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.

Step-by-step 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} (\sin B)/(AC)=(\sin C)/(AB) \\ (\sin32)/(9)=(\sin107)/(AB) \\ AB=(9*\sin107)/(\sin32) \\ AB=16.2in \end{gathered}`

Hence, the value of AB is 16.2in.