll],
four functions of 45°
.
VSsiny
cosy +
-y)
_ cos y + VB sin y VS cos y — sin y _ ^V3coty + 1 cot y — \ Vs 18. sin
Find
3x =
sin 3 a; in
sin (2a;
= sin 2
a;
+
terms of sin
x.
x)
cos X
+ cos 2 a; since.
TEACHERS EDITION. sin
2x
cos 2 a;
= 2 sin x cos x. = cos^o; — sin^rc.
cot
73
+cosa;
Ja;='Y-
Substituting, sin
3x
=
2 sin re
+
= But
sin X cos^a;
3 sin X cos^aj
cos^a;
— sin^a;.
3
= 1 — sin^a;. 3
a;
= 3 sin x — 3 sin^x — sin'a; = 3 sin x — sin^a;. 4:
+
=
a;
cos X.
^l- cos
— sin^x
Substituting, sin 3
i
V3
cos^a;
cos
1
+
1
— cos X
cos
a;
X
— 3 cos x = 1 + cos X. — 4 cos X = — 2. cosx
= -• 2
19. Find cos 3 a; cos
3x
sin 2 a;
cos 2 a;
in terms of cos a;,
= cos {2x + x) = cos 2 X cos X — sin 2 x sin x. = 2 sin x cos x. = cos^a; — sin^a;.
sin^x
= 1 — cos^x
= 1- 1^3 4
sm
X
4'
-4-iVs.
Substituting,
— sin^a; cos x — 2 sin^x cos x = cos^a; — 3 sin'^a; cos x. sin% = 1 — cos^a;.
cos 3 a; == cos^a;
But
Substituting, cos 3 a;
= cos^a; — 3 cos + 3 cos'a; = 4 cos% — 3 cos x. a;
22. Given and cos^x.
Given tan | a; =
1
;
tan
find cos x.
ix =
^-
a;
1-Va96
=V} + cosa;
1 —cos 1 = 1 + cos X + cos = 1 — cos 2 cos X = 0. cos x = 0.
1-oaVq
•
a;
= 0.10051 a;.
cos J X
= ^ - + cos X
=^
I'l
21. Given sin
a;.
— cosx
cos X
a;
1
find sin ^ x
= VOM.
+ cosa; 1
;
= 0.2. cos^x = 1 — sin'^x = 1 - 0.04.
sin cos
= 0.2
sin X
cos X
20.
sin x
cot |
a;
= V3
;
+ 0.4 VS
find
= 0.99494.
—
TRIGONOMETRY.
74 23.
Given cos
and tan
cos 2 sin
a;
= 0.5
;
find cos 2 x
2
2 x. .<;
a;
cos 2a;
^%+V2
= cos'^a; — sin^a;. =
W'-(iy-' =
0.25
Vs.
multiply by
2-V2'
-0.75
_ 2
tan
a;
tan 2 a;
X
COS
a;
1
= \/2- 1 = 0.4142.
2\/3
a;
— tan^a;
1
—3 cot 2 ^
= -V3.
tana;
=
=1
a;
^x
^V2+V2 ;
find the
iV2-\/2
sin X
cos .'.
= cos I sin
Given tan 45° functions of 22° 30^ 24.
(2-V2f
= ^ V(2 - V2)2 X V2 = (1-JV2)XV2
^
2 tan
=
|
\ 4-2
= _ 0.50 = - 1.
= sm
2-V2
I2+V2
a;
= cos X. %va?-x + cos'^a; = 1. sin^a; + sin'^a; = 1. 2sin'^a; = 1. sin^a; = \. = ^ \/2 = cos sin
2-V2
sin X
a;
=
\/2
+1 =
25. Given sin 30° functions of 15°. a;.
sin 30°
= 0.5
;
find the
= 0.5 = -•
sin J a; or sin 22° 30^
_
2.4142.
2
ll-^\/2
= aV2-V2 2 = 0.3827.
cos 30°
=V-^V! = iV3.
sin ^
a;
= ^-
cosa;
cos A a; or cos 22° 30' l
^
+ iV2
sin 15°
2
= jV2-V3 = 0.25885.
= aV2+V2 2 = 0.9239. tan \ x
l-iV^
=
sm 2i''_ * A a;
cosja;
AV2-V2 f 1V2+V2
cos 15°
= -v
\
—^ 2
= iV2 + V3 = 0.96592.
TEACHERS EDITION.
2V3
=
tan 15°
75
2 tan a;
= z sma cos
^1 + ^ 1
2- V3 2 + V3
cos^a;
2
But
+ V3
.-.
1
cos^a;
+
+
tan^a;
+ ^V3
= -1-.
2sina^cosaj
= 2siT^^cos^^
2 sin X cos
=
cos •.
28.
a;
a;
1
^^^13o^sjn33^+sin3_°. cos 33°
+
cos 2
a;
tannic
+
sin-'a;
cos'^aj
15°.
cos 2 a;
= cos^a; — sin^a;.
1- ein^a? cos^a;
1
+
sin^B cos^a;
= cos (x+y) + cos {x—y). by
cos'-^a;
— sin^a; =
2 cos a; cosy
(1)
_ cos% — sin^a;
(2).
cos^a;
tan
a;
cos x.
cos^a?
1
2 sin X cosy
Divide
+
=
18°,
= sin (a;+y) + sin {x—y). (2)
a:
1
cos 3°
Then (1)
2 sin
= 1 — tan^a;
26. Prove that
= y=
1
a?
Prove the formula cos 2
X
1.
iV3
= 2 + V3 = 3.7321.
Let
=
sin^a;
cos^a;
4-3 = 2- \/3 = 0.26799.
-V;
+
_ cos^a; + sin^a;
(2-V3)2
cot 15°
1
a;
cos^j;
2-V3 ^ 2-V3
-42 + V3
+ tan^j; =
;
= sm (a;+y) + sin (a;-y) cos {x+y)
+
sin^a;
_ cos^a; — sin^a; ~ i
+ cos {x—y)
Substitute values of x and y,
= cos^a; — sin^aj.
^^^^go_sjn33^J:_sin3f_ cos 33°
+ cos 3° 29.
Prove the formula
27.
=
sin 2 a;
= 2 sin x cos x.
o a;
tan i
a;
1
+
tan^a;
sin
= 1
2 tan x
sin 2
•
Prove the formula
tan 2 a;
=
+
a;
cos X
vT^ cos
a?
vr + cos
a;
76
TRIGONOMETRY.
1
1
•
TEACHERS EDITION.
= ^ \/2 COS + J \/2 sin = 2"\/2 (cos + sin x). Ko — = cos X — sin (45°
cos (7= cos
a;
a;
tc
4-
tan
/
.'.
—
\
1
x)
+
sin X
By
Dividing numerator and denominator by cos X,
-
tan (45°
a;)
= 1 — tan X 1
cos
=
a?
-.
cos X
[180°- (^ + ^)]
= — cos (A +
A + cos B + cos C + COS B — cos {A + 5).
cos J.
[22],
= 2 cos J (^ + B) cos J (^ - B) — cos (J. + B). By
+ tan X
B).
[17],
= 2 cosK^ + 34. If A, B, C are the angles of a triangle, prove that sin
A + B\n B
-{
sin
C
By
= 4 cos J J. cos ^ ^ cos I C. Bin
=
H^ + B) cos
^-
cos
1(^ +
ButK^ + Then, by
-S)
.'.
+ B) cos | J.
C= cos
[90°
-1{A +
B)]
tan
+
C
are the angles of
a triangle, prove that
A + cos B + cos C = 1 + 4 sin ^ sin J ^ sin | C. -J
tan
tan
B).
C= tan [180° -(^1 + B)] = - tan {A + ^).
A + tan 5
= tan {A + B){1 — tan A tan P) = tan {A + B) .•.
cos
are the angles of
Again,
cos J ^.
= ^mi{A + B). sin A + smB + sin = 4 cos ^ -4 cos ^ ^ cos J C. If ^, 5,
1
B + tan C = tan ^ X tan B X tan C. Since ^ + 5 + (7= 180°,
.-.
4 sin I
(J.
C
C=1S0°-{A+
==
35.
A
[22],
2 cos ^ -4 cos I 5.
cos J
If A, B,
36.
tan
5)].
= But
[23],
a triangle, prove that
= ^^
cos A^-{- cos J5^
.-.
[2cos^(^ + ^)] X [cos i Ia-B)-co& ^{A+B)]-{-h
= (2 sin J C) (2 sin ^ ^ sin ^ 5) + = 1 + 4 sin ^ ^ sin J .B sin ^ C.
{A + 5)
= 2sini(^ + B) [cos ^(^ - B) +
-5)
+
X [2 sin ^ J. sin ^ 5] +
= sin^+sin ^+sin [180°-(^+5)] = sin ^ + sin B + sin {A + B) = 2 sin ^ (^ + B) cos 1{A--B) 2sin
^)
= [2cosH^+^)]
^ + sin j5 + sin C
+
H^ -
B) cos
- 2 cos2 ^ (^ +
— tan {A + ^) tan ^ tan B, tan J. + tan B + tan Q = tan (^ +
^)
- tan {A +
^)
— tan {A + 5) tan J. tan B = - tan {A + 5) tan ^ tan B = tan j4 tan B tan C.
TRIGONOMETRY.
78
C
If A, B,
37.
a triangle, prove
+
cot ^ J.
cos
cot J i?
X
a;
cos a;
2 cos 2 a; 2 sin X cos
cot f C.
J5 + iC=90°,
Since^^ +
2x [13]
sin
^ + cot J C
cot J
= cot J J. X
are the angles of tliat
[12] sin 2
..cotiC=ta,n^{A +
H^
+
cot J J.
cot f
= tan J
40. Change to a form more convenient for logarithmic computation
^ + cot J C
cot X
+ 5) + i{A+ + t&n}{B+C) = tan J ( A + -g) X tan J (^ + C) X tan K^ + QBy substitution, cot -4. + cot J ^ + cot J C = cot^ilX cot ^ 5 X cot J C tan
( J.
C)
cot
form more conlogarithmic computation
Change
+
tan
aj
_ cos X
sin
a;
sin X
cos
a;
sin X cos X
_
2
(cos^a;
2 sin
+ a;
sin^o;)
cos x
2 sin 2
a;
39. Change to a form more convenient for logarithmic computation cot X — tan X.
— tan X _ cos K _ sin X
cot X
sin
a;
cos x
_ cos% — sin% sin X cos a;
y.
a;
= ^^^^^^, a;
sin
y tan?/=^^^^-^J.
[2]
cosy
Adding,
_ cos X cos y + sin x sin y sin
a;
cosy
Substitute for cos x cos ?/-l-sin x sin y its
— y), _ cos (x — y)
equal cos {x
sin X cos
[9]
y
41. Change to a form more convenient for logarithmic computation cot X — tan y. ,
_ cos^a; + sin^g;
tan
cos
to a
venient for cot X + tan X. cot X
+
sin X
2"
38.
a;
2 cot 2 a;,
B),
and cot J ^ = tan + Q^ and cot J J. = tan J (^ + C). .'.
a;
2 cos 2 a;
tany
v = sm ^• cosy
TEACHERS EDITION. 1 1
— cos 2x
1
+
cos
2x
79
'
TBIGONOMETEY.
80
Page
Exercise XII. 1.
What
do the formulas of
§
become when one of the angles right angle
By
36
is
53.
division,
AD _ sin B DB sin A
a
?
B
?iB4 =
^.
sin
A
a
" BD
a
But
What
3.
come when
Formula
does
^ = 90°
when ^-180°? triangle cases
If angle
C is
a right angle,
a -
sin
=
sin
c
.
When
_ sin C _
c
c
A ^
b sin
sin
B
=
_Sines
c;
= c.
and
A
4.
'
DB ^ sin ^ (7 B
b'^
cos
= BC. = AC
A = l.
c=AB. =b—
a
Ba b^AC
AD _ sm^O sin
A = 0°.
c.
B
bisect angle C.
CD
cos
B
b
a
sin
=
A
C
adjacent sides.
CD
A = 90°,
a2
.-.
side into parts proportional to the
Then
— 2bc cos A.
0°,
of a triangle divides the opposite
CD
c^
When A =
Prove by means of the Law of that the bisector of an angle
Let
+
c\
a 2.
b^
is
[26]
= 62 + c2-26c. When A = 180°, cos ^ = - 1. .•.a2 = 62 + c2 + 2Jc.
;
C
c
sin
?
become in each of these
+
a^
.-.
sin
a sin
^ = 0°
does the
1
B sin B a _ sin A = tan A sin B b b
=
a^
C
What
?
Formula
A = sm A
[26] be-
when
?
B
is
c=BA.
=
a
=
b
+
c.
Prove that whether the angle acute or obtuse
What
c
=
a cos
B
two symmetrical formulas obtained by
+
b cos A.
are
the
TEACHERS EDITION. changing the letters? What does the formula become when B = 90° ?
A
81
THIGONOMETRY.
82 The
limit of
^-5
the limit of the
.'.
is
180°.
maximum value
Since
angle
is
C
is
ofH^-^) 180°
= 90°.
(ii.)
2
The /.
limit of
^ + 5 is
the limit of the
maximum value
of^{A + B)
= 1^0! =90° 2
(ii.)
C=90°;
A- B = 90°, and B=0.
a— a + (i.)
A
b
^
tan
^{A — B) + B)
isin^{A
b
When C= 90°. + B = 90°.
B=90°-A.
Q-5^ tann^-(Q0°-^)1 a +
tan 45°
b _^
tan
or
b
(^-45°) 1
= tan (^-45°).
tri-
^-5 = 90°, and 5=a
_ tan ^{A — B)
b
tan ^ ( J.
+ 5)
A + B + 0^180°, A + 2B = 180°
A-B
7. Find the form to which Formula [27] reduces, and describe the nature of the triangle when (i.)
When
a— a +
180°.
a right angle, the
a right triangle.
=90°
;
TEACHERS EDITION. 2.
Given
a= J.
C=
795,
55° 20^
c ==
a
=
663.986.
c
79° 59^
B=
44°
C=
.-.
=
rt
= 2.67111 = 468.93.
log c
795.
^= A-\-B=
a.
colog sin
6=567.688,
B = 44° 4F;
log
= 2.90526 A = 0.00654 log sin Q = 9.75931
log
find
= 79°59^
83
4F
124° 40^ 55° 20'.
4.
Given
find
a = 820, A = 12° 49',
5 =141°
c
a
= 820.
2.90037
^ = 0.00667 log sin B = 9.84707
A=
= 2.75411 = 567.688.
b
12° 49'
^ = 141°
59'
^ + .5 =154° C= 25°
48'
.-,
= 2.90037 colog sin A = 0.00667 log sin C = 9.91512 = 2.82216 log c = 663.986. c log a
3.
a
log
= 2.91381
a
A = 0.65398 log sin B = 9.78950
log 6
=3.36729
b
= 2276.63. =
find
= 804,
C=35°4^ 6 = 577.31, c = 468.93.
^ = 99°55^ ^ = 45° 1'
A = 0.65398
sm
= 9.62918
log
= 3.19697 = 1573.89.
log c
= 804.
A= ^=
99° 55' 45°
V
5.
Giv^n
find
C= 47°
= 1005, A = 78° 19', ^ = 54° 27'; c
A + B = 144° 56' C= 35° 4'. .-.
= 2.90526 colog sin A = 0.00654 log sin B = 9.84961 = 2.76141 log 6 = 577.31. 6 log
2.91381
colog sin
c
a
12'.
colog sin
log a
Given
12',
= 2276.63, = 1573.89.
6
59';
colog sin
log b
C= 25°
c
a
1340.6,
=
1113.8.
,6
= 1005.
^=
78° 19'
B = 54° 27' ^ + ^ = 132° 46' C= 47° 14'. .-.
14',
a=
TRIGONOMETKY.
84 logc
TEACHERS EDITION. In order to determine the disA from a place
9.
tance of a hostile fort
and the angles ABC measured, and found to be 322.55 yards, 60° 3V, and 56° 10^, respectively. Find the disB, a line
and
BC
BCA were
tance
AB. a
=
322.55.
85
86
TRIGONOMETRY.
the results
when
Id" V, and y
= 42°
cZ
=
11.237,
5V.
d= 11.237. x= 19° V 2/= 42° 54^ x
+ y= .-.
z
\ogd colog sin z log sin X
log a
=
61° 55^
118°
5^
= 1.05065 = 0.05440 = 9.51301
x
TEACHERS EDITION. Given
15.
C= 105°
h
=
A = 30°,
7.07107,
a and
find
;
c
87
= ^V2xiV3 + ^V2xi
without
using logarithms.
Let of c
p and
made by
q denote the segments the dropped from C.
a
±
= 9.562 x*V2 (v/6+V2)
5 = 45°. sin A= —
_
xV2
19.124
V6+V2
2 sin
_ (19.124xV2)(\/6-V2) 6-2 = 9.562 (VS-l) = 6.999 = 7.
5 = iV2. a_ i b~ iV2 a
= _b_ V2
b
7
X iVS
sin J.
7.07107
f V.4J
7V3
= 5.
1.41421 cos
= a sin 5
7\/6
2 V2 = 3.5 V6 = 8.573.
^ = |\/3 = 0.86602.
b
P = bx
= 7.07107 X 0.86602 = 6.12369. ?
