Question

In a triangle where side a = 15 inches, side b = 11 inches, and angle A = 30 degrees, what is angle C?

Answer 1

Angle C is 128.49°

Explanation:

Recall the Law of Sines

`(sinA)/(a)=(sinB)/(b)=(sinC)/(c)`

Set up a proportion with the given information to find angle B

`(sin30^\circ)/(15)=(sinB)/(11)`

`11sin30^\circ=15sinB`

`11(0.5)=15sinB`

`5.5=15sinB`

`(5.5)/(15)=sinB`

`B=sin^(-1)((5.5)/(15))`

`B\approx21.51^\circ`

Use angles A and B to find angle C by the Triangle Angle Sum Theorem

`180^\circ=A^\circ+B^\circ+C^\circ`

`180^\circ=30^\circ+21.51^\circ+C^\circ`

`180^\circ=51.51^\circ+C^\circ`

`128.49^\circ=C^\circ`

Therefore, angle C is 128.49°