8.4. Given a right triangle, the height dropped to the hypotenuse is 4 times smaller than the hypotenuse. Find the acute angles of this triangle.
Answer: $15^{\circ}$ and $75^{\circ}$. Solution. Let $A B C$ be the given triangle, $C M$ the height from vertex $C$ of the right angle, and $C O$ the median. Consider the right triangle $C M O$. By the property of the median from the vertex of the right angle, $C O=\frac{A B}{2}$, and therefore (by the condition) $C O$ is twice as long as $C M$. Therefore (by the property of a right triangle with a leg equal to half the hypotenuse) $\angle C O M=30^{\circ}$. Then, considering the isosceles $\triangle C O B$ with the external angle $C O M$, we get $\angle O C B=\angle O B C=15^{\circ}$, $\angle C A B=90^{\circ}-15^{\circ}=75^{\circ}$.