Question

In triangle ABC, m angle A = 41 degrees, m angle B = 32 degrees, and AC = 9 in. What is AB to the nearest tenth of an inch?thank you :)

Answer 1

Given

In a ΔABC, ∠A=41°, ∠B=32°, and AC=9in.

To find the value of AB.

Explanation:

It is given that,

Therefore,

$\begin{gathered} \angle C=180-(41+32) \\ \angle C=180-73 \\ \angle C=107\degree \end{gathered}$

Then,

$\begin{gathered} \frac{\sin B}{AC}=\frac{\sin C}{AB} \\ \frac{\sin32}{9}=\frac{\sin107}{AB} \\ AB=\frac{9\times\sin107}{\sin32} \\ AB=16.2in \end{gathered}$

Hence, the value of AB is 16.2in.