10.365 Determine the acute angles of a right triangle if the ratio of the radii of the circumscribed and inscribed circles is $\sqrt{3}+1$.
Solution. Let $a, b$ be the legs of the given triangle, $c$ be its hypotenuse, and $R$ and $r$ be the radii of the circumscribed and inscribed circles. Then $R=\frac{c}{2}, r=\frac{a+b-c}{2}$ (Theorem 1) $\Rightarrow$ $\Rightarrow \frac{r}{R}=\frac{1}{\sqrt{3}+1}=\frac{\sqrt{3}-1}{2}=\frac{a+b-c}{c}==\frac{a}{c}+\frac{b}{c}-1=\sin \alpha+\cos \alpha-1$, where $\alpha-$ is one of the acute angles of the triangle.

$$
\text { Hence } \frac{\sqrt{3}+1}{2}=\sin \alpha+\cos \alpha \Rightarrow\left(\frac{\sqrt{3}+1}{2}\right)^{2}=1+\sin 2 \alpha \Rightarrow \sin 2 \alpha=\frac{\sqrt{3}}{2} \text {. }
$$

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