The base of a triangle is 600 and the angles at the base are 30° and 120°. Find the other sides and the altitude without using log17.
0.86602
= sin 5 = JV2 = 0.70711.
= ax 0.70711 = 5 X 0.70711 = 3.53555. c=p + q = 6.12369 + 3.53555 = 9.65924.
q
feet,
arithms.
AB = 600. A = 30°. 5=120°. .-.
C= 30°. A=C. a 7
16.
Given
B = 60°
c
=
a and
find
;
A = 45°,
9.562, b
without
using logarithms.
= c = 600 _ a sin 5 sin
A
_ 600 X sin (180°- 60°) sin 30°
C= 75°. a
=
e sin
sin
sin
feet.
^ 600x|a/3
A
1
C
a= sin (45° + 30°) = sin 45° cos 30° + cos45°sin30°.
= 600x1.732051 = 1039.2. h = b sin A == 1039.2 X J = 519.6 feet.
TEIGONOMETRY. 18. Two angles of a triangle are, the one 20°, the other 40°. Find
a
out using logarithms.
=
X
20°,
c
=
and a and
5.43775.
=
a sin
X
3
0.9659
0.4226
40'',
= 6.857.
_a
Given one side of a triangle
20. sin
nat sin
y
equal to
a;
nat sin y
a
.-.
C
A
h be opposite sides.
sin X
Then
0.766
0.4226
=
sin 37
X
3
sin J.
the ratio of the opposite sides with-
Let
5
sin
:
6
the
adjacent angles
= 0.3420.
equal each to 30°.
Find the radius
=
0.6428.
of the circumscribed circle without
3420 6428.
using logarithms.
:
:
:
:
27,
:
855
:
2E
1607.
sin^ The angles of a triangle are 5 10 21, and the side opposite
19. as
:
sin
the smallest angle
Find the other
is
sides
equal to
3.
without using
logarithms.
Since the angles A, B,
C
are as
5:10:21,
^ = ^6 ^=10
of 180°
=
Determine the number of solu-
tions in each of the following cases
=
a
80,
a=
but
and .*.
two
6
=
100,
A = 30°.
< 6 100, > 5 sin ^ = 100 X A < 90°.
80
Page
(iii.)
.'.
|,
(iv.)
.'.
a
ft
one solution.
but
40,
6
A< a = 13.4, a
(v.)
59.
90°.
no solution.
solutions.
= 50, = 100, ^ = 30°. a = 50 = 6 sin il = 100 X h
15.588.
= 100, ^ = 30°. a = 40< bsmA^lOOxh =
and
=
.*.
(ii.)
a
:
3
18 Vs.
i2=9V3 =
Exercise XIV.
(i.)
V3
^V3
= 25°. of 180° = 50°. C=|iof 180° = 105°.
1.
A = sin 120°
= sin (180° - 60°) = sin 60°. sin 60° = |\/3. 27 _54 _ 54xV3 •.2i2 =
:
6
= 11.46, ^=77° 20'.
= 13.4 >
5=11.46.
one solution.
a
= 70, 5 = 75, A = 60°. a = 70 < & = 75, >6sin^ = 75xiV3,
;
89
TEACHERS EDITION.
^ = 60° < 90°.
and .*.
(vi.)
two a
=
log a
log sin
solutions.
134.16, b
=
colog sin 84.54,
B = 52° b
.*.
2.
logc
9^ 11^^.
B< 90°,
c
B = 0.7897.
< a sin B. 84.54 < 134.16 X 0.7897.
&
no solution.
Given
find
4.
Given
= 91.06, b = 77.04, A = 51° 9^
find
B = 41° 13^
a
C= 6^^
a
B=
12°13^34'^
colog a
b
C=146°15'26^^
log 6
= 840, = 485, ^ = 21°31^ Here .'.
a^
b,
c
=
1272.15.
and log
B <.0.
sin
one solution, colog a log b
log sin log sin
B .-.
= 0.97308 = 8.54761 A = 0.06505 = 9.58574 - 10 = 0.38525.
= 7.07572 - 10 = 2.68574 A - 9.56440 B = 9.32586 = 12° 13^ 34^^ = 146° 15^ 26^^
.-.
log a
colog sin
C= A=
0.43560
=1.95933
log a log sin
colog sin
=
3.10454
=
1272.15.
2.92428
c
c
c
3.
a 6
Given
find
B = 57° 23^ 40^^ C= 2° 1^20^^
= 9.399, = 9.197,
^ = 120° 35^ colog a log h
log sin
log sin
B
c
=
0.38525.
= 9.02692 = 0.96365
9.99963
A = 0.10857 = 2.06753 = 116.82.
57° 23^ 40^^ 2°
Given
a = 55.55, 6 = 66.66, B = 77°
W
Here .•.
find
4.0''
b^ a,
;
A = 54° 3F 13^^, 7^^, (7 = 47° 44^ c = 50.481.
and log
sin
A < 0.
one solution.
= 1.74468 log sin B = 9.98999 = 8.17613-10 colog b log sin A = 9.91080 = 54° 31^ 13^^. A log a
10
A = 9.93495 B = 9.92552 =
C=
9.74466 5.
log
c
87° 37^ 54^^
= 116.82.
= 8.04067 - 10 = 1.88672 log sin A = 9.89143 log sin B = 9.81882 = 41° 13^ B = 87° 37^ 54^^
log c
log sin
(7
F20^^
.\G
=47°
44^
7''.
TRIGONOMETRY.
90
=
1.74468
C=
9.86925
log a log sin
= 1.70313 = 50.481.
log c c
a
=
Given a h
find
.5
Given 309, 6
=
360,
B=
find
6'^,
.•.
c
c^= 55.41.
=
log b log sin
colog a log sin
B .-.
C log a log sin
colog sin log
c
c
2.55630
A = 9.55904 = 7.51004 - 10 B = 9.62538 = 24° 57^ 54^^. = 133° 47^ 41^^. = 2.48996 C = 9.85843 A = 0.44096 = 2.78935 = 615.67.
Second Solution.
^/=
=
180°
-B
155° 2^ 6'\
C'=B-A = 3°
\
log a
43^ 29^^
= 2.48996 = 8.81267 A = 0.44096
log sin C^
colog sin
two
log c^
= 1.74359
c^
= 55.41.
44°
V
28^^
135° 58^ 32^^
e=
97°44^20^^
(7=
5° 47^ 16^^
=
13.954,
solutions.
3° 43^ 29'^
= 615.67.
38° 14:^12^^;
Here a
C=133°47MF^ C'=
=
9.787,
c^= 1.4203.
24°57^54^^ 2'
=
5^=
c
^ = 21°14'25^^
^^=155°
= 8.716,
^=
A = 0.08920
colog sin
6.
7.
colog
sin
5 < 0.
;
TEACHERS log sin
A=
log b
= 0.71684 = 9.35655 - 10 B = 10.00000 = 90°. = 32° 22' 43'^
colog a log sin
£ .-.
C log b log cos log c
9.
log a
9.92661
= 0.71684 ^ = 9.72877 = 0.44561 =2.791.
c
Given a
=
34,
6
=
22,
B = 30^ 20' A = 51° 18'
find
^/=
27",
128° 41' 33",
(7= 98° 21' 33", (7'= c
=
20° 58' 27", 43.098,
c'= 15.593.
Here
b <^a, but >»
a
sin
J5,
^ < 90°. .•.
two solutions log
=
a
log sin
colog b log sin
1.53148
B = 9.70332 = 8.65758 - 10 A = 9.89238 A = 51° 18' 27". A^=
128° 41' 33".
.-.
C=
98° 21° 33".
.-.
C'= 20° 58' 27".
log a
=1.53148
log sin
(7=9.99536
colog sin
A = 0.10762
log c c
= 1.63446 = 43.098.
EDITION".
and
91
TRIGONOMETEY.
92 log a
colog 6
log sin log sin
= 1.87506 = 8.53760 - 10 -8 = 9.44715 A = 9.85981 A = 46° 2V A5'\ A'=
133° 36^ 15^^
(7= 117° 20^ 39^^
C'= 30° log a
8^
=
9'^.
1.87506
colog sin
A = 0.14019
log sin
= 9.94854 = 1.96379
log
c
= sin A =
0.14019
(7''=
9.70075
log a
colog
log sin log c^
1.87506
TEACHERS EDITION. log (5
-
a)
colog (a
+
6)
10
= 0.13675 ^{A + tan UB-A) = 8.66129 i{B-A) = 2° 37^30^^.
log tan
log
= 0.73239 = 7.79215 -
log 6
B)
93
94 logc
TRIGONOMETRY.
TEACHERS EDITION. 8.
Given
95
96 Since e
TRIGONOMETRY.
TEACHERS EDITION. 16. In order to find the distance between two objects A and B separated by a swamp, a station C was chosen, and the distances CA = 3825
CB =
yards,
In triangle
3475.6 yards, together
b
^
.
+a=
49^
^ sin
34° 51^
sin 41° 49^'
_ 562 sin
7300.6 log 562
log sin 34° 51^
58° 44^ 30^^ colog sin 41° 49^ 4° 30^ 30^^
B= A=
log tan
log b
63° 15^
b
54° 14^
= 2.54332 = 6.13664-10
log(&-a) colog(6 + a)
^(5 + ^) =
In triangle
10.21680
(5 - ^) = 8.89676 H^ - ^) = 4° 30^ 30^^
a
^ sin 95°
C
9.94799
B
0.04916 3.57978
c
3800.
log 562
colog sin 43° 12^ log a
17.
and
Two
B
562 cos 5° 40^ sin 43° 12'
log cos 5° 40'
logc
inaccessible
objects
A
40^
sin 43° 12^'
_ log sin
= 2.74974 = 9.75696 = 0.17604 = 2.68274 = 481.65.
CBD
562
3.58263
colog sin
5V
B = 1S0°-{C+D) = 43° 12^.
log tan J
log 6
34°
sin 41° 49^
b-a = Sid A B+ A = 117° 29^ i{B + A)=
= 41° 562
to B.
b
ACD
A = 1S0°-{C+D)
with the angle ACB =^ 62° SV, were measured. Find the distance from
^
97
= 2.74974 = 9.99787 = 0.16460 = 2.91221 = 816.98.
are each viewed from two
In triangle ACB C and D 562 yards apart. The angle ACB is 62° 12^ BCD "*—^ x tan 41° 8^ ABB 60° 49^ and ADC tan }iA-B)= a+b stations
34°
5V
;
required the distance e
AB.
|
(A+B)
H^ + ^) = Hi80°-C) = 58° 54'. a- 6 = 816.98 -481.65 = 335.33. a + 6 = 816.98 +481.65 = 1298.63.
TRIGONOMETRY.
98
= b) = colog log tan i(^ + ^)= log
(a— {a +
b)
2.52547 6.88651
<
-10
10.21951
logtani(.4-5)= 9.63149
i(^-^)=
23° 10^26'^.
^ = 82° log a
4^26^A
2.91221
log sin
C
9.94674
colog sin
A
0.00418
logc
2.86313
c
729.68.
18.
Two
trains start at the
same
time from the same station, and move along straight tracks that form an angle of 30°, one train at the rate of 30 miles an hoar, the other at the rate of 40 miles an hour. How far apart are the trains at the end of half an hour ?
A i{A
teachers' edition. log a
99
TRIGONOMETRY.
100
= 8.16749-10
colog
s
colog
(s-b)
= = =
log (s- a) log (s-c)
9.52288
^=]36°23^50^^
- 10
^
1.23045
.-.
2 )20.60206-20
165° 45^.
(7=
14° 15^
78, 6
= 101, c = 29
.-.
2.
Given a
=
3.
10.30103
& c
63° 26^
Given a
6'\
;
78
colog
= 208 s = 104.
s-h= s — c= = - a) = log(s-5) = log \s-c) =
26.
5
145
s
= = = =
- 10
7.82974
8.43180-10 0.47712 2.03342
2)18.77208-20 75.
logtanJ^=
9.38604
= 13°40^16^^ A = 27° 20^ 32^^.
^J.
7.98297-10
(s
8.58503
- 10
0.47712 colog
1.87506
9.46009
= =
log
- 5) = {s-c) =
log
tan|5=
colog
J^ = 16° A = 32°
s
log(s-a)
2)18.92018-20
= colog (s 5) = log(s-a) = log (s - c) =
111
=
3.
s
s
= 145, c = 40;
=
(s- a) log(s-5) log {s-c)
colog
colog
5
a
colog
colog
logtanJ^=
25^ 16^^
c= 40 2 s = 296 s = 148. s — a= 37. s - 5 = 3. = s-c 108.
= 101 = _29
— a=
= 111,
126° 52^ 12^^
2s
s
34^ 44^^.
find the angles.
find the angles.
a=
5 = 168° C= 11°
1.68124
5= J ^ = B= A^- B =
log tan 1
+
(s
7.82974-10 1.56820 9.52288
- 10
2.03342
5^27^^ 2 )20.95424-20
10^ 54^^
^B=
7.98297-10 9.52288
10.47712 71° 33^ 54^^.
^ = 143° 7M8^^. 5 + ^ = 170° 28^ 20^^. C= 9° 3F 40^^.
- 10
1.41497 1.87506
.-.
2 )20.79588-20
logtan
1^ = 10.39794
^B=
68° 11^ 55^^
4.
Given a
= 21,
find the angles.
6
= 26,
c
=
31
;
TEACHERS EDITIOX
= 21 6=26 c = 31 2s = 78 s = 39. s-a=18. s-6 = 13. a
s-c= colog
s
colog (s-a) log
(s-h)
log
(s
- c)
= = = =
.-.
colog log
{s-a)=
(s - 6) = log {s-c) =
colog
8.40894-10 8.74473
- 10
1.11394 0.90309
9.58535
6^ 13^^
1.25527 8.88606
-
10
0.90309
2 )19.45836-20
logtaiiJ^= 9.72668
B = 56° ^ + ^ = 98° .-.
.-.
5.
Given a
C=
= 19,
a= &= c
5
12M9^^
= 34,
19
34
= _49
2s=102 s= 51. s-a'= 32.
s-h= 8-C=
6^ 36^^
81° 47^ 11^^
find the angles.
17. 2.
colog
(s-a)
-
log
(s
log
(s-c)
5)
M
= 8.40894-10
s
s
= = = =
logtanJ^=
8.
A = 42°
colog
8.29243
8.49485
- 10 - 10
1.23045 0.30103
2)18.31876-20
2)19.17070-20
logtani^=
101
49
=
9.15938 8° 12^ 48^^.
TRIGONOMETRY.
102
= 8.12494-10
cologs
log(s-a)=
1.50515
- 6) =
8.60206
{s-c)=
1.25527
2)
19.48742
logtanJ^=
9.74371
colog log
(s
8.
Given a =73,
- 10 -
(7= 75° 10^ 41^^
Given a
=
31, 5
= 58,
c
=
79
find the angles.
a=
37
6=
58
colog
s
{s-a) log(s-&) log (s-c)
colog
= = = =
50. 29. 8.
8.06048 8.30103
- 10 - 10
1.46240 0.90309
2 )18.72700-20
logtan^^=
9.36350
J^ = 13°0M4^^. A = 26° 0^ 29^^. =
8.06048
log(s-a)=
1.69897
colog
s
colog
(s
log
(s
- 6) = - c) =
8.53760
-
10
- 10
0.90309
2)19.20014-20
logtanJ5=
^B= .-.
;
colog
log
= 174 s= 87.
s-b= s-e=
82
41.
— c=
32.
s
(s-a) log(s-6)
2s
— a=
73
&=
colog
c=_79
s
a=
s-b= s
7.
9.60007
21°42M0^^
B = 43° C= 110°
= 82,
c
= 91;
c= 91 2s = 246 s = 123. — = 50. a s
20
J5 = 28°59^52^^ B = 57° 59^ 44^^. .-.
6
find the angles.
25^ 20^^.
34^ 11^/.
(s - c)
= = = =
7.91009
8.30103 1.61278
1.50515
- 10 - 10
TEACHERS EDITION. cologs
colog(s-a) log(s-6) log (s-c)
= 8.16481-10 = 8.26819-10 = 1.11344 = 0.17846 2)17.72490-20
log tan J
A
8.86245
A = 8°
20^
= 8.16481-10 log(s-a) = 1.73181 colog -b) = 8.88656 - 10 cologs
(.s
log (s-c)
-
0.17846
2)18.96164-20
logtaiiJ5= 9.48082
^B=
16° 50^
B=
33° 40^
.-.(7=138°.
10.
Given a=y/5,b = V6,
c
= V7]
find the angles.
=V5 = 2.2361 6 =V6 = 2.4495 c=V7 = 2.6458 a
2s =7.3314 s
= 3.6657.
s-a = 1.4296. 8-6 = 1.2162. s-c = 1.0199. = = log (s-c) = colog s colog (s-a) = log (s-b)
0.08500 0.00856 9.43585 9.84478
- 10 - 10
2 )19.37419-20
logtanJ^=
9.68709
|^ = 25°56^36^^ A = 51° 53^ 12^^
colog
(s -
103
104
TRIGONOMETRY. s = — a= s - 6 = s — c =
colog
5.
colog {s-c) log
(s
- 6)
log
(s
-
5. 1.
