3B. The radii of the circumscribed and inscribed circles of a right-angled triangle are in the ratio $5: 2$, respectively. Determine the acute angles in the triangle.
Solution. Let $R$ be the radius of the circumscribed circle, and $r$ the radius of the inscribed circle in the right-angled triangle. If $\frac{\alpha}{2}+\frac{\beta}{2}=45^{\circ}$, then

$$
\operatorname{ctg}\left(\frac{\alpha}{2}+\frac{\beta}{2}\right)=\frac{\operatorname{ctg} \frac{\alpha}{2} \cdot \operatorname{ctg} \frac{\beta}{2}-1}{\operatorname{ctg} \frac{\alpha}{2}+\operatorname{ctg} \frac{\beta}{2}}=1
$$

It follows that

$$
\begin{aligned}
\overline{A B} & =\overline{A D}+\overline{D B}=\overline{O D} \operatorname{ctg} \frac{\alpha}{2}+\overline{O D} \operatorname{ctg} \frac{\beta}{2} \\
& =r\left(\operatorname{ctg} \frac{\alpha}{2}+\operatorname{ctg} \frac{\beta}{2}\right)
\end{aligned}
$$

Since $\frac{\alpha}{2}+\frac{\beta}{2}=45^{\circ}$, we get:

![](https://cdn.mathpix.com/cropped/2024_06_05_ae7d00f0d51ae2d35791g-07.jpg?height=579&width=450&top_left_y=845&top_left_x=1015)

$$
2 R=r\left(\operatorname{ctg} \frac{\alpha}{2}+\operatorname{ctg} \frac{\beta}{2}\right)
$$

From this, $\operatorname{ctg} \frac{\alpha}{2}+\operatorname{ctg} \frac{\beta}{2}=2 \frac{R}{r}=5$. The triangle $A B C$ is a right-angled triangle, so $\alpha+\beta=90^{\circ}$. Thus, $\frac{\alpha}{2}+\frac{\beta}{2}=45^{\circ}$. From equation (1), it follows that $\operatorname{ctg} \frac{\alpha}{2} \operatorname{ctg} \frac{\beta}{2}=6$. Taking $\alpha<\beta$ and solving the system of equations $\operatorname{ctg} \frac{\alpha}{2}+\operatorname{ctg} \frac{\beta}{2}=5, \operatorname{ctg} \frac{\alpha}{2} \operatorname{ctg} \frac{\beta}{2}=6$, we get $\operatorname{ctg} \frac{\alpha}{2}=3$ and $\operatorname{ctg} \frac{\beta}{2}=2$. Therefore, $\alpha=2 \operatorname{arcctg} 3, \beta=2 \operatorname{arcctg} 2$.