4.1. Point $O$ is the intersection of the altitudes of an acute-angled triangle $ABC$. Find $OC$, if $AB=4$ and $\sin \angle C=\frac{5}{13}$.
Answer: 9.6. Solution. From the similarity of triangles $A B B_{1}$ and $O C B_{1}$ (here $B B_{1}$ is the height), it follows that $\frac{A B}{O C}=\frac{B B_{1}}{C B_{1}}=\operatorname{tg} C$. Given $\sin C=\frac{5}{13}$, therefore $\cos C=\frac{12}{13}, \operatorname{tg} C=\frac{5}{12} \cdot$ Thus, $O C=\frac{A B}{\operatorname{tg} C}=\frac{4 \cdot 12}{5}=9.6$.