= = = =
s
a)
- 10
8.95861
0.00000 0.69897 0.69897
^C= 56° C= 112° an
- 20
6'\
53^
isosceles triangle,
= 33° = 6,
Given a
&
C)
33^ 27^^.
=
6,
c
=
6
;
find the angles.
The
triangle
equilateral
is
and
also equiangular.
A = B=C=^of 180° = 60°.
.-.
14.
Given a
= 6,
6
= 5,
c
sides
a and
= 12
;
find the angles.
The sum is less .•.
the triangle
15.
two
of the
than the side
Given
VS — 1
;
a
is
=
impossible.
h
=V6,
find the angles.
6
a=2 = V6 = 2.4495
c= VS- 1 = 0.732 2 6 = 5.1815 5 = 2.5908. s-a = 0.5908. s- 6 = 0.1413.
s-c=
b
c.
2,
log tan ^ J.
=
iA=
26^ 33^^.
^ = ^=1(180°-
13.
= 9.38869-10
1.8588.
s
c
9.61725
logtan|5 = 10.23855 logtan^(7= 9.11946
log tan^ (7= 10.17828
is
logr
colog
2 ) 20.35655
Since this
log r2
= 9.77144-10 = 9.15014-10 = 0.26923 = 9.58656 - 10 = 18.77737 - 20
log(s-a) log(5-&) log {s-c)
ll.
s
=
22° 30^
TEACHEES EDITION.
hA =
105
TEIGONOMETRY.
106 have but one value
;
II. the side opposite
may have two
not known, and ues
;
while in Case the angle is
therefore the angle also
val-
may
have two values. 20, If the sides of a triangle are 3,
4,
and
6,
find the sine of the
largest angle.
= h= c =
_6
=
13
a
2s 5
3
4
= 6.5.
107
TEACHERS EDITION.
Exercise XVII. 1.
Given a
= 4474.5,
b
=
C= 116° 30' 20^' find the F= Ja&sin C.
= 3.65075 = 3.33536 = 9.69897 -
log a log b
colog 2 log sin
log
C =
9.95177
=
6.63685
F
F 2.
Given
A = 66°
10
=4333600. b
4' 19^'
= ;
21.66,
c
=
36.94,
find the area.
F=^bcsmA. log 6
2164.5,
area.
;
Page
69.
TRIGONOMETRY.
108
s=
637.
— a= s-b= s-c=
432.
s
192.
= 2.80414
log
s
log
{s-a) = 1.11394
log Is-b) log {s-c)
2 log i^ log
F 7.
13.
F
= 2.63548 = 2.28330 = 8.83686 = 4.41843.
= 26208.
Given h = 149,
^ = 70° 42^ 30^^
TEACHEES EDITION. log
109
TRIGONOMETRY.
110
a Obtain a formula for the area
13.
of an isosceles trapezoid in terms of
the two parallel sides and an acute angle.
AB = a.
Let
F=^{a + h)
c.
- = tan A P c
.-.
=p
tan A.
F= l{a^h)x^{a—h) tan A = \{a?- ¥) tan A.
14.
Two sides and
included angle
and and included
of a triangle are 2416, 1712,
30°
;
and two
sides
angle of another triangle are 1948, and 150°; find the sum of
2848,
their areas.
Let a
=
2416, c
= 1712, B =
F= h ac sin B. log a
=3.38310
30°.
;
Ill
TEACHERS EDITION.
Exercise XVIII. 1. From a ship sailing down the English Channel the Eddy stone was observed to bear N. 33° 45^ W.
and
after
the
ship
miles S. 67° 30^ 15^ E.
Find
its
had
sailed
18
bore N. 11° distance from each
W.
it
position of the ship.
a
= 18
miles.
ACE^ DCB=
33° 45^
ABF=
11° 15^
67° 30°.
ACB = 1S0°~{ACE + DCB) = 78° 45^ CBD = 90° -DCB = 22° 30^ ABC= 90° - {CBD + ABF) = 56° 15^ .^^C=45°. 6_sin_S a sin J.
c
_ sin C sin
A
Page log a
70.
TRIGONOMETRY.
112 S'A
sin
ASS'
TEACHERS EDITION.
colog sin
By
1.00000
log a
A
Table VI.
0.30103
logc
1.30103
c
20.
113
:.
B = Z. area O = 28.274. 301036
BOQ-
X360°
1296000 75259
In a
with the radius 3 find the area of the part comprised 6.
between
circle
parallel
lengths are 4 and
chords 5.
(Two
whose
area sector
BOC 75259
solu-
324000
tions.)
log 75259 log 28.274
colog 324000
log area
Area sector
F=Vs{s-
log(s-6) cologs(s-a)
= = =
0.30103
0.30103
9.30103-10
2 )19.90309-20
logtanJ-BOC= 9.95155-10
J50C= 41° BOQ= 83°
48^ 38^^ 37^ 16^^
X
28.274.
= 4.87656 = 1.45139 = 4.48945 - 10 = 0.81740
50C= 6.5675.
In triangle
log (s-c)
X360^
324000
BOO
TRIGONOMETRY.
114 log
{s-d)
log
(s
colog
=0.39794
- a)
= 0.39794 = 9.56067 - 10
s{s-o)
40^391^
DOA
1296000 sector
DOA =
2)0.35655
logtanJDOJ.
= 0.17828
^DOA= 56° DOA = 112°
26^35.5^^
F= log log log
Area
sector
= 0.94772 = 8.8658.
DxA = 4.72.
Area segment
\^s{s-o){s-a){s-d). s{s-o)
=
- d) (s - a)
= 0.39794 = 0.39794
(s
0.43933
2)1.23521
= 0.61761 = 4.1458.
\ogF
F
A
and B, two inaccessible obsame horizontal plane, are observed from a balloon at C and from a point D directly under the balloon, and in the same horizontal plane with A and B. If CD = 2000 yards, Z ^ CD = 10° 15^ 10^^ 7.
Area segment
DACB
= areaO-[%C+i)a;^] = 21.4587. Area segment DAC^B^ = DxA - B'xC^
= 2.6247.
AD = 361,76.
jects in the
Z BCD =
6° 7^ 20''^
34^ 50^^ find
Z ADB = 49°
AB.
AD = DCxt&nACD, logtanAOD =9,25739 log
X 28.274.
=5.60894 =1.45139 log 28.274 colog 1296000 = 3.88739-10
53^ IV^,
DOA
360°.
log 406391
log area
In triangle
406391
1296000
X
DO
log^D
=
3.30103
2.55842
DB = DC xts^n BCD.
TEACHEES EDITION.
U^ + ^)
=
65° 12^ 35^^.
115
.
SPHEEICAL TEIGOIsrOMETET. Page
Exercise XIX. 1.
70°,
104.
The angles of a triangle are! 4. Prove and 100°; find the sides' three right
80°,
of the polar triangle.
Given to find
If angles A, B,
Q
respectively,
are right angles, the side b the meas-
c^.
= 110°. 180°- 80° = 100°.
a^=180°6/=
have
angles, the sides of the
triangle are quadrants.
A = 70°, B = 80°, C= 100°;
a\ y,
that, if a triangle
c the measure of and a the measure of angle
ure of angle B,
70°
angle
C,
:
A, are each
c/=i80°-100°=
80°. .•.
sides
= 90° of
;
ABC
triangle
are
quadrants. 2. 90°,
The sides of a triangle are 40°, and 125° find the angles of ;
the polar triangle.
Given a
= 40°, b = B\ C
90°, c
=
125° ;
required A^,
^/=180°- 40° = ^/=180°- 90°=
140°.
side.
90°.
- 125° =
(7= 180°
5. Prove that, if a triangle have two right angles, the sides opposite these angles are quadrants, and the third angle is measured by the number of degrees in the opposite
In spherical triangle
55°.
B = C=
let
3.
Prove
that
the
quadrantal triangle
is
polar of a a right
tri-
Let the triangle
ABC be
a quad-
rantal triangle.
Then Let
A^B^C
But .-.
triangle
angle.
= 90°.
a right
BC.
B + b^= 180°, .-.
;
c^= 90°.
90°. is
of degrees in
b'= 90° C-fc^= 180°,
= 180°. 5= 90°.
h
A'B^C
AC and AB A is measured
ABC.
Now
be the polar triangle.
^^=
angle.
Let A^B^C^ be the polar triangle of
I
B'^
.'.
We are to prove quadrants, also that by the number
angle.
rt.
ABC
.•.
tri-
and
A^B^C^ is 5^=90°, (7=90°.
triangle
But
B^+
b
=
180°.
isosceles,
TRIGONOMETRY.
118
h=
90°.
C'+ c = c=
180°,
.-.
.-.
5
.*.
Now
and
90°.
are quadrants.
ABCi?, bi-rectangular.
.'.
A
.'.
J. is
is
the pole of side BC.
measured by
side
BC.
How can the sides
6.
cal
c
triangle
length,
when
of a spheribe found in units of
the
radius of the sphere
By
length is
of
the
known.
using the formula 2 irB
=
C.
For instance, if the sides of a triangle were 40°, 90°, 125°, the sides in terms of R would be
;
TEACHERS EDITION. than 90°, their cosines have opposite and their product will be
C=c= 90°.
Also
signs,
cos
negative. If cos c
then
=
a negative
c is greater
B = cos 6. B = h.
quantity,
than 90°.
a
(iv.) If
Prove, by aid of Formula
[40],
.-.
C= 90°.
.-.
[40],
A = cos a X sin B. sin B is positive. B < 180°. Hence sign of cos A is same as cos
and both
sign of cos a,
Deduce from [37]-[42] the
4.
tan^ ^ 6
From
must be
either greater or less than 90°
;
that
= tan ^{c — a) tan ^{0 + a).
[37],
cos 6
page
74,
= cos c
1
1
sec
1
1
-I
a
= 90°
= 90°. = cos a X
cos
.'.
Hence a (ii.)
;
(iv.) if
.
and 6-90°?
If c
(i.)
= 90°
cos h.
= 0. = 90°.
= 90°. = X sin B. cos ^ = 0. A = 90°. cos -B = COS 6 X 1 = cos 6. B=h. If 6 = 90°, B = 90° and A = a.
If
.
^^„ cos
a
cos J.
COS c
— cos c
c os
+
cos a
+
page
47,
cos 6 [18],
a
cos c
— cos 6 = tan^ ^ 6.
+ cos 6 And if in [23] and [22], page 48, we write a and c in place of A and 1
B and
divide [23] by [22],
cos a — cos cos a
+
we
get
c
cos c
= — tan ^{a e) tan i{a — c) = tan J (c + a) tan J (c — a). -j-
.'.
tan^ J 6
= tan ^ (c + a) tan f (c — a). 5.
= 90° and a = 90°. cos A = COS a sin B. A = a = 90°.
+
a
— cos 6 _
.-.
If c
= cos
1
1
.-.
(iii.)
7>
1
But by
a or cos 6 or 6
+
a
cos a
;
and a
;
J.
What inferences may be drawn
respecting the values of the other parts (i.) if c = 90° (ii.) if a = 90°
= 90°
a
„^o — cos = cos a — cos c
COS
:
we have
whence
alike in kind.
(iii.) if c
for-
mula
.-.
3.
= 90°.
= X 0. c = 90°.
always alike in kind.
is,
6
cos c
a right spherical triangle each leg and the opposite angle are that in
Formula
= 90°,
A = 90°, B = 90°.
Then 2.
119
Deduce from [37]-[42] the for-
mula tan2 (45°
=
-^A) tan J
(c
—
a) cot J (c
+
a).
TRIGONOMETRY.
120
From
[38],
sin
= sm a sm c
A
— sin A _
1
+
sin
A
sin c
— sin
sin c
+
sin
c
— tan a _ sin
(c
— a)
c
+
(c
+
tan a
sin
a)
— g) 1 p _ sin (c t&n^B
when, operating as in Example we have 1
tan tan
4,
sin (c
+
sin (c
— a) esc (c + a).
a a
a)
7. Deduce from [37]-[42] the forpage 47, we substitute mula 90° -r A for z, and remember that tan^ J c = — cos {A+B) sec {A—B). cos (90° + A) = — sm A, [19] reduces By [42], cos c = cot ^ cot B to the form
If in [19],
_ cot
= cot2(45°+J^) ^^^^ + sm ^
tan
1
= tan2(45°-J^), (since 45°
+
and 45°
J J.
— ^A
sin
sin c
+
are
By
Deduce from [37]-[42] the
'
By
B = sin (c — a) esc (c by operating
+
for-
a).
as be-
1
— cos B _ tan c — tan a
1
+
[18],
cos
B
page
tan
c
+
+
And we sin G
— cos c
1
+
m place oic tan
place of tan
i.
B + cot A
47,
tan^ I
c,
~ cos{A-B) tan^ J c = - cos {A+B) sec {A—B). Deduce from [37]-[42] the
8.
c,
A and
for-
mula
sin
a
m
tan2 J J.
= tan [J ( J. + ^) - 45°] tan [J (^-5) + 45°].
•
cos a a,
cos c
tan
47,
write on the second side
-,
page
1
cos i?
cos c
obtain
[18],
c
'
.-.
tan a
l^^^^ = tanH5. 1
+ cos
B — cot A B + cot A sin ^ cos A sin J. cos ^ sm B cos ^ sin J. cos B _ sin Asm B — cos J. cos ^ sin Asm B + cos J. cos B _ — cos (^ + B)
mula
[39],
1
tan
= tan }{c — a) cot J (c + a).
From
— cos c _ tan B — cot A
tan
and
.•.tan2(45°-J.4)
tan^ ^
1
^4
sin J.
= tan ^{c — a) cot J (c + a).
6.
.
B'
whence, as before,
complementary angles). And by dividmg [21] by [20], page 48, and writing c for A and a for B, we have c — sin
J.
and reducing, we
From cos
[40]
a
= cos A CSC 5 = cos A sin
B
TEACHERS EDITION. whence, by proceeding as before,
By
1
— cos a _ sin B — cos A
1
+
cos
a
sin
.-.
tan (a;-45°) tan (45°
B + cos A
(sin
cos
1
+
If in
cos
a a
tan^ h a.
[6], p. 44,
we make x =
45°,
And,
=
in
like
tan (x -45°)
= \{A^-Bl
x-y = B, tan [^(^
+
tan2 ^ a
a;
Exercise XXI. 1.
Show
that Napier's Rules lead
sin
a
[38], [39], [40],
and
[41].
sin b sin
a
sin h
= cos co. c cos co. A = sin c sin A. = cos co. c cos co. B = sin c sin B.
2.
come, sin CO.
5
B = tan a
tan co.
c.
cos J.
= tan a cot A = tan 6 tan co. c. = tan b cot
sin CO.
A = cos a cos co.
cos
sin CO.
cos
A
sin CO.
cos
B
c.
the
cos
- 5) +
45°]
A
cos -4
= tan [} {A + B) - 45°] tan [i {A-B) + 45°].
Page
to the equations contained in For-
mulas
+
sin J? .-.
A,'
+ ^)- 45°]
_ sin B —
making
we have
cos
we
+y=
tan [^ (^
= sm X — cos X
which x
values,
x
the equation becomes
sin X
+ y) + y)
= l{A-B\
cosy
manner,
— sin y)
y
whence
45° in [10], page 46,
(a^
(cc
may have any
^^
_ cos y + sin y cos y — sin y y
— y) — cos — y) + cos
(a?
y
make x
=
cos x) (cos
If in this equation, in
and y
we have tan (45°+ 2/)
+
a;
sin {x
sin
1
+ y)
(sin X — cos x) (sin y + cos y)
page 47,
[18],
121
109.
= tan co. B tan 5 = cot B tan 6. = tan a tan co. A = tan a cot -4,
What if we
will Napier's Rules be-
take as the five parts of
triangle,
the hypotenuse,
the
two oblique angles, and the complements of the two legs.
c.
= cos
a
5 = cos b =
^.
sin B.
cos co. J.,
cos 6 sin A.
(i.) Cosine of middle part equals product of cotangents of adjacent
parts.
(ii.) Cosine of middle part equals product of sines of opposite parts.
TRIGONOMETRY.
122
Exercise XXII. 1.
Solve the right triangle, given 5 = 43° 32^ ?>V\
a =36° 27^
Taking a
as the
have, hy Eule sin
whence 3
and
a
tan h ^
middle part,
I.,
-r,
= tan h cot B, — sin a tan B, tan h
we
Page
114.
123
TEACHEES EDITION. Solve the right triangle, given 10^ b = 150° 59^ U^\
4.
a
= 120°
6.
c
Solve the right triangle, given 49' 51", a = 14° 16' 35".
= 23°
= COS a cos h. t&nA = tana esc b. tan B = tan b esc a. cos c
log cos a log cos & log cos c c
log tan a log esc 6
log tan
cos b
=9.70115
= =
log cose
=9.96130
log sec a
= 0.01362
log cos &
=9.97492
b
= 9.91180 = 9.64295 = 63° 55' 43.3'^
sin
log sin
0.31437
A = 10.55002 cos
cot
log
Solve the right triangle, given 55° 9' 32'^ a = 22° 15' 7''.
7.
5.
c.
c
Solve the right triangle, given 33' 17", a = 32° 9' 17".
