Question

I need help with my pre-calculus homework, please show me how to solve them step by step if possible. The image of the problem is attached. This is parts A and B of the same question.

Answer 1

We know that the law of sines states that:

$\frac{\sin\alpha}{a}=\frac{\sin\beta}{b}$

For simplicity, let:

$\beta=m\angle A_1BC$

In triangle A1BC this leads to:

$\begin{gathered} \frac{\sin31}{5}=\frac{\sin\beta}{6} \\ \sin\beta=\frac{6}{5}\sin31 \\ \beta=\sin^{-1}(\frac{6}{5}\sin31) \\ \beta=38.174 \end{gathered}$

Therefore:

$m\angle A_1BC=38.174$

Now, triangle A2BC is isosceles which means that both the base angles are equal, since angle CA1B and A2A1B are supplementary we have:

$m\angle CA_2B=180-38.174=141.826$