Solving quadrantal spherical triangle :

Solving quadrantal spherical triangle : In case giving a spherical triangle in which one side is 90° , we use the following fundamental rules ,

sin m iddle = tan( adj .) × tan ( adj .) sin m iddle = cos(opp .) × cos (opp .)

Example : 1. Solve the following right angle spherical triangle ABC given that : c = 90° , a = 115° 25' , B = 60° 18'

c = 900

Solution :

A

In the first sin B = tan A . tan (90 − a ) ∴ sin B = tan A . cot a sin B ∴ tan A = = sin B .tan a = −1.827961 cot a ∴ A = 180o − tan −1 1.827961 = 118D 40′′ 52′′

In the second

sin (90 − b) = cos B . cos (90 − a ) ∴ cos b = cos B . sin a = 0.4475034 D ∴ b = 63 24′ 59′′

In the third sin (90 − a ) = tan B . tan [− (90 − C )] ∴ cos a = − tan B . tan (90 − C ) ∴ cos a = − tan B . cot C ∴

tan C = −

tan B = 4.0847979 cos a

D ∴ C = 76 14′ 38′′ 11

90o - b

B 90o - a

-(900 – C)

Sheet (4) Solving quadrantal spherical triangle : Solve the following quadrantal spherical triangle ABC given that : sin m iddle = tan( adj .) × tan ( adj .) sin m iddle = cos(opp .) × cos (opp .) Lecture 1. 2. 3. 4.

a = 90° , A = 45° , c = 90° , c = 90° ,

B = 80° 10' , c = 72° , a = 116° 44' 48" , a = 115° 25' ,

C = 50° 20' b = 90° b = 44° 26' 21" B = 60° 18'

, , , ,

c = 49° 23' , B = 100° , A = 121° 20' , c = 60° 35' ,

b = 76° 41' c = 50° 10' B = 42° 01' B = 122° 18'

, , , , ,

a = 60° 10' , B = 80° 10' , c = 108° , C = 60° 20' , a = 110° 11' ,

b = 80° 20' C = 48° 50' B = 45° A = 115° 40' B = 14° 20'

Section 1. 2. 3. 4.

a = 90° b = 90° c = 90° a = 90°

Home work 1. 2. 3. 4. 5.

c = 90° a = 90° a = 90° a = 90° c = 90°

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