= 44°
cos b
= cos c
sec a.
tana =9.61188
log cose
=9.85283
log cote log
= 9.84266 cos B = 9.45454 B = 73° 27' 11.16".
log sec a log cos b
= 0.07231 = 9.92514
= sin a tan B. sin a = 9.57828 tan B = 10.52709
tan h log
log
log tan 6 =10.10537 b
cos
=51°
A =
sin
c.
= 10.10537 log cote = 9.84266 log cos A = 9.94803 28' 25.71".
=32°
41'.
A = sin a esc
log sin a
c.
=9.72608
= 0.15391 A = 9.87999 A = 49° 20'
log CSC e log sin
log tan b
A - 27°
b
53'.
tan b cot
c.
= 9.40562 = 10.33488 cos B = 9.76050 B = 54° 49' 23.3".
log tan
log
«
log tan a
= 0.06320 B = 9.80703 5 = 147°19'47.14'^
cot
= 0.39358 — —
B = tan a
log esc a
B = tan a
c.
A =9.78557 A = 37° 36' 49.4'^
log cote
cos
19° 17'.
A = sin a esc
log ^ CSC e
10.23565
log tan 5 =9.74383
=
=
log sin a =9.39199
J..= 105°44'21.25''.
c
= cos c sec a.
COS
B = tan a cot c. =
9.79840
= B = B=
10.00675
log tan a log cot c
log cos
16.3".
9.80515 50° 19' 16'^
TRIGONOMETEY.
124 .
c
8.
Solve the riglit triangle, given 13^ 4^^ a = 132° 1¥ 12'\
b
= 97°
cos b
= cos
c
or
sin
=9.09914
log sec a
= 0.17250
log sec
=9.27164
log cos
=
h
esc
c.
= 9.86945 = 0.00345 c A = 9.87290 A = 48° 16^ 10^^
B = sec a cos A. a =0.66004
A = 9.02323 log sin B = 9.68327 B = 28° 49^
79° 13^ 38.18^^
A = sin a
sin
28° 14^ 31.3^^.
151° 45^ 28.7^^
sec a.
log cose
log cos 6
= =
=
or
151° 10^
57.4^^,
2.6^^
log sin a log esc
log sin
But
A
10. Solve the right triangle, given
a
=
77°
and a must be of the same
A = 131°
50^^ sin c
log sin a
kind, .-.
2V
log CSC J.
43^ 50^^
log sin c
B = tan a cot
cos
log tan a
=
a
=
9.10259
log cot
c
log cos
B =
9.14455
sin c
=
sin
a
esc
log sin a
= 9.98935
^
=0.00243
= =
log sin c c
sin c
>
1,
= 92°
Since
c
A.
tan
log tan a log log
tan b log sin a
78° 53^ 20^^.
log tan b b
=
101° 6^ 40^^
sin h
=
tan a cot A.
cos
log
tana =10.64939
log cot
A =
9.02565
=
9.67504
log sin b
impossible.
B = 50° V V\ =
tan a sec B.
=11.31183
=
tan B.
9.99948
= 10.07619 =50°.
A = cos a sin B.
log cos a log sin
= sin a
B = 10.07671
its sine, it
values which are
c
also
is
sec B = 0.19223 tan c = 11.50406 c =91°47M0^^
supplements of each other.
Hence
e
9.99178
found from
is
which
47^ 32^^
log tan
may have two
= sin a esc A. = 9.98935 = 0.18588 = 10.17523
11. Solve the right triangle, given
a
58^ 53.3^^.
Solve the right triangle, given 77° 21^ 50^^ A = 83° 56^ 40^^
log CSC
.".
40^ 40^^
10.04196
B = 81° 9.
c.
A = 40°
= 8.68765
B = 9.88447
log cos .4
=8.57212
A = 92°
8^ 23^^.
TEACHERS EDITION. 12. Solve the right triangle, given
B = 12°
o = 2° 0^ 55^^
tan h
a tan B.
=8.54612 -9.35170
log sin a log tan
= sin
40\
5
log tan &
= 7.89782
h
=0° tan
log tan a
log sec
^
log tan
c
c
cos
c
27^ 10.2^^
= tan a sec B. = 8.54639 -0.01070
= 8.55709 = 2° 3' 55.7^^
A = cos a sin ^.
log cos a
= 9.99973
B = 9.34100 cos ^ =9.34073
log sin
log
A
=
77° 20^ 28.4^^
13, Solve the right triangle, given
a
=
20° 20^ 20^^
tan b log sin a
B = 38° = sin
10^ 10^^
a tan B.
125
TRIGONOMETRY.
126
16. Solve the right triangle, given c
= 112°
48^
A = 56° V 56^^ B = cos c tan A. 1
cot
= 9.58829 tan J. = 10.17427
log cose log
log cot
^=
9.76256
£ = 120°
= sin c sin A. sin c = 9.96467 sin ^ = 9.91958 sin
log
log
3^ 50^^
a
log sin a
a
=9.88425
= 50°.
log cos
= cos A tan c. A = 9.74532
log tan
c
tan b
=10.37638
log tan 5 =10.12170 b
17. Solve the
ri^
=
127°
4^ 30^^
TEACHERS EDITION. 20. Solve the right triangle, given
127
22. Solve the right triangle, given
A = 116° 43^ 12^^ B = 116° 3V 25'\ A = 90°, B = 88° 24^ 35^^. cos c = cot A cot B. cos a = cos A CSC £. But cot J. = 0. log cos A = 9.65286 log CSC B = 0.04830 cos c = 0. .•.
log cos a
= 9.70116
a =120° 10^ 3^^
= cos J. .A = 0. a = 0. a
cos
= cos -B CSC A. cos 5 = 9.64988 CSC J. = 0.04904 cos h = 9.69892
cos b
log log log
b
But
COS .•.
cos
= cos B CSC J.. A = l. b =B. b = 88° 24^ 35^^.
cos b
=119°59M6^^.
log cot log cot
log cos
= cot A cot B. ^ = 9.70190 5 = 9.69818 c = 9.40008 c = 75° 26^ 58^^
c
21. Solve the right triangle, given
A=
CSC B.
a =90°.
CSC
log cos
=90°.
c
Define a quadrantal triangle,
23.
its solution may be reduced to that of the right triangle.
and show how
A
quadrantal triangle
is
a
tri-
46° 59^ 42^^, B = 57° 59^ 17^^. angle having one or more of its sides equal to a quadrant. cos a = COS CSC B. Let A^B'Q^ be a quadrantal tri-
A
log cos
log esc
A = 9.83382 B = 0.07164
log cos a
a
= 9.90546 = 36° 27^
= cos B CSC A. B = 9.72435 A = 0.13591 b = 9.86026 b = 43° 32^ 37^^
cos b
log cos
log CSC log cos
cos c
= cot A
^ = 9.96973 cot B = 9.79599
log cos
c
c
A'B'=
90°, or
a
quadrant.
Let ABQ\)^
Then
its
polar triangle.
since
.-.
A'B'^ (7=180°, C=90°. ABQ\% a right triangle.
.•.
all parts of
may
the polar triangle
be found by formulas for right
The parts
triangle.
of
A^B^C^ may
then be found by subtracting proper parts of
A^Cfrom
180°.
cot B.
log cot log
angle with side
= 9.76572 = 54° 20^
24. Solve the quadrantal triangle
whose
sides are
a
=
b= c
=
:
174° 12^ 49.1^^ 94° 90°.
8^20'^,
TRIGONOMETRY.
128
Let A^, B^, C, a\ ¥, c' repre- log tan 0° 49^ 25.5^^= 8.15770 sent the corresponding angles and log tan 4° 57^ 45.4^/= 8.93867 sides of the polar triangle. 2 ) 7.0963 7
Then A'^
5=>
B'=
85°
C^=
90°.
tan^
47^ 10.9^^,
bV
log tan
W\
^a^= 8.54819 ia^= 2° 1^25^''. 4°
a^=
2^50^^.
^ = 175°57^10^^
\ c'
= - cos {B'+A') sec {B^-A'). B'+ A^=
91° 38^ 50.9^^
B^- A^=
80°
25. Solve the quadrantal triangle
4^ 29. F^.
in
which
= 90°, A = 110° c
log cos log sec
(5^+ J-O =8.45863 (^^- .40 = 0.76356
47^ 50^^
5=135°35^34.5^^
2 )9.22219 log tan Jc^
A\ B\
(7,
a',
¥,
c^ repre-
sent the corresponding angles and
Jc^= c^ =
22° 12^561^^.
sides of the polar triangle.
44° 25^ 53^^.
Then
C==135°34^ tan2 J b'= tan [J
Let
= 9.61110
{B'+
7^^
A^ - 45°]
a'= 69° 12^
10^^.
&/= 44° 24^ 25.5^^
(7=
90°.
tan[45°+H^^-^0]-
- 45° = 49^ 45° + i {B'- A^) = 85°
J (^^+ .SO
log tan
0° 49^ 25.5^^=
log tan 85°
tan
25.5^^ 2^ 14.6^^
A^=
log CSC
= 10.42043 = y 0.15505
log tan
A^=
8.15770
A^= log tan 1 6^=
a
9.60952
6^= 44° 17^
5 =135° (^^+
45°
6^58^^
75°
42^ 50^^
^0 - 45°]
= 104°
53^
tan
B^=
tan h^ csc
log tan
5^=
9.99101
=
0.02926
10^^.
tan[45°-H^'-^0]-
H-S^+^0 - 45° = 0°
10.57548
9.21903
J&^= 22° 8^35^^
tan2 J a'= tan [J
h^.
log tan a'
2^4.6^^= 11.06133 2)
tan a^ esc
log CSC a^ log tan
2^^.
a^.
B^= 10.02027 B^= 46°20M2^^ b
= 133°
39^ 48^^
49^ 25.5^^.
- J {B'-A') = 4° 57^
45.4^^
cos c'
= cot J.''
cot B^.
TEACHERS EDITION.
Given in a right spherical A = 42° 24^ 9^^ ^ = 9° 4^
log cot A'=^ 9.42452
28.
B^=
triangle,
log cot
log cos c^
9.97973
W^
= 9.40425
129
solve the triangle.
;
= cot J. cot B. tI = 10.03943 B = 10.79688 c = 10.83631
cos c c/
=
75° 18^ 2V^. log cot
(7=104°4F39^^
log cot log cos
26.
A,
Q
Given in a spherical triangle which and c = 90° solve the tri-
is
impossible.
;
.".
angle.
a
sin
.'.
= sin c
sin
= tan
impossible.
is
A.
= 1x1. a = 90°.
tan b
triangle
c cos
a
29. In a right triangle, given 119° IF, = 126°54^ solve
=
^
the triangle. tan c
A
log tan a
= cox0.
log cos .-.
cot
.-.
b
= 45°.
27.
=
90°
Given ;
= 0xco. B = 45°,
^ = 60°, C= 90°,
tan cot
= =
10.47452 71° 27^ 43^^
= sin a tan 5. sin a = 9.94105 tan B = 10.12446 tan b = 10.06551
tan b log
and
log
solve the triangle. sin
0.22154
c
log
c
B=
log tan c
5 = cos c tan A.
= tan a sec B. = 10.25298
6
= sin c sin A. b = tan c cos yl. 5 = cos c tan J.. a
cos .A ,
log cos a
=130°4F42^^
= cos a sin 5. =9.68807
B = 9.90292 log cos A = 9.59099 ^ = 112° 57^ log sin
sin .•.
sin
= 1. a = sin A. a = A = 60°. c
30.
= 00. tan 5 = 00. 6 = 90°. cos e = 0. .'•.cot^ = 0. tan
c
.".
B = 90°.
c
2^\
In a right triangle, given
= 50°,
b
= 44°
18^ 39^^
;
solve the
triangle.
cos a
log cos e colog cos b
log cos a
a
= cos c sec b. = 9.80807 = 0.14535 = 9.95342 = 26° 3^ 51^^.
TRIGONOMETRY.
130 sin
A=
TEACHERS EDITION.
131
sinM— cos'^B :
—
tan^a
sin^a
sin^a
cos^c
sin^c
cos^a
sin^c
sin% cos^a
cot^'c
— sin^a
cos^c
sm^c
cos'^a
_ sin% (cos^a — cos^c) (1) cos^'a sm'^c
Now cos c
= cos a
cos^a
= cos^c
cos
6.
But
cos^c
cos^S cos^c
= cos^a
sm
= sin
c
— cos% cos*6
cos^c sin^5
-Q^2
cos^5.
0~D'
b
0E\ DE'
sin^
= s in^Z)
sm^c
0E\ OD^ - OE^ 0E\ ED''
Substitute these values in sin^a
on^~OE^
(1),
ED"
— co&'^B
=
1.
'cos-'c
cos^a cos^S cot^^c
_ '"
cos'-^i
cos^c
X
sin-6
sin^^
.•.
sinM = cos^^ +
sin (6
+
c)
sin (b
+
g)
cos^6
=
2 cos^ J a cos 6 sin
—sm— +
sin & cos c ;
cos'^c— cos^acos*&
cos
sin^i?
cos^c sin^6
2
•
9
D
sin^a sm^ij
/cos^c f
c.
= sin 5 cos c + cos h sin c
cos^6 sin^5
•
sin^a sin^^.
36. In a right triangle prove that cos^5
cos^c sin^6
= sin^a
sin^^.
sin^^
— cos^g
= sin^a
— cos^^ = sin^a
.•.
cos^6
= sin'^a
_ ODx OE^
— cos'^acos^SX :
)
cos^c sin^6
J
:
c
,
T
1
\
7
cos
J
= (tan b cot c+1) cos b sin But tan b cot c = cos A, tanJ cote +1 ^ cos J. + 1. .•.
•
sm c c.
(1)
(2)
:
TRIGONOMETRY.
132 cos
4
^+
1
COS
Substitute in
(2),
(tan bc^e + 1) =
Substitute in sin (6
+
hA.
2 cos^ J A.
(1),
= 2 cos^ ^A cos h sin
c)
c.
37. In a right triangle prove that
— b) = 2 sin^ ^a cos 6 sin sin (c — b) = sin c cos b — cos c sin b sin {c
= sm c cos
7
A-,
1
—
cosf sin b\
5
^^s
(1)
=
cos
=
2
,
6).
(2)
In triangle
A. 1
— cos A
tan
CBD
5(7= tan 5Z)
sinM = '\~
= tan ni sec B
cos
A
_
tan cos
1
(3),
— cot c tan 5 = 2 sin^ | A.
,
sin (c
o
— 5) = 2 sin^ J a sin e cos b.
w
hypotenuse, and n the segments of the hypotenuse made by this arc adjacent to the legs a and b respectively,
prove that (i.)
tan^a
(ii.)
sin^p
= tan c tan m, = tan m tan n.
= tan m ^, X
B = tan m
In triangle
2d.
sinp
and
=
5
cos
cot c
cos
(2),
38. If, in a right triangle, p denote the arc of the great circle passing through the vertex of the right angle and perpendicular to the
m 5
Multiplying the two equations, tan''a
Substitute in
sec 5.
(3)
tan a
Substitute in
c.
cot c
= sin c cos 6(1— cote tan
1— cote tan
BCA cos B = tan a cot cos B :. tan a =
In triangle
sin c cos oy
\
cot c tan b
c.
tan c.
CBD tan
m
cot
ili";
CAD sinp = tan n
cot iV!
Mx cot i\^=
1.
in triangle
But
cot
.-.
if+iV"=90°. sin^p
=
tan
m
tan w.
TEACHERS EDITION.
Exercise XXIII. 1.
In an isosceles spherical
tri-
and the
side
angle, given the base h
a
A
find
;
the angle at the base,
133
Page
Also, cos -4
= tan ^ a
cot a
1— cosa
B
V l-hcosa
the angle at the vertex, and h the altitude.
cos%
1— cos^a
= cos a X
ABA^
be an isosceles triangle, A and A^ being the equal angles, a and a^ the equal sides. Let h the arc of a great circle be drawn from B perpendicular to
Let
116.
1
-I-
a
cos
= cos a J sec^ J a.
AA'. Let
h
and
6
=J6
c
=
co. c
be the given parts.
ABA\ ABA\ ABA\
in triangle
a in triangle
B = ^B in cos
triangle
A = cot a
= sin a sin^ B = esc a = cos a cos A sin
2"
&
tan |
b.
sin ^ B.
Given the side a of a regular
3.
spherical polygon of
sin | b.
the
sec J &.
angle
distance
A
n
sides
;
find
of the polygon, the
H .from
the centre of the
In an equilateral spherical polygon to one of its vertices, and triangle, given the side a find the the distance r from the centre to the middle point of one of its sides. angle A. In the equilat. triangle AA^A^^ draw &rc AC ± to A^A^^. 2.
;
Then given
in
right
triangle
AA^C,
a.
sin ^ J.
= sin
J a esc a cos
a
1
=V^
sma
^'2 (1
— cos a - cos^a)
'2(1
-(-COS a)
/
1
ABDE
In the regular polygon arcs from the vertices A, B, etc., through the centre C, and from Cto M, the middle of one side.
draw 2
^ 1 4- cos
^ sec I
a.
a
134 Then
TRIGONOMETRY.
TEACHEES EDITION. Let AC^ be a cube.
