We know that the law of sines states that:
[tex]\frac{\sin\alpha}{a}=\frac{\sin\beta}{b}[/tex]
For simplicity, let:
[tex]\beta=m\angle A_1BC[/tex]
In triangle A1BC this leads to:
[tex]\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}[/tex]
Therefore:
[tex]m\angle A_1BC=38.174[/tex]
Now, triangle A2BC is isosceles which means that both the base angles are equal, since angle CA1B and A2A1B are supplementary we have:
[tex]m\angle CA_2B=180-38.174=141.826[/tex]
