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VikaD [51]
1 year ago
9

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 i

s attached. This is parts A and B of the same question.

Mathematics
1 answer:
Anna [14]1 year ago
8 0

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

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