E,equired
AB^D^. The lines AA^ and BB^ determine the plane ABB^A\ and the lines A^D^ and B^C^ determine the the dihedral angle
plane A^D^C^B^- and as
BB'
are
±
to
AA^
and
A' B' and B'C\
spectively, the planes
must be
re-
per-
pendicular. .".
the angle required
is
90°.
E-ABCD-F
be an octaheRequired the dihedral angle
Let dron.
E-BA-F,
Draw ^^and
FK 1. to AB, also
0.2" from intersection of axes. is
Then
EKF the plane angle required. Let
AB=\,
Then
0K= 0.5
(since
0^, OB, and and equal),
0^ are
dicular
and
0E{= OA) = sin
45°
= 0.7071. cot
log
EKO = ^• OE OiT
- 9.69897
OE =0.15051 log cot EKO = 9.84948 colog
perpen-
135
EKO = 54° EKF=^ 2 EKO = 109° Let
AE and DE
44^
9^^.
28^ 18^^.
be two faces of
,
TRIGONOMETEY.
136 log
AC
=
log sin 108°
0.51002
log sin
= 0.23078 = 0.71901 AD ^Z) = 5.2361. J ^Z) = 2.61805.
colog sin
colog sin 36° log
log sin J J. Oi) = log
AD AO
log I colog
= 0.00000 AB ABC = 9.93753
log
=9.97821
DCB = 0.30103 '
ADC = 0.23856 ADC = 1.73204:. AD=^iADC
log
But
=
IAD + colog J. 0.
AC gon
=0.41798
.-.
= 9.51177
a diagonal bf plane penta-
is
ABCDE. sin FCB -.AB:: AC^
|AOi)= 9.92975
log sin
IA0D=
0.86602.
sin
ABC: AC.
AB^mABC sin
58°16^52^^ log
AC>i)=116°33M4^^
AB
=
FCB
0.00000
log sin
ADC= 9.97821
colog sin
FBC = 0.23078
log
= 0.20899
AC-
AC= 1.61804.
AF=^AC
But
= 0.80902. In right triangle sin
Let
AOB, BOC, COD,
be
etc.,
equilateral triangles forming five of
or log sin
the surfaces of a regular icosahe-
log
and let AB, BC, CD, etc. = 1. Kegarding ABCDE as a plane
colog
dron,
pentagon, each angle .-.
ADF=
AFD
— AD
ADF= log Ai^+colog AD. = 9.90796 = 0.06247
Ai^
AD ADF= 9.97043
log sin
= 108°.
ADi^= 69° 5^ 48^^ But ADi^=JA-OD-C. A-OB~C= 138° IF 36^^
A5C=108°, BAC=2>Q°.
.-.
In a triangle of which sides are AB, BC, and ADC (regarding
ADC
as a straight line joining centres of
bases of triangles sin
AOB and
BOC),
DCB :AB:: sin ABC: ADC
A spherical
5.
cal
sides
and equal
diagonals divide right triangles.
AB sin ABC .-.ADC^ s'm DCB
'
square
is
a spheri-
quadrilateral which has equal
of the square, side a.
angles. it
Its
two
into four equal
Find the angle A having given the
TEACHERS EDITION.
B '""fl
137
TRIGONOMETRY.
138
Substituting for cos (180° etc.,
their equals
— cos J.',
-^0-
cos
A^= —
and
— cos A^= cos B^ cos C^ — sin B^ sin C^ cos a^.
cos
Write formulas
and
A
are given,
when
the side b
and a,
when
by
2.
h,
a=
for finding
and
c,
B
are
given. (i.)
By Rule
A = cot 6 tan m = tan 5
By Rule
cosp cos 5
whence .".
cosj9
cos
a seen
Or, since
?i
cos a
(li.)
By
5''.
A^
.5''
sin
cos
c^.
124. find
^=
Q3°15nV^, = 124° 7a7^^ ^=132°17^59^^ 88° 12^ 20^^,
C=
50° 2^
1^^;
c=
4^8^^
59°
17° 57^ 28.5^^.
^{a + b)^10Q°
cos A.
JC=
= cos 71 cosp, = cos a sec n. = cos m cosp, = cos 6 seem. = cos 6 seem. ={c — m),
25°
9M8.5^^
F
0.5^^
logcosJ(&-a) =9.97831 log sec J (a + 5) =0.55536 =9.33100 log cot J (7 log tan J (^ log sec
log sin J
B = tan n cot a,
tan n
= tan
a cos
-S.
II.,
= cos m cosp, cosp = cos b sec m. whence cos a = cos n cosp, cos p = cos a sec w. whence cos b sec m = cos a sec n. m = {c— n), Or, since = cos a sec w cos (c — n). cos 6
+ ^)=
0.86467
H^ + ^) = 0.86868
log cos J (a (7
log cos ^ c
cos b
.•.
sin
Given
I.,
cos
Rule
C^ cos
cos 5''
tan m,
= cos b sec m cos (c— m).
By Rule
whence
5
C^ cos a^;
II.,
cos a
whence
— cos A^
H& - a) =
I.,
cos
whence
(7=
XXV. Page
for finding,
B^ sin
B^= ~ cos J.'' cos (7 + sin J.'' sin
+
Napier's Rules, the side a c,
cos
cos C^
sin
similarly,
- 1, Exercise
1.
B^ 4-
obtain
Multiply by
cos
we
etc.,
+
=9.44464
6)
= 9.62622 = 9.93954 c J = 29° 32^ 9'^
logsinJ(&-a) =9.48900 log CSC J (a + 6) =0.01751 =0.33100 log cot iC log
tanJ(^-^) =
^{B-A)= i{A + B)= A=
9.83751 34° 31^ 24^^
97° 46^ 35^^ 63° 15^ 11^^
.5=132° 17^59^^ c
=
59°
¥ 18'^
TEACHERS EDITION, 3.
Given
l{h
find
= 129° 58^
a =120° 55^ 35^^,
J.
5=
B=
88°12^20^^
3^^
63° 15^ 9^^
c= 55°52^40^^
(7= 47° 42^ V^;
i(a-5)= 16°2F37.5^^ J (a
+
5)
=
104° 33^-57.5^^
J (7= log cos ^ (a log sec
23° 51^
= = =
0.59947
+ B)=
10.93600
-5)
Ha + 5)
log cot J
C
log tan 1{A
0.5^^
9.98205
0.35448
J(^ + ^)=96°36^36^^. log sin
Ka-^)
logcsc-^a + log cot
=9.44976
=0.01419
J)
^C
=0.35448
log tan |(J.-^)
=
9.81843
K^-^) = 33°2F27^^ HA+^)=96°36'36^^.
A = 129° B = 63° log sec J (J.
+
logcosKa + log sin ^
log cos
,5)
= 0.93890 =
9.60675
=9.94618
^-c
c
= 27°56^20^^. = 55° 52^ 40^^.
Given
5
= 63°15a2^^
c
= 47°42' V\
^ = 59°
4^25^^;
15' 9^^.
=9.40053
&)
C Jc
4.
58^ 3^^
find
= 88° 12^ 24^^ C = 55°52'42^^
.5
a = 50° F40'^
+ c)^
139
TRIGONOMETRY.
140 (c- 6) = Kc + ^) =
0.00001
log sin J
colog sin
log cot J
^
=
10.28434
log tan J
(C- .5)=
9.82205
^{Q-B)=
colog cos
log sin h lo g cos
B=
A
;
= cot y cot B, = tan B cos a.
y
II.,
A = cosp sin x, p = COS A CSC X. cos B = cos JO sin y, cos p = COS B CSC y. COS J. CSC = COS B esc y. = C— y, since
.'.
.'.
Or,
cos
= cos^csc?/sin(C—
By Rule
^{B + A) = f c =
33° 21^ 58^^.
2/).
=10.65032
+ a) =10.72585
= =
2-
1
ar,
p = COS ^
.'.
COS
B
cscy
y
=
(7—
9.20915
(7= 80° 41^5.4^^.
J(5-74)=
9.04625
log CSC
H^
-4) =
0.25965
+
=10.65032
log tan Jc' log tan J (J
-a) =
9.95622
-a) = 42° 7^ i(b + a) = 79°21M0^^ 1(6
CSC y.
= COS .4
9.33908
logsin II.,
cos
77° 23^ 24^^.
log cos J c log cos
.'.
COS
6° 23^ 12^^.
i(^ + ^y= 9.74035 logsecJ(&-a) =0.12972
A = C0SJ9 sin cos p = cos ^ CSC X. cos ^ = cos jD sin y,
Or, since
i{B-A)=
log sin
I.,
cos .*.
6=121°28aO^^ C=161°22ai^^
c=154°46^48^^
a;
cot X
Rule
39° 45^0^^,
= cot cot A, = tan A cos 6.
cos b
By
find
a= 37°14a0^^
log tan J (5
a;
.'.
Given
26°58M6^^
log tan I c
a?
cos J. (ii.)
126.
logcosi(^--^)= 9.99730 log sec |(^ + ^) = 0.07823
cos
.'.
2.
^= B=
Page
I.,
cos a
By Rule
= 9.92056 J a = 33° 36^ 30^^ a = 67° 13^
^a
56°1F57^^
1. What are the formnlas for comwhen B, C, and a are puting given and for computing B when A, C, and b are given?
cot
= 9.66376
21^ 12^^
Exercise XXVI.
.'.
A
33°34^37.8^^
C= 123°
By Rule
+ 5) =7.84843 1{C+ B) = 2.40837
logcos J(c
9.53770
esc
= = a
a;.
&
a;,
5 = cos ^ esc X sin (C—
a;).
121° 28^ 10^^ 37° 14^ 10^^.
(7= 161° 22^ 11^^
TEACHERS EDITION. 3,
Given
log cos J
find
^ = 128°4F49^/, a=125°4F44^^ 5= 107° 33^ 20^^ h= 82°47^34^^ c= 124° 12^ 31^^;
1{A-B)=
141
(5 -(7)
log
i(6
10° 34^ 14.5^^.
logsin
=9.97159
log tan J a log cos
9.99257
1{A-^B)=
0.32660
log tan J (a
-c) =9.80021 lih-c) =32° 15M5/^
log tan 1(6
=10.27624
log tan J c
+
5)
log sin J (5 +(7) log sec J (6 -c)
=10.59541
l{a+h) =104° 14^ 38.5^^. log sin i (J. colog sin ^{A log tan
-5)= + B)=
log cos ^ a log cos J
9.26351
= = l{a-l)
logtan|(a-6)
0.05457
9.59432
=
82°47^34^^
^(^ + 5) = 9.94543 colog cos J (a -h) = 0.03118 = 9.67012 log cos ^ c = 9.64673 log cos J a 1(7= 63° 4P.
C=127°22^.
88° 12^ 21^^ 78° 15^ 48^^
5.
J.
Given
find
= 125°4F44^^,
5=
52° 37^ 57^^;
a=
86° 15^5^^;
find 5
= 107°33^20^^ 55°47^40^^' 14^39.5^^.
i(A-C)= =26° J6
18^ 58.5^^
4.5^^.
(^-(7) = 9.96883 i{A + C) = 0.60896
log cos J log sec
= 9.69424
log tan J (a
i{a
+ +
c)
=0.27203
c)
=118° 7^33^^
+
C)
=
88° 12^ 21^^,
A== 78°15M8^^.
i(^+C') = 116°
c
5=
21° 27^
= 152° 43^ 5K^
c=
a = 128°31M6^^,
H^ + C) = 104°
log tan J 6
5 = 153° 17^ &^, C= 78°43^36^^
7^ 54^^.
152° 43^ 51^^.
21° 27^ 5^^
log sin
Given
9.95364
=0.07283 =9.86322
^ = 39°
(7= 82° 47^ 35^^,
4.
=
=9.88969
= = c A= I
a =125°41^44^^. h
^ J
=10.27624
^-c
6/>'.
J(5-C) = 9.78226 log CSC ^5 + C) = 0.04636
1{A + B) = 118° 7^ 34.5^^ =62° 6M5.5^^. Jc
colog cos
9.90074
+ c) =0.23040 + c) = 120° 28^
log tan J (5
(7-127°22^
J (J. - 5) =
=
(^+(7) = 0.35807 tan J a =9.97159
log sec*
0^21^^.
i(5-C)= 37°16M5^^ =43° 7^37.5^^ Ja
log sin J (J. log sec
f(a
log cos J 5
log cos J
9.98644
-c) =0.00743 =9.95248
5
=9.94635 i
^ = 27°
53^ 50^^,
^
;
TRIGONOMETRY.
142
J(A-C) = 9.563i3 log CSC (^+ (7) = 0.01356 = 9.69:124 log tan 16 logsin
Ha - c) =
10° -ZV IS'\
=
128° 31M6^^.
a
-I
log tan i (a
= 107° = B 55°
-c) =9.27093
Exercise XXVII. 1.
Given
Page find
find
a= 73°49^38^^ 5=116°42^30^^ 6
= 120° 53^ 35^^,
A=
c
^1
=
B.^
=
Co
A
log sin
log sin b log CSC a
B
log sin
.5= [180°-
= 116° 102° = ^ {B+A) ^{B-A)= 13° i(5
l{b-a)=
59° 12' 15'^ 55° 42'
=
23° 57' 17.4",
97° 42' 55",
C^=
29°
(a
+
&)
8' 39".
> 180°,
sin -Sj
I.
54' 54^^.
log sin
3V
A
log sin b
58.5^'.
colog sin a log sin 5i
= 9.76524 = 9.85498 = 0.31377 = 9.93399
9.98908
log CSC
0.61892
^1=120°
log tan
Hb-a) =
9.63898
B^=
60°28'43.5'^
c=120°57'27'^
+ a) = 9.99641 logcscH&-a) =0.39873 log tan ^ {B-A) = 9.39401
log sin ^{b
C
9.78915 58° 23^
C=116°47'
= 150° 6 = 134° A = 144°
Given a
47' 45".
59° 12' 15".
=10.24698
^c=
log cot J
>6
= sin .A sin 6 CSC a.
{B+A) = J (5-^) = c
a
42^ 30^^
log sin J
log tan J
8''.
hence two solutions.
47^ 36^^.
+ a)= 97°2F36.5'^ 23°
47M5'^
C,
A > 90°,
(63° 17^ 30^0]
120°
Ci=
¥\
= 9.99992 = 9.93355 = 0.01753 = 9.95100
47^ 40^^
128.
= 120°57^27^^
C=116°47^
88° 52^2^^;
33^ 20^^.
c
57'
4''.
42"
8° 20' 35.5".
=
142° 36' 28.5".
i(^-^i)= 1{A-B^)=
11° 47' 28.5".
\{a
+
b)
;
42° 35' 13.5".
+ 6) = l{a-b) =
log sin i (a
9.78338
log esc
0.83833
log tan J (.4-5i)= 9.31963
log cot J
5",
15' 54",
22'
^ia-b) =
(7i
= 9.94134 }Ci = 48°51'27.7". Ci = 97° 42' 55.4".
TEACHERS EDITION. log sin i{a
+
b)
=
9.78338
143
TRIGONOMETRY.
144 ^ {B
+ A) =
121°
^{B-A)= 1(5- a) = ^{b + a) =
log sin
colog sin
3°59^53^^ 151°
logsmH^+^) colog sin
colog sin ^{b
—
a)
= =
log cot J
(7
(7= 79°5F47.7^^ = 159° 43^ 35^^
(7
Given 5 c
65° 39^ 46^^;
find
= 100° 2^1.3^^ 5=
90°,
B=
98°30^28^^
a=
95°20^38.7^^;C=148° 5^33^^
c=147°41M3^^
= 9.99811 = 9.99519 = 0.00670 = 0.00000 b = 90°.
= 124° 7^20^^ = 159°53^ 2^^
log CSC
5 A
log sin b
1{A + B)^
99° 16^ 19.7^^.
i{A-B)= Ha - &) =
0° 45^51.7^^ 2° 40^ 19.4^^.
(7= 159° 43^ 35^^.
B
=0.03812
colog sin J.
UA+B) = U^-B) =
9.99428
colog sin log tan
^{a-b) =
8.66904
log tan
^c
log sin
= 9.95959 = 9.92024
log sin a
= 9.91795 b = 124° 7^ 20^^
log sin b
J.
Given
log sin
find
= 123°40nS^^
3.
log sin a
= 9.15200 = 35° 10^ 20^^. C J C=70° 20M0^^
^ = 113°39^2K^
1.87484
=10.53816
Jc=
73° 50^ 51. 7^^
c=147°4F43^^
(5 + ^) = 118° 39^ 49.5^^
iiB-A)= - a) = = (5 + a)
i (5 1
= 9.25234 1
0.15663
log tan J
i
C
=9.92968
9.68433 + «) (5-^) =9.31104
log sin
log cot I
log
5^ 25^^.
= 9.47198 c ^ = 16° 30^ 48^^ 1^36^^. c = 33°
log sin J (5
a=
(5
-a) =8.84443
log tan J c
.S
^-
H^-^) = 0.69787
log tan I (6
2.
+ a) = 9.99842 i{b-a) = 0.31128 tan i(^-^) =8.94264
4:4:\
11° 34^
5°
log tan
colog sin
29° 13^ 52^^. 94° 53^ 33^^.
