10.022. Determine the acute angles of a right triangle if the median drawn to its hypotenuse divides the right angle in the ratio 1:2.
Solution.

$A O=O B, \frac{\angle B C O}{\angle A C O}=\frac{2}{1}$ (Fig. 10.22). In $\triangle A B C, \angle A C B=90^{\circ}$; thus, $O$ is the center of the circumscribed circle, meaning $A O=O B=C O$, $\triangle A O C$ is isosceles, $\angle A=\angle O C A, \angle B=\angle B C O$. Let

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-566.jpg?height=368&width=354&top_left_y=122&top_left_x=239)

Fig. 10.23

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-566.jpg?height=394&width=470&top_left_y=93&top_left_x=701)

Fig. 10.24

$\angle O C A=x$, so $\angle B C O=2 x, x+2 x=90^{\circ}, x=30^{\circ}$, thus, $\angle A=30^{\circ}, \angle B C O=\angle B=60^{\circ}$.

Answer: $30^{\circ}, 60^{\circ}$.