Answer to Question #146188 in Trigonometry for Raffy

Question

Using Napier’s rules solve the triangle ABC if:
1. a = 45° 20′
c = 63° 10′ C = 90°
2. A = 48° 26′ B = 100° 40′ C = 90°
3. a = 103° 15′ b = 58° 23′ C = 90°

Answer 1

sin a = sin A sin c = tan b cot B

sin b = sin B sin c = tan a cot A

cos A = cos a sin B = tan b cot c

cos B = cos b sin A = tan a cot c

Note 60' = 1° . To covert from ' to ° , we divide by 60

1. a= 45.33° c = 63.17° C = 90°

sin a = sin A sin c

sin A = sin a/sin c = sin 45.33/ sin 63.17 = 0.7970

A= 52.84°

sin b = tan a cot A

sin b = tan 45.33 cot 52.84 = 0.7667

b= 50.06°

CosB = cos b sinA

Cos B = cos 50.06 × sin 52 84 = 0.5116

B= 59.23°

2. A= 48.43° B = 100.67° C=90°

Cos A= cos a sin B

cos a = cos A/ sinB = cos 48.43/ sin 100.67 = 0.6752

a= 47.53°

CosB = cos b sin A

cos b = cos B / sin A = cos 100.67 / sin 48.43 = -0.2475

b= 104.3°

sin a = sin A sin c

sin c = sin a / sin A = sin 47.53/ sin48.43 = 0.9859

c= 80.37°

3. a = 103.25° b = 58.38° C = 90

Sin a = tan b cot B

CotB = sin a/ tan b = sin 103.25 / tan 58.38 = 0.5993

B = 59.07°

Sin b = sin B sinc

sin c = sin b / sin B = sin 58.38 / sin 59.07 = 0.9927

c= 83.07°

Cos B = cos b sin A

sin A = cos B / cos b = cos 59.07 / cos 58.38 = 0.9804

A = 78.64°