9.99953 1.33144
log tan J
(^-5) =8.12520
log cot J
C
=9.45617
(5+^) = H-S-^) =
9.94422
1(7= 74°2^46.3^^
1.05901
H6 - a) =
C=148°5^33^^
9.7478 9
log sin 1
colog sin
+ 5) = i{a-b) =
log sin ^ (a
0^ 28.5^^.
log tan J
c
=10.75112 79°56^51^^
Jc= c =
159° 53^
2^^.
4.
Given
38° 0^ 12^^
A = 24° 33^ 9^^ B = a= 65° 20^ 13^^; show
that the triangle
is
impossible.
TEACHERS EDITION, cot X
log cos a log tan
B
=
cos a tan B.
log cos log sec
= 9.62042 = 9.89286 = 9.51328 X = 71° 56^
(c
sin (c
30^^
Since sin is
sin {c
A ^
= 9.95884 = 0.10349
log sin X
log sin log cot X
145
— x) = cos A sec B sin x.
--
9.97806
— x) = 0.04039 — x) = 1.0974. {c — x)';> 1, the angle C
impossible. .-.
Exercise XXIX.
the triangle
Page
131.
is
impossible.
TRIGONOMETEY.
146
^ 5
log tan J
log
A=
=9.75330
= 0.03328 = 0.23927 tan i (7 1 A = 29° 32^
log tan ^
C=
47°
J C = A= B=
60°
2' 26^^
59°
4^ 28^^
36^^
94° 23^ 12'r
C= 120°
4.
3,
Given
A=
20° 9' 54",
= 56° c = 66°
B=
55° 52' 31",
6
19' 40",
b c
= 108°30a4^^ ^=110°10^40^^ c= 84°46^34^^; C= 99°42^24^^
2s
6
5 c
= 131° = 108° = 84°
35^
44"
20'
;
C = 114° 20' 17".
4^ 52^^,
find
a
find
a = 20°16'38",
2l = 132°14^2F^
4^^
99° 42' 24".
Given
a a = 131° 35^
10' 40".
Wr
^B=
IK
132° 14' 21".
^=110°
= = =
= s =
20° 16'
38"
56° 19'
40"
66° 20'
44"
142° 57'
2"
s-a= s-b= s-c=
4^^
30^ 14^^
46^ 34^'
= 324°51^52^^ s = 162° 25' 56'^
71° 28' 31".
51° 11' 53". 8' 51".
15° 5°
7'
47".
2s
log sin
— (s — (s
s-a=^
30° 50' 52''.
log sin
s-b=
53° 55' 42".
log sin
(s
s-c =
77° 39' 22".
log CSC
s
log sin
— a) (s — b) (s — c)
log CSC
s
log sin
log sin
9.70991
(s
9.90756
—
a)
9.89172
b)
9.41715 8.95139
c)
0.02311
log tan^ r
8.28337
log tan r
9.14168.
9.98984
M
log tan^ r
10.12754
log tan J
5
= 9.24996 = 9.72453
log tan r
10.06377.
log tan i
(7
=
10.19029
=
10° 4' 56.8".
0.52023
log tan J log tan J
A 5
log tan ^
(7
= 0.35386 = 0.15621 = 0.07393
iA=
66°
7^ 10.6".
I5=
55°
5' 20".
I
C=
49° 51/ 12".
log tan
J J.
J 5 = 27°
56' 15.5".
f (7= 57° 10' 8.6". A = 20° 9' 54". .5
= 55°
52' 31".
a= 114°
20' 17".
TEACHERS EDITION.
•
1.
Given
Exercise
XXX.
Page
147
132.
148
TRIGONOMETRY.
8-A-
TEACHERS EDITION.
= 69° = 120° c = 159° 2 s = 349° s = 174° s-a = 105°
15^
6
42^ 47//
s-b= s-c= -^
-
23^
6.5^^
52° 41^ 33.5^^ 26° 57^ 43^^
=
c)
16^ 26^/
38^ 13^^.
87° 19^
Hs-6)= 1 (s
18^
15° 19^ 40^^.
(s - a) =
7° 39^ 50^^.
=11.32942 ^s log tan |(s- a) =10.11804 log tan
J(s-6)=
9.70645
logtan ^(s-c)=
9.12893
log tan
logtan^J.E'
=10.28284
log tan
=
i~E
IE=
10.14142.
54° 10^
-£;=216° 40^ 3.
Given
4.6^^ 18^^.
find
a= 33° 1^45^^ ^=133° 48^ 53^^. 6 = 155° 5a8^^ = 110° 10^; m = tan a cos C. cose = cos a seem cos (6— m). (7
tan
log tan a log cos log tan
c
m
6-m =
log cos log cos
54)
(^54)
167° 22^ -(12°16^42^0-
= 9.92345 = 0.01064 m (6 - m) = 9.98995 = 9.92404 c c = 147° 5^ 30^^
log cos a log sec
(^
= 9.81300 = 9.53751 = 9.35051
m=
33°
1M5^^
= 155° c = 147° 2s =
5^ 18^^
h
53° 55^ 26^^
=
^s
a=
6'^
a
149
5^ 30^^
TRIGONOMETRY.
150
Exercise XXXII.
Page
148.
What angle 1. Find the dihedi-al angle made rod upon the plane. by the lateral faces of a regular ten- does this line make with the rod ? sided pyramid given the angle A = 18°, made at the vertex by two ;
adjacent lateral edges.
CO
Let
the angle
IF
Sb
be a straight line, making with the plane OH;
A
straight line passing through
B
the vertex of the pyra- the foot of CO, making the angle of CO upon It will with the projection describe a sphere.
About mid,
DO
intersect the lateral surface, forming the plane
a regular spherical decagon, of which the sides
=
18°,
being measured by
the plane angles at the centre.
and
C,
forming an
cal triangle
angle
A
isosceles spheri-
ABC.
Then (by Prob. given side
A
and
Pass a plane through
a
Ex. XXIII.)
3,
= 18°, n =
It
required to find the angle
is
COI==
GH.
X.
AVith
from
a radius equal to unity, a centre, construct the
as
spherical triangle
Then
c=B,
10, to find
d=COI=x
of the polygon. 1
•sm ^
A= A
log cos 18^
colog cos 9°
log sin
A
sec ^ A cos
MiA
I
180°
CEK=-Tt.
-•
Since
= 9.97821 = 0.00538 = 9.98359
OE
is
.'.
2. Through the foot which makes the angle
plane, a straight line
B
the projection of
CO
on the plane OH; CE drawn from C to E, is perpendicular to OE.
CDI =rt.
42^
By Formula cos
angle
angle.
triangle.
74° 21^
A = 148°
the plane.
DCI.
i = A,
This line
is
of
A
a rod with a
.•.
[37],
d=
cos X
cos
i
cos
= cos A
3, Find the volume drawn in makes the lique parallelopipedon
with the projection of the three unequal edges
c.
cos B.
V of ;
a, b,
an ob-
given the c,
and the
TEACHERS EDITION, three angles
I,
m,
make with one /\
n,
which the edges
another.
151
TRIGONOMETRY.
152
= 5.36570
logi:
= 4.68557
log'6-18000
logi?2
logi^
=
7.07312
=
7.12439
i^=
A
5.
latitude
13316560.
thip sails from a harbor in I,
and keeps on the arc of a
great circle.
Her
course (or angle
between the direction in which she sails and the meridian) at starting Find where she will cross the is a. equator, her course at the equator,
and the distance she has
Let
NESW be
the equator,
the earth,
iVand Sthe Let
south poles.
A
from which the ship the
parallel
starts from,
of
its
Then
BAE=a
WCE
nortli
and
be the point starts,
of latitude
and
course.
sailed.
AB the
the
AFD ship
great circle
TEACHERS EDITION.
K
= a X cos = iaV2 = 90°\/2. 90° V2 - 90° = 90°( V2 - 1). Arc
Given the latitudes and longi-
8.
Z
153
tudes of three places on the earth's
and also the radius of the show how to find the area the spherical triangle formed by
surface,
earth of
;
arcs of great circles passing through
The shortest distance d be- the tween two places and their latitudes Find the difI and V are known. 7.
places.
ference between their longitudes.
Let
A
C
Then
position of the other.
m
and
if
one
city,
Let A, 5, and
represent the north pole,
the position of one city,
B the AB = d,
represent the longitude of
mV that
of the other,
I
G
represent the
on
three places
the
earth's surface. I
the
of
positions
how
62 shows
to find the dis-
tance between two places
when
the
and V that of latitudes and difference in longitude the other, angle Cwill be equal to are given. {m-m^), and 5(7 will equal 90°- Z, In this case we have the latitudes and ^Cwill be equal to 90° -^^ given, and also the longitudes so Therefore we have an oblique that we can find the difference in latitude of one city,
;
spherical triangle with three sides longitude. given to find the angle C.
GH = equator
Let
Now from Formula [44], cos c = cos a cos h + sin a sin
h cos C;
tan
then by substituting, cos
d=
cos
(90°-
I)
cos (90°
or cos
sin
cos
I)
(90°— V) cos (m — m^)
;
d = sin I sin V
+ .'.
and from
- V)
+ sin (90° -
X
cos (w
cos
I
cos
then, from
V cos (m — m^).
;
| 54, in triangle
ABC,
m = cot a cos 6, § 62,
BC= sin a sec m sin (5+m),
and the same with the distance between the other places. Therefore,
we have
given
the
— m^) = (cos — sin
Z
sin V)
three sides of a spherical triangle,
X sec
I
sec V.
to find the area.
(i
TRIGONOMETRY.
154
By
[51],
tan^ \
E= tan I
s
tan i (s
— a)
X tan i{s — h) tan J (s — c). Then, since we have the radius of sphere given (= R) and the spherical excess = E, from Formula the
log CSC
s
log sin
(s- a) (s- 5)
log sin
log CSC (s-c) log
tannic tan J
= = = =
0.16409
9.00210 8.56391
0.23716
=17.96726
=
(7
8.98363
1(7=5° 30^ 2^^ (7-11° 0^4^^
F=^TrR\ 180=
-
660.
Difference of time
9. The distance between Paris = Y^5(660) minutes and Berlin (that is, the arc of a = 41 min. great circle between these places) is Time Berlin, at 12 h. 44 mm. eq'ial to 472 geographical miles.
The latitude of Pans is 48° 50^ 13'^; that of Berlin, 52° 30^ 16^^. When
10.
noon at Paris what time
ing 45°,
it is
at Berlin
is it
The I
altitude of the pole besee a star on the horizon
and observe
?
its
azimuth
to be 45°
;
find its polar distance.
AO
Let
of Paris, Berlin.
represent
and Then
BK C
the latitude
the latitude of
represents the dif-
ference in longitude.
iVP=45°.
6M = 6 = 41° 9M7^^ CB = a=2>T 29M4^^
AB==c= 7° 2s= 86° s = 43°
s-a= s-b= s
tan'^
52^
.:PZ = 4:5° = (472
60)
15^ 45.5^^
5° 46^
5^58.5>'>'.
- c = 35°
23^ 45.5^^
C= cscs
sin (s
We
— a) sin (s — b) esc (s — c).
FM^p. PZM^a.
have given two parts of the
triangle,
= 45°. = 45°. cosp = cos a cos a = \/|. cos I = V^. cosp = J. p = 60°. a
(90° -
1.5^^
2°
J
H-
31^31'/
1.
.'.
cos
I.
TEACHEES EDITION. Given the latitude I of the and the declination d of the sun find the local time (apparent solar time) of sunrise and sunset, and also the azimuth of the sun at these times (refraction being neglected). When and where does the sun rise on the longest day of the year (at which time d= -\- 23° 27^) in Boston {I = 42° 2V), and what is the length of the day from sunrise to sunset? Also, find when and where the sun rises in Boston on the shortest day of the year (when (^=- 23° 270, and the length of
Also to find azimuth
11.
observer, ;
this day.
To sun
find the
is
Then, by cos
t
when
a.
MQ measQZM= 180° - a.
PMQ,
In triangle ured by angle
is
[37],
PM= cos PQ cos MQ.
or
- d) = cos (180° - Z) cos (180° - a). sind = — cos (— cos a), cos a = sin sec
Now
cos
Time
of sunrise
cos (90°
I
c?
I,
= — tan d tan
t
I.
= 12
o'clk a.m.
15
Time hour angle
155
of sunset
o'clk P.M.
15
the
on the horizon.
= 9.63726 = 9.95977 = 9.59703 t = 66° 42^ 26^^
log tan
d
log tan
I
log cos
t
12
- — = 7 h.
33 min. 10
sec.
15
= 4 h. 26 min. 50 sec. 15 .*.
shortest
day
= 2 X 4 h. 26 min. 50 sec. = 8 h. 53 min. 40 sec. Again, cos
.-.
PM:= 90° - d. ^Q = 90°. PQ = 90° - -h = 180° Z
log sin d log sec
log cos a
a .-.
cos
or
PMQ, by
in triangle
[39],
QPM== tan PQ cot Pif, cos t = tan(180°-r)cot(90°-tf), cos t = ~ tan tan d. I
I
90°
1.
Then
a
= sin sec = 9.59983 = 0.13133 = 9.73116 = 57° 25^ c?
I.
15^^
a^=122° 34^45^^
Longest day = 12 hrs. -h[(7h. 33min. lOsec.
-4h.
=
26 min. 50
15 h. 6 min. 20
sec.
sec.)]
TRIGONOMETKY.
156 12.
AVlien
is
the solution of the
problem in Example 11 impossible, and for what places is the solution impossible
d has
?
for
maximum
its
value
23° 27^.
Suppose
=
I
66° Sy.
Then tan (180° -
Z)
= - tan
^
= — cot d. = Formula cos t — tan tan d becomes cos t^ — cot d tan d I
= that
is,
-1.
i=180°;
.-.
By
Napier's Rules,
the sun just appears in the sin
south on shortest day.
For places within the Arctic cirI > 66° 33^, and —tan I numerically greater than — cot d.
cle,
Hence
— tan
I
tan
c?
h =&hxl sin
= 9.87209 log sin d = 9.59983 log sin h = 9.47192 Altitude = A = 17° 14^ log sin
>—1
d.
I
35^^
(numerically), or cos
which sun
is
may
t
=
By
± 1+,
= cos 7 tan d. = 9.82424 log cos log tan d = 9.63726 log cot a = 9.46150 Azimuth = a = 73° 5V W\ cot a
not possible. Hence the during 24 hours.
fail to rise
I
13. Given the latitude of a place and the sun's declination find his altitude and azimuth at 6 o'clock ;
A.M. (neglecting refraction).
Napier's Rules,
Com-
14. How does the altitude of the pute the results for the longest day sun at 6 a.m. on a given day change of the year at Munich {I = 48° 9^. as we go from the equator to the a. pole ? At what time of the year is
FZM=
FZ=dO°-l PJf=90° -d=p.
ZPM= ZM= 90° - h. t.
I
= 48°
9^
Sun's declination on longest day, 23° 27^.
a maximum at a given (Given sin A = sin Z sin d.) it
The
farther
'place ?
the place from the
equator, the greater the sun's alti-
tude at 6 A.M. in summer.
equator it is
it is 0°.
At
At
the
the north pole
equal to the sun's declination.
At a given
place, the sun's altitude
TEACHERS
157
EDITION".
And from ^ 65, the times of bearat 6 A.M. is a maximum on the longest day of the year, and then ing due east and west are sin h = sinl sin e (where e = 23° 27''). 12 A.M. and p.m., 15 15 15. Given the latitude of a place
—
north of the equator, and the dec- respectively. lination of the sun find the time of Since the day given is the longest day when the sun bears due east day of the year, the declination of and due west. Compute the results the sun = 23° 27^ ;
for
the longest day at St. Peters{I = 59° 560I
d=
we have given
.*.
burg
= 59°
56^
to find
cos
t
=
log cot
I
= 9.76261
Now
cot
23°
27''
and
t.
tan d.
I
log tan cZ= 9.63726
log cos
t t
.*.
= 9.39987 = 75° 27^
=6
12
24^^
hrs. 58 min. A.M.,
15
— = 5 hrs.
and
2 min. p.m.
15
Let zenith,
NESWhQ the horizon, ^ the NZS the meridian, WZE
WAE
the prime vertical, noctial,
P the
elevated pole,
position of the sun
MB
its
the equi-
declination,
M the
when due and
east,
ZPM
its
16. Apply Example 15
the case
the general result in (cos
when
=
t
cot
I
tan d) to
the days and nights
are equal in length (that
is,
when
d = 0°). Why can the sun in summer never be due east before 6 a.m.,
hour angle. Then we have the right spherical or due west after 6 p.m. ? How triangle PZM, with and PZ does the time of bearing due east and due west change with the decliknown, to find ZPM. nation of the sun ? Apply the genGiven I and c?, to find U eral result to the cases where I
PM
PM=
PZ=dO°-l
From cos
^ 48,
Case
the north pole
II.,
5= tana cote.
Substitute in this equation,
or .'.
cos
t
cos
t
cos
t
= tan PZ cot PM, = tan (90°- cot (90°- d). = cot tan d. 1)
I
When
?
and nights are and ^ = 90°;
the days
equal, c?=0°, cosi=0°,
that
is,
the sun
and due west at d must both be
is
due east at 6 a.m.
