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:

`(\sin\alpha)/(a)=(\sin\beta)/(b)`

For simplicity, let:

`\beta=m\angle A_1BC`

In triangle A1BC this leads to:

`\begin{gathered} (\sin31)/(5)=(\sin\beta)/(6) \\ \sin\beta=(6)/(5)\sin31 \\ \beta=\sin^(-1)((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`