5. In a triangle $A B C, \sin A=\frac{3}{5}$ and $\cos B=\frac{5}{13}$. Find the value of $\cos C$.
(A) $\frac{56}{65}$ or $\frac{16}{65}$
(B) $\frac{56}{65}$
(C) $\frac{16}{65}$
(D) $-\frac{56}{65}$
(E) $\frac{56}{65}$ or $-\frac{16}{65}$
5. Answer: (C) $\sin A=\frac{3}{5}$ implies that $\cos A=\frac{4}{5}$ or $-\frac{4}{5}$, and $\cos B=\frac{5}{13}$ implies that $\sin B=\frac{12}{13}$, since $0<B<180^{\circ}$.
If $\cos A=-\frac{4}{5}$, then $\sin (A+B)=\sin A \cos B+\cos A \sin B=\frac{3}{5} \cdot \frac{5}{13}-\frac{4}{5} \cdot \frac{12}{13}<0$, which is not possible since $0<A+B<180^{\circ}$ in a triangle. Thus we must have $\cos A=\frac{4}{5}$. Consequently, since $C=180^{\circ}-(A+B)$, we have
$$
\begin{aligned}
\cos C & =-\cos (A+B)=-\cos A \cos B+\sin A \sin B \\
& =-\frac{4}{5} \cdot \frac{5}{13}+\frac{3}{5} \cdot \frac{12}{13}=\frac{16}{65} .
\end{aligned}
$$