find c in an oblique triangle wih a= 10 b= 6 and angle A= 30 with law of sines

Question

You need to use the law of sines and at first find all the angles in the triangle with the law of sines, and later find side c

Answer 1

Given sides a = 10, b = 6 angle A = 30

Use law of sines

a/SinA = b/SinB = c/SinC -----> (1)

Now ignore c/SinC in (1)

a/SinA = b/SinB

10/Sin30 = 6/SinB

10/0.5 = 6/SinB

20 = 6/SinB

Dvide to each side by 6.

20/6 = 6/6SinB

3.33 = 1/SinB

Cross multiplication.

SinB =1/3.333

SinB = 0.30003

B = Sin^-1(0.3003)

B = 17.45

Now we know two anglesof triangle ,let's find out C

C = 180-(A+B)

C = 180-(30+17.45)

C = 180-47.45

C = 132.55

Now ignore a/SinA in (1).

b/SinB = c/SinC

6/0.3003 = c/Sin(132.55)

19.98 = c/0.7366

c = 19.98(0.7370)

c = 14.71