3. The altitudes of an acute-angled triangle $A B C$ intersect at point $O$. Find the angle $A C B$, if $A B: O C=5$.
# Problem 3.

Answer: $\operatorname{arctg} 2$ (in variant II-1); $\operatorname{arctg} 3$ (in variant II-2); $\operatorname{arctg} 4$ (in variant II-3); $\operatorname{arctg} 5$ (in variant II-4).

Solution. Let $k=A B: O C$. If $B B_{1}$ is the height of $\triangle A B C$, then $\triangle A B B_{1} \sim \triangle C O B_{1}$ (they are right triangles and $\angle B A B_{1}=\angle B_{1} O C$ ). Then $k=\frac{A B}{O C}=\frac{B B_{1}}{B_{1} C}=\operatorname{tg} \angle A C B$.