I3.2 Let $A B C$ be a triangle. If $\sin A=\frac{36}{\alpha}, \sin B=\frac{12}{13}$ and $\sin C=\frac{\beta}{y}$, find the value of $\beta$, where $\beta$ and $y$ are in the lowest terms.
$\begin{array}{l}\sin A=\frac{36}{45}=\frac{4}{5}>0 \Rightarrow 0^{\circ}0 \Rightarrow 0^{\circ}<B<90^{\circ} \text { and } \cos B=\sqrt{1-\left(\frac{12}{13}\right)^{2}}=\frac{5}{13} \\ \sin C=\sin \left(180^{\circ}-(A+B)\right)=\sin (A+B)=\sin A \cos B+\cos A \sin B \\ \quad=\frac{4}{5} \cdot \frac{5}{13}+\frac{3}{5} \cdot \frac{12}{13}=\frac{56}{65} \\ \beta=56\end{array}$