[ Inscribed and Circumscribed Circles ] [ Pythagorean Theorem (direct and inverse).]

In a right triangle, the ratio of the radius of the inscribed circle to the radius of the circumscribed circle is $\frac{2}{5}$. Find the acute angles of the triangle.
The radius of the circle inscribed in a right-angled triangle with legs $a, b$ and hypotenuse $c$ is $\frac{a+b-c}{2}$.

## Solution

## First Method.

Let the radius of the inscribed circle in a right-angled triangle with legs $a, b$ and hypotenuse $c$ be $2x$, and the radius of the circumscribed circle be $-5x$. Then $c=10x$. Therefore, $a^{2}+b^{2}=100x^{2}$.

Since the radius of the circle inscribed in a right-angled triangle is $\frac{a+b-c}{2}$, we have another equation

$$
a+b=14x
$$

From the obtained system of equations, we find that

$$
a=8x, b=6x, \text{ or } a=6x, b=8x.
$$

Therefore, the tangents of the acute angles of the triangle are $\frac{3}{4}$ and $\frac{4}{3}$.

## Second Method.

Let the legs of the given right-angled triangle be $a$ and $b$, the hypotenuse be $c$, and the angle of the triangle opposite the leg $a$ be $\alpha$ (we will assume that $\alpha \leqslant \frac{\pi}{4}$). Then the radius of the circumscribed circle is $\frac{c}{2}$

$$
\sin \alpha=\frac{a}{c}, \cos \alpha=\frac{b}{c}
$$

$$
\begin{gathered}
\frac{\frac{a+b-c}{2}}{\frac{c}{2}}=\frac{2}{5} \Rightarrow \frac{a}{c}+\frac{b}{c}=\frac{7}{5} \Rightarrow \\
\Rightarrow \sin \alpha+\cos \alpha=\frac{7}{5} \Rightarrow \cos \left(\alpha-\frac{\pi}{4}\right)=\frac{7}{5 \sqrt{2}},
\end{gathered}
$$

from which we find that

$$
\alpha=\frac{\pi}{4}+\arccos \frac{7 \sqrt{2}}{10}.
$$

Then the second acute angle is

$$
\frac{\pi}{2}-\alpha=\frac{\pi}{4}-\arccos \frac{7 \sqrt{2}}{10}
$$

## Answer

$\frac{\pi}{4} \pm \arccos \frac{7 \sqrt{2}}{10}$, or $\operatorname{arctg} \frac{3}{4}, \operatorname{arcctg} \frac{3}{4}$.