Question

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

Answer 1

Answer:

Angle C is 128.49°

Step-by-step explanation:

Recall the Law of Sines

$\frac{sinA}{a}=\frac{sinB}{b}=\frac{sinC}{c}$

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

$\frac{sin30^\circ}{15}=\frac{sinB}{11}$

$11sin30^\circ=15sinB$

$11(0.5)=15sinB$

$5.5=15sinB$

$\frac{5.5}{15}=sinB$

$B=sin^{-1}(\frac{5.5}{15})$

$B\approx21.51^\circ$

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°