G p.m. less
Since
I
and
than 90°, cos
t
cannot be negative therefore t cannot be greater than 90°. As d ;
TRIGONOMETRY.
158 increases,
decreases
t
;
that
both
times in question
the
is,
.'.
approach
sin
d^sinl
sin
I
sin h.
= sin dcsch.
noon.
lfl
this
cos
i
>1
therefore
;
If
impossible.
is
= d,
1
i
the
pole,
= 90°;
Z
=
90°,
cosi=
therefore the sun
in
mer always bears due east and due west at 6 p.m.
0°,
sum-
OB is due north. OB make with the
direction
angle will
shadow
OA
of
on the plane, at 1
P.M.
at 6 a.m.
Given the sun's declination when he bears due
17.
and
MI^
What
;
a point
in a horizontal a staff OA is fixed, so that its angle of inclination AOB with the plane is equal to the latitude of the place, 51° 30^ N., and
then cost=l, and t = 0°; that is, the times both coincide with noon. The explanation of this result is, that the sun at noon is in the the zenith hence, on the prime vertical, at
At
18.
plane
his altitude
east; find
the latitude
of the ob-
server.
N Given direction of OB due north, 30^= I, and plane
MN
AOB = 51° horizontal
;
Produce
BOC.
to find
OA
;
it
will pass
through
The sun being on the equinoctial, FOS = 90°, and the shadow 00 will lie in the plane of the pole.
this angle.
MJSF;
it
Draw OZ
JL
to
plane
will lie in the plane of
OB
and OA. In the figure of observer,
let
F the
Z be
the zenith
elevated pole,
M
the position of the sun.
ZM= 90° - h. FM= 90° - d.
Then
..CAB=\b°.
ABO= 90°, jection of
FZ=dO°-l. Since the sun
MZF /.
is
M bears
due
OA
since
by Napier's Kules,
FM= cos FZ cos MZ.
30^,
ure of plane angle
Then
is
the pro-
MN.
being the meas-
AOB.
in right spherical triangle
ABC, by tan
OB
on plane
Arc AB = 51° east,
a right angle.
cos
SFZ = hour angle of sun at 1 p.m. = 15°. SFZ = CAB, being vertical angles.
[41],
BC^ tan BAGs'm
AB.
159
TEACHERS EDITION. log tan 15°
=9.42805
log sin 51° 30^
= 9.89354 = 9.32159 = 11° 50^ 35^^
BQ BO
log tan
Arc
20. At a certain place the sun is observed to rise exactly in the northeast point on the longest day of the
year
When
Arc -SC measures plane angle BOC. .:
BOC= 11°
50^ 35^^
east
a
the sun rises in the northon the longest day of the year,
= 45°, To
What
the direction of a wall in latitude 52° 30^ N. which casts no shadow at 6 a.m. on the 19.
find the latitude of the place.
;
c?=23° 27^
find
is
a
cos
log sec
In the formula
I.
I
= sin d sec = log cos a + log esc d. I.
= 9.84949
log cos 45°
longest day of the year.
log CSC 23° 27^ log sec
I
= 0.40017 = 0.24966 I = 55° 45^ 6^^.
Find the latitude of the place which the sun sets at 10 o'clock
21. at
on the longest day. Since the sun sets at 10 o'clock, is equal
the hour angle of the sun to 15° X 10 = 150°.
The declination The wall must
lie in
ing through the sun, in order that
may
cast
In the
it
.'.
we have
no shadow.
parts given
figure,
and
PZ= 90° -I PM= 90° - d.
By sin
cot
V)
cos
Or,
I
cot re
= tan {90° -d) cot MZP. = cot d cot X. = cos tan Z
I
log tan
e
= 9.78445 = 9.63726
log cot
a;
=9.42171
X
=
75° 12^ 38^^.
the triangle two
in viz.,
I
the angle
ZPM,
= cos cot d. = 150°.
d= log
log
e.
log cos
;
t
22.
t
23° 27^
= 9.93753 cot £? = 0.36274 cot I = 0.30027 I = 63° 23^ 41^^
log cos
[41],
(90°-
equal
In the formula,
irZP=a;.
find
is
ifP= 90° -d
ZPD = 6 X 15°= 90°. To
of the sun
the line pass- to 23° 27^.
t
To what does the general
for the hour angle, in ^ 67, become when (i.) h = 0°, (ii.) I = 0° and c? = 0°, (iii.) Zor c? = 90°.
formula
TEIGONOMETRY.
160 In the general formula, /i
= 0.
§ 67, let
(11.)
Then sin
sin I
i
=±
[cos J(Z
X sin i
^-
sin J (Z
-4
+p) sec I cscp]^.
h—
^
2-
p)
cos
= v/sin(s— J)sin(s— c)csc h csc Z = 0.
c.
d=0.
t
t
FZ =90° -I =90° = Pif - dO°-d = 90° = 6. e.
cos
.^I±£2^(U£). j(;+rt =
sin I
sec
Z
ii+p)^^^j '-^°f^pl
sinp
I
Substitute these values in the
first
equation,
— cos ^ ^
/
+cos(? +
l
~^
p)
sin ^
i
2
^ Jl_-cos_(Z_+^) X cos
I
t.
i
— cos i = sin
I
= Vsin J (90°- h) sin J (90°-;^) = sin J (90° -/i). = sin (90° - h). sin t=90°-h = z. ^
sinp
— cos = Vl — cos^ COS
1
= a.
Substitute
2
1
90° -?i
s-b = ^ (90° - A). s-c = ^ (90° - ^). csc b = csc 90° = 1. csc c = csc 90° = 1.
=
cscp
cos
1
^ir= A=
(iii.) {I
+p)
Zor(?=90°.
PZ=90°-2 = 0° = a.
smp
(I
+ p)
.•.
no triangle will be made.
.•.
answer indeterminate.
Pir= 90° -cZ=0° = COS
= (sin
Isinp I
cosp
+ COS I sinp)
x^L_ smp COS
I
= tan coip + 1. = — tan cot j3. I
COS
and
Z
cot j3 cos
i
= tan d, = — tan
no triangle formed.
.*.
result indeterminate.
23.
What
mula
for the
6.
does the general for-
azimuth of a celestial 68, become when t = 90°
body, in § = 6 hours ?
When t = 90°, m = 0, and we have
_p-90°-d
But .*.
t
/.
a right spherical triangle with the two legs given to- find the angle I
tan
d
opposite one of the legs.
161
TEACHERS EDITION. Then by
substituting the values
of the given parts in the formula,
II.
If (2=0°.
From
B = tan h esc a,
tan
§ 69,
cos
I
=
we have tana =cot(Zsec^.
or
tancZ
= tan (90°-c?) esc (90°-Z)
tan a
cot a
= tan
and cot a 24.
Show
§ 69, if
t
d
sec
Multiply
cos
i
that the formulas of lead to the equation
= 90°,
and that Z = sin A esc cZ they lead to the equation sin
;
cos
I.
I
= sin h
sec
if
cZ
= 0°,
t.
If< = 90°.
From
g 69,
sin
^
MQ.
(1)
sin
d= cosm cos MQ.
(2)
Divide
but
= cos n
(1)
sin
^
sin
cZ
now
by
cos
(2),
n cosm cos
n = ZP=90°-
m = 0°.
and sin
h
sin
d
sin
Z
.'.
.'.
= sm
Z.
= sin
A esc
•
Z
cos
tanm
Z.
7
cZ.
m sin h esc
cos
t
=
tanm
cos
tan a
COS
t
(3)
by cos
(4),
m sin h
c?.
(3) (4)
TRIGONOMETRY.
162 log cot
fZ
log cos
t
log tan
m
= = =
0.10719
10.05196
m = 48° sin
h
c?
=
t
= 22° = 15°
9.94477
25^ 10^^
= sin (Z + m) sin d sec m, ^
(Z+m)= 9.99206 = 9.78934 sin d seem =0.17804
90°
log
log sin h
h
d t
log tan
m
=10.37792
= 9.95944 - 10 = 65° 37^ 20^.
log sin
d
= 9.94351 = 9.57387 = 0.41302 = 9.93040 h = 58° 25^ 15^^.
log sec
m
polar distance of star 67° 59^ 5^^ its hour angle 15° 8^ 20'''',
12'''';
find its altitude
16^ 22^^
log sin (l+m)
Given latitude of place 51°
26.
0^ 55^^. 8^ 12^^.
m = 67°
log sin h
19''
59^ 5^0-
= 10.39326 = 9.98466
log cot
log cos
log sin
log
- (67°
and
its azi-
muth.
log sec {l+m) log tan
t
log sin
m
log
= 0,32001 = 9.43218 = 9.96490 =9.71709
tana a
= 152°
28^.
Given the declination of a 54^ its altitude 22° 45^ 12^^ its azimuth 129° 45^ 37^^ find its hour angle and the latitude of the 27.
star 7°
;
observer. sin
t
= 8ma
PM= polar distance of star.
ZPM= hour angle of star. PZ = co-latitude of observer. Find PZM= azimuth of
and
DM
its
log sin
d t
star,
altitude.
t
d=90°- PM. Let PQ = m. tan m = cot d cos sin h = sin {I + m) sin d sec m. tan a = sec {I + m) tan sin m. = 51° 19^ 20^^. t.
t
sec d.
= 45° 42^
m = cot d cos cos n = cos m sin h esc d. tan
Let
I
log cos h colog cos
h
= 9.88577 = 9.96482 = 0.00414 = 9.85473
log sin a
Given
cos
t.
I
=90°-(m±7i).
= 10.85773 = 9.84411 m = 10.70184 m = 78° 45^ 45^^.
log cot
d
log cos
t
log tan
TEACHERS EDITION.
=
log cos TO
log cos
29. Given the obliquity
9.28976
= 9.58745 = 0.86187
log sin h
log CSC
d
a star 51°, its
W
39^^.
m-n = 12° V
Q'\
90° _ (m
- w) = 67° = 67° .-.
Z
= 23°
ecliptic e
= 9.73908 n = 56°
n
163
longitude 315°; find
its
declination and
its
right ascen-
sion.
In Fig.
given
47,
VT= 315°
58^ 54^^
or
- 45°,
TM= 51°,
58^ 54^^.
i2Fr= 23° 27^
d,
= 23°
27^
;
BM=
r.
In right triangle cos
and
VM= cos
log cos 51°
log cos
P represent the equinoctial A VB, S the let
position of the sun,
and
Q
the pole
EVF. VS = u.
of the ecliptic
Then
VB = r.
Then by [38],
in the right triangle
sin
or
Also by cos
or
d
= tan r cot u. tan r = tan u cos e.
= 10.09163 = 0.15051 (n) log CSC 315° log tan MVT = 10.24214 in) MVT- - (60° 12^ 14.5^0-
B VM,
= 23° 27^-(60° 12^ 14.5^0 = - (36° 45^ 14.5^0By
[38],
sin
log sin
[39],
cos e
VM VM= 63° 34^ 36^^.
log tan 51°
B VS,
sm u sm e.
BVS= tan B Fcot
FT.
BVM=BVT+ TVM
SB = sin VS X sin B VS,
sin
FT'cos TM,
MT esc
= 9.84949 = 9.79887 = 9.64836
In right triangle
8R = d. RVS=e.
VTM,
tan ili'FT^ tan
log cos 315°
pole of the
d.
find the declination
and the right ascension
In the figure
,
VB^r
Given the longitude u of the to find sun, and the obliquity of the eclip- and 28.
tic e
of the
27^, the latitude of
VS,
BM= sin FJf sin B VM. VM = 9.95208
log sin i? log sin
FIT =9.77698
BM
= 9.72906
BM=d=2>2°2¥l2^'.
«
TRIGONOMETRY.
164 Also,
by
sin .
=9.80257
logcoti^FJlf
= 0.12677 (n) = 9.92934 {n)
log
sm
VB
Fi^= -(58° Fi2
{A-B) =
9.93528
%m^{A+B) =
0.00735
log sin J
[41],
VB = tan BM cot E VM.
log tan i^lf
.-.
;
11^ 43^0-
= 360° -58° IIMS^^ = 301°48M7^^
colog
=9.61897
log tan J c log tan
J(a- 6) = ^(a-6)
.-.
a
90°
=
9.56160
= 20°
1'21.5".
69° 11' 48".
- 69°
11'
48"=
20° 48' 12".
31. Given latitude of place 51° 30. Given the latitude of a place 31' 48", altitude of sun west of the 44° 50^ 14^^ the azimuth of a star meridian 35° 14' 27", its declina138° 58^ 43^^ and its hour angle tion +21° 27' find the local ap20°; find its declination. parent time. ;
Byi67,
P^=90°-?, Pir= 90° -(Z = p,
ZM= 90° - A required
t
= ZBM.
p = 68° 33'. J(Z + A+^) = 77°39'37.5". J(Z-/i+p) = 42° 25' 10.5".
= 90° - 44° 50' 14^' = 45° 9' 46^'. = A 138° 58' 43''. B = 20°. 1{A-B)== 59° 29' 22". 'l{A-\-B) = 79° 29' 22". Given
J
c
=22°
c
log
cosJ(^+P + ^)= 9.32982
log sin
J(Z+_p-A)=
9.82901
= =
0.20614
colog cosZ
colog sinp
2 )19.39614
=
log sin I
J
^
34' 53". t
log
cosJ(^-5) = 9.70560
colog cos
jU+^) = 0.73893
=
= 29° = 59°
9.69807
55' 55.5". 51' 51".
3 h. 59 min. 27f sec. p.m.
15
=9.61897
log tan J c
= 0.06350 Ha + &) = 49° 10' 26".
log tan J (a
0.03117
+
5)
Given latitude of place ?, the p of a star, and its altitude h find its azimuth a. 32.
polar distance :
I
TEACHERS EDITION. cos
165
^A= Vsin s sin (s— a) esc b esc c.
A^ FZM or a,
Let
a=p, b c
= 90°- h, = 90°- I
Then
= sm[90°-} (l+h-p)] = cos J (h + l—p).
sin s
sin (s-a)
Altitude Co- latitude
Polar distance
Azimuth
= ZM= 90°- h. = FZ = 90°= FM = 90° -d=p. = FZM oi a.
CSC b
1.
= esc
esc c .'.
cos I a
= sin [90°- H^+^+P)]
= cos J(/i + ^+p). = CSC (90°— A) = sec h. (90°—
l)
=
sec
I.
=
Vcos|(_p+A+^)cos|(/i+^-p)sec^secA
SUEYETI.N-G.
Exercise 1.
Page
I.
Required the area of a triangular
and 15
13, 14,
143.
whose
field
sides are respectively
chains.
= Vs{s — a) (s — 6) (s — c). s = i(13 + 14 + 15) = 21, s-a = 21 -13 = 8, Area
Area
2.
= V21 x 8 x 7 X 6 = VS^ x 7^ X = 84 sq. ch. = 8.4 A. = 8 A. 64 p.
Required the area of a triangular
20, 30,
s- 5 = 21 -14 = 7, s-c = 21 -15 = 6.
and 40
field
2*
=3X7X
whose
2^
sides are respectively
chains.
= V45 x 25 x 15 x 5 = V3=* X 5^ = 3 x 5 V3 x 5 = 75 VTS = 290.4737+. 290.4737 sq. ch. = 29.04737 a. = 29 a. 7.579 p. = 29 a. 7f p., nearly. Area
3.
and
Required the area of a triangular
Area = J base X altitude. Area = J X 12.6 X 6.4 = 40.32 4.
whose base
is
12.60 chains,
sq. ch.
=
4.032 A.
= 4 A. b^^j p.
Required the area of a triangular
and 3.70
chains, respectively,
Area Area
5.
field
altitude 6.40 chains.
field which has two and the included angle 60°.
= ^bc sin A. = ^ x 4.5 x 3.7 X 0.866 = = 115/oP., nearly.
Required the area of a
field in
7.20945 sq. ch.
= 0.7209 a.
the form of a trapezium, one of
and the two perpendiculars upon onal from the opposite vertices 4.50 and 3.25 chains. whose diagonals
Area .
=
is
J
9 chains,
X
= 3 A.
9 (4.5 78
p.
+
3.25)
sides 4.50
= 34.875 sq. ch. = 3.4875 a.
this diag-
SURVEYING.
168 6.
Required the area of the
chains,
FF^=6A0
chains,
CC^= 4
chains,
field
ABCDEF (¥ig.
BE= 13.75
chains,
and AA^=4:.15
chains.
chains,
2area^i^.£'
=
6.4x9.25
BDFA = 13.75 (4.75 + 2zxQ2iBDC =10x4 2 area ABCDBF area ABCDFF 2 area
130.38125
7.
FF'=
=
sq. ch.
Required the area of the
=
13.038125
field
DD^=
= 7) =
19), if
^^=9.25
7 chains,
DB= 10
59.2
161.5625
=40
a.
= 260.7625 = 130.38125 = 13 a. 6j\ p.
ABCDEF (Fig. 20), if AF^--^ 4 chains,
EE'= 6.50 chains, AE^= 9 chains, AD = 14 chains, AQ' AB^= 6.50 chains, BB^= 7 chains, CC^=Q.7o chains.
6 chains,
10 chains,
2 area
Tli^i^''
2&YesLEE'B 2 area
ABB^
=24
=4x6
2areai^''^^^i^=5(6 +
= = =
6.5)
=6.5x5 = 6.5 x 7
62.5
32.5 45.5
BCC'B^ = 3.5 (7 + 6.75) = 48.125 = 27 2areaaDC^ =6.75x4
2 area
2 area area
ABCDEF ABCDEF
119.8125
8.
sq. ch.
= 11.98125 a.
Required the area of the
AC= 5,
BB^
dicular from
AGBCD (Fig. 15), if the diagonal from B to AC) = 1, DD^ (the perpen-
field
(the perpendicular
D to AC)^ 1.60, EE'= 0.25,
= 0.52, KK'=OM, AE^ = 0.2, S^K'^'-Om, and ^^^ = 0.40.
= 239.625 = 119.8125 = 11 a. 157 p.
FF'= 0.25, GG'=
=13. =5(1+1.6) = 0.05 = 0.25 x 0.2 2 3.vesiEE'F'F =0.5(0.25 + 0.25)= 0.25 2&reiiFF'G'G =0.45(0.25 + 0.6)= 0.3825 2sirea,ADCB 2 area
AEE'
=
0.504
2 area ^.ff^ir'ir= 0.6 (0.52 + 0.54)=
0.636
=
0.216
= =
15.0385
2 area (?G^i7^^ 2 9^vesi,KK'B
2 area Sivesi
0.60,
HM'
E^F^^O.bO, F^G^=0A5, G^H^=0A5,
=
0.45(0.6 +0.52)
=0.4x0.54
ADCBKHGFE ADCBKHGFE
7.51925.
TEACHERS EDITION. Required the area of the
9.
field
169
AOBCD (Fig.
16),
iiAD = Z,AC
= 5, AB=Q, angle i)J.a= 45°, angle ^4C=30°, ^^^=0.75, AF^=2.2b, AE= 2.53, AG'= 3.15, EE'= 0.60, FF'=^ 0.40, and 0Q'= 0.75. 2 area
^1)05 =
2q.xq2.EGB
3
X
5
X
0.7071
+
5
x
6
x
= 25.6065 = 2.6025
0.5
=0.75x3.47
2s,Yea,AI)CBGE
=28.2090
AEFH= 0.75 x 0.6 + 1.5(0.6 + 0.4) + 0.4 x 0.28 = 2.062 = 26.147 2 area ADCBGHFE = 13.0735. area ADCBGHFE 2 area
Determine the area of the
10.
F and P^
if
PP'=
PP'G = PP^P =
angle
Area
ABCD from
89° 35^
angle
185° 30^ 309° 15^
PP^A = PP^i) = 349°
•
field
two
interior stations
1.50 chains,
P^PB^ = P^PD = P'PC = P''P^
45^
3°
35^
113° 45^, 165° 40^ 303° 15^
= A PAD + A PCD + A PPC+ A PAB.
ZPP'D= ZPDP'=
10° 15^ 4° 5^
ZPP^P =
174° 30^
PD = PP' sin
Z PPM = 50° 45^ ZP^P^ = 15° 30^
ZPBF=
PP^P>
PDF' = 0.17609 log PP^ log sin PP^D = 9.25028 colog sin PDP^ = 1.14748
ZPP'C-
ZPCP^-- 33°
PD
= 0.57385
PC = PP'smPP'C sin
PCP^
=
0.17609
log sin
PP^C =
9.99999
colog sin
PCP^ =
0.25621
log
log
PP^
PC
= 0.43229
40^.
1°55^
PA = PP^ sin
sin
log
89° 35^,
sin
log
PP^
log sin
PPM
PAF^
= 0.17609
PPM =9.88896
PAP^= 0.57310 = 0.63815 PA
colog sin log
PB
PP' sm PP'B sm PBP^
= log sin PP^B = colog sin PBP^ = log
log
PP^
PB
0.17609
8.98157 1.47566
= 0.63332
Z ^Pi) = 51° 55^ ZZ>PC= 137° 35^ ZPPC=60°20^ Z^PP = 110°10^
SURVEYING.
170
FAD = PDxPA sin APD. = 0.57385 log PD
log
log
APD = 9.89604
log sin
=
log 2 area
P^i) =
2 area 2 area
log Pi)
= 0.63815
PA
log
log sin
= 0.57385 = 0.43229 DPC = 9.82899
1.10804
log 2 area
12.825.
2
2 area
Sires.
log
APB = 9.97252
log
log 2 area
=
2areaP^P
=17.538.
log
1.24399
= 0.83513
PCD
=6.8412.
PBC= PCx PB sin PPC.
= 0.63815 = 0.63332
P^ PB
PC
log sin
P^P = PAxPB sin ^PP.
log
= PD x P(7sin PPC.
2 area PCZ)
2 area
= 0.43229 = 0.63332 sin PBC = 9.93898
PC PB
=
log 2 area
1.00459
2areaPP(7 =10.106.
A P^P> = 12.825 A PCD = 6.841 2 A PBC = 10.106
2 2
2APAB
=17.538
2ABCD =47.310 ABCD = 23.655 sq. 23.655 11.
sq. ch.
=
2.3655 a.
Determine the area of the
P and P\
PP^= 1.50 angle P'PB = P'PA = P^PC = if
P^PD =
field
= 2 A.
ch.
58^- p., nearly.
ABCD from
two exterior
stations
chains.
41° 10^ 55° 45^
angle
77° 20^
104° 45^
PP'D = PP'C = PP^B = PP^A =
66° 45^ 95° 40^ 132° 15^
103°
0^
= (A P^CP + A P'CD) - (A P^^P + A P'AD). ZP^PC= ZP^PP = 104° 45^ ZP''PP = 41° 10^ ZPCP'= Z PDP' = 8° 30^ ZPBP'= 6°35^ ZPAP' = 21° 15^ Z P^P^ = 55° 45^ Area
P'B =
PP'
log
= 0.17609 P^PP= 9.81839 PBP^ = 0.94063
PP^
log sin
colog sin log
P^PP PBP^
sin
sin
P'B
0.93511
P'D =
PP^
PP^
P^PD PPP^
= 0.17609
P^PD = sin PPP^ =
9.98545
=
0.99184
log sin
colog log
7°
sin
sin
log
77° 20^
P'D
0.83030
0'.
i
TEACHERS EDITION. P'G log
FF' sin P'PQ sm PCP^
colog sin log
2 area
= 0.17609 P^PC= 9.98930 PCP^ = 0.91411
PP'
log sin
P^C
1.07950
P^CP
log
P^P
PM
2
P'PA
log
P'A
= 0.53415
P^AB
A P^CP A P^CP
= P^Cx P^Psin CP^P. log P'C = 1.07950 log P'B = 0.99184 log sin CP^n = 9.68443
1.78985
log 2 area
2 area
=0.93511
15^
2 area P'^CP
61.639.
= 0.53415 sin ^P^P= 9.68897
log 2 area
2
log
colog P^P''
= 0.17609 = 9.91729 = 0.44077
Z ^P^i) = 36°
log
2 area
PP^
Z^P''P = 29°15^
PMP =P^P x P'A sin ^P^P. log
log
ZPP^C=36°35^
P^CP = P^Cx P^B sin PP'^O. = 1.07950 log P^C = 0.93511 log P'B log sin PP^ (7 =9.77524 2 area
p^^ _ PP' sin P'PA sm P^P-'
ZCP''i) = 28° 55^
log 2 area
2 area
=
171
=
1.75577
P^CP = 56.986.
172
SURVEYING.
Exercise
II.
1.
Page
152.
TEACHERS EDITION.
3.
irs
174
SURVEYING.
5.
TEACHERS EDITION. 6.
175
176
SUEVEYING.
7.
TEACHEES EDITION.
9.
1 2
3 4 5 1
177
178
SUHVEYING.
Exercise HI. 1.
Page
163.
179
teachers' edition.
Page
Exercise IV. From
1.
by a
line
ABCB,
the square
EF parallel
to
E.
3 a.
= 64 =30
AB From
1 E. 32 p.
line
2
24
p.,
part off 3 A.
R.
32
A. 3 R.
25
;
Voi =
8 ch.
= AB.
sq. ch.
n. 24 P., part off 2 A. Then, from the remain-
containing 8 a. 1
AI)=7
to
off 2 a. 3 r.
8 A. 1 R. 24
2 A. 1
sq. ch.
ABCD,
^i^ parallel
der of the rectangle part
r.
8
the rectangle
by a
containing 6 a. 1
AB.
6 A. 1 R. 24
2.
161.
25
p.
ch.
by a
line
GE parallel to IJB.
= 84 sq. ch. = ABCB. p. = 24.5 sq. ch. = AEFD. p. = 29.0625 sq. ch. = EBHQ. p.
AE = ^^^='-^^3.5ch. AD 7 A^CD^84^^2ch.
AD EB
7
=AB-AE =12- 3.5 = 8.5 ch. ^ EBHG ^2jm25 ^ 3 ^^
EB
Part
3.
by a
line
off
6 A. 3 e. 12
EF parallel 6
A.
to
3 R. 12
p. ;
=
containing 15 A.
10 ch.
= ABFE. ch. = ABCD.
68.25 sq. ch.
= 150 sq.
15 A.
nearly.
ABCD,
from a rectangle
AB AD being p.
^h..
8.5
= '-^ = 15ch. AB=^^^ 10 AD
^^ = ^^^=68^ = 4.55ch. AB
From a square ABCD, whose
4.
which side
15
shall contain 2 a. 1 r. 36
AD. 2
A. 1 R. 36 p.
AB
p.,
side
by a
= 24.75 sq. 9
is
line
ch.
9 ch., part off a triangle
BE
drawn from
B to
the
;;
SURVEYING.
180
5. From ABCD, representing a rectangle, whose length is 12.65 ch., and breadth 7.58 ch., part off a trapezoid which shall contain 7 A. 3 e. 24 p., by a line BE drawn from B to the side DO.
7 A. 3 E. 24
=
p.
79 sq. ch.
ABCD = 12.65 X 7.58 = 95.887 sq. A BCE= 95.887 - 79 = 16.887 sq. .r^55. = 2BCE = 2x16.887 = 4.456 CE 7.58 BC .
6.
part
2 E. 16
p.,
ch. .
ch.,
,
nearly. ^
AB = 12 ch., AC= 10 ch.,
In the triangle ABC, off 1 A.
ch.
by the
line DjE" parallel to
and
5C= 8
ch.
AB.
1 A. 2 E. 16 p. == 16 sq. ch.
CAB = Vl5x 3x5x7 = 39.6863 sq. ch. CDE = CAB - ABED = 39.6863 - 16 = 23.6863 CAB CDE :: CA^ CD'
sq. ch.
:
:
:
39.6863
:
:
CB^
23.6863
:
:
:
:
CE\ 10^ CD\ 82; CE\ :
:
.'.
CD =
.-.
C^-6.18
7.725 ch. ch.
AD = CA- CD =10- 7.725 = 2.275 ch. BE = CB - CE = 8 - 6.18 = 1.82 ch. 7.
In the triangle ABC,
part off 6 A. 1
E.
6 A. 1
24
E.
p.,
24
by
= 64
p.
AB=2Q ch., AC = 20
the line
ch.,
and
:
:
16 ch.
sq. ch.
CAB = V31x5x 11x15 = 159.9218 sq. ch. CDE^ CAB - ABED = 159.9218 - 64 = 95.9218 CAB CDE CA" CD" :
BC=
DE parallel to AB.
:
sq. ch.
teachers' edition. Since the
tr'
angles have the same altitude, they are to each other as
Hence
their bases.
181
it is
only necessary to divide the base 10 into the
three parts, 2 ch., 3 ch., 5 ch.
9.
Divide the five-sided
ABCHE
field
among
three persons, X, Y,
and Z, in proportion to their claims, X paying ^500, Y paying! 750, and Z paying % 1000, so that each may have the use of an interior pond, at P, the quality of the land being equal throughout. Given AB = 8.64 = 6.S2 ch., and ^^ = 9.90 ch. ch., ^(7=8.27 ch., CZr= 8.06 ch., The perpendicular FD upon ^^-5.60 ch., FB^ upon ^C=6.08 ch., FB^^ upon CH^ 4.80 ch., FB''^ upon HE^bA^ ch., and FB^^^^ upon as the divisional fence between X's and Z's EA = 5.40 ch. Assume shares it is required to determine the position of the fences and and and shares Y's Z's shares, PiV between X's and Y's respectively.
HU
FH
FM
;
If
Pbe joined
to the vertices, the field
is
divided into triangles, whose
bases are the sides, and the altitudes the given perpendiculars sides
AFB = 8.64 X 2.80 =
24.1920
BFC = 8.27 X
25.1408
= CFH = 8.06 X 2.40 = JTFE = 6.82 X 2.72 = EFA = 9.90 X 2.70 = 3.04
ABCHE 2, 3, 4.
9
FH'is assumed
FHE
is less
19.3440
18.5504
26.7300
must be divided as the numbers 500,
sq. ch.
+
113.9572
as the line
3
+
=
4
9.
:
:
2
:
25.3238
sq. ch.
:
:
3
:
37.9857
sq. ch.
:
:
4
:
50.6476
sq. ch.
between X's and
than X's share by 25.3238
= X's = Y's = Z's
share. share. share.
Z's shares.
- 18.5504 =
Since the
6.7734
then 6.7734
,.
tri-
sq. ch.,
must be taken from the triangle FEA. The area of sq. ch., and the altitude Pi)^^/^ = 5.40.
this difference is
:
2
sq. ch.
=113.9572
The whole area 113.9572 750, 1000, or as
angle
upon the
from P.
FEM
^J/=2P^^=2xa773_4^ 2.5087 ch. FB''^^
5.40
PMA = FEA - FEM= 26.7300 - 6.7734 = 19.9566. sq. ch. Since Y's share
+ FAB (44.1486),
is
greater than
the point iV^is on
FMA (19.9566) AB.
and
less
than
FMA
SURVEYING.
182
PMA equals PAN;
Y's share diminished by
that
FAN= 37.9857 - 19.9566 = 18.0291
sq. ch.
^^^2P4i^^2xiM291 = 6.439 FD 5.60 10.
Divide the triangular
ABO, whose
field
ch.
sides
AB, AC, and
and 10 ch., respectively, into three equal and Di^ parallel to BC.
are 15, 12,
EG
is,
ABC =
X
\/18.5
ADF = ^ AEG = I
of 59.81169
:
AB''
:
AE''
:
AC''
:
Zg'.
:
ABC ADF:
:
:
:
59.81169
11.
:
8.5 ==
:
:
15^
:
:
122;
:
5U81169
by
sq. ch.
sq. ch.
sq. ch.
AE= 12.247 ch.
AE^.
:.
A^.
.',AG=
9.798 ch.
:
:
:
:
15^
:
:
122
A&. Jp\
:
.
.-.
AD = 8.659 ch.
,-,AF
= 6.928
ch.
ABC, whose sides AB, BC, and AC among three persons, A, B, and C, AB,. so that A may have 3 a., B 4 a., and
Divide the triangular
field
are 22, 17, and 15 ch., respectively,
by
C
fences parallel to the base
the remainder.
CAB CDG CEF CAB
= V27 X 5 X 10 X 12 = 127.2792 sq. ch. = CAB - ABGD = 127.2792 - 30 = 97.2792 sq. = CAB - ABFE = 127.2792 - 70 = 57.2792 sq. CDG M^ CG^ :
:
:
127.2792
:
:
:
ch.
ch.
:
OT
97.2792
:
:
:
C&.
:
17^
:
:
152
.
C^.
:.
^2
cj)\
...
CF^.
.-.
CG = 14.862 CD = 13.113
ch. ch.
CAB:CEF::CB':CF^ :
127.2792
:
:
BC
fences
Tb" AD" ZC' AF\
19.9372
:
X
::
39.8744
:
6.5
= 19.9372 = 39.8744
of 59.81169
ABC AEG 59.81169
X
3.5
parts,
GZ'
57.2792
:
:
:
GF.
:
I72
:
152
:
:
CF= 11.404
Oil ..CE=
ch.
10.062 ch.
^-
183
TEACHERS EDITION.
Exercise V. 1.
notes
Find the :
difference of level of
back-sights, 5.2, 6.8,
and
4.0
;
two places from the following fore-sights, 8.1, 9.5, and 7.9.
8.1
+
5.2
4-6.8+4
9.5
+
7.9
=
field
25.5
=16 9.5
2.
Write the proper numbers
following table of field notes, and Station.
in the third
make
and
fifth
columns of the
a profile of the section.
SURVEYING.
184
Station.
+ S.
B
6.000
H.I.
-S.
H.S.
H.G.
Remarks.
Bench on rock
25 10.2
20,8
0.0
30
feet
stake 1
5.3
20.4
5.3
2
4.6
20.0
6.4
3
4.0
19.6
7.4
4
6.8
19.2
5.0
7.090
18.8
5.1
6
3.9
18.4
6.2
7
2.0
18.0
8.5
8
4.9
17.6
6.0
9
4.3
17.2
7.0
5
10 10.25
572
4.5
16.8
7.2
11.8
16.7
0.0
10
west of
1.
10.25
Date Due
BOSTON COLLEGE
3 9031 01550242
