In a right triangle $A B C$, from point $E$, located at the midpoint of the leg $B C$, a perpendicular $E L$ is dropped to the hypotenuse $A B$. Find the angles of triangle $A B C$ if $A E=\sqrt{10} \cdot E L$ and $B C>A C$.
## First solution method.

Let $\angle A B C=\beta, E L=x$. From the right triangles $E L A, B L E$ and $A B C$, we find:

$$
A L=\sqrt{A E^{2}-E L^{2}}=\sqrt{10 r^{2}-r^{2}}=3 x
$$

$$
B L=E L \operatorname{ctg} \beta=x \operatorname{ctg} \beta, B E=\frac{E L}{\sin \beta}=\frac{x}{\sin \beta}
$$

$$
A B=B L+L A=x \operatorname{ctg} \beta+3 x, B C=2 B E=\frac{2 x}{\sin \beta}
$$

$$
B C=A B \cos \beta, \text { or } \frac{2 x}{\sin \beta}=(x \operatorname{ctg} \beta+3 x) \cos \beta
$$

Solve this equation:

$$
\begin{gathered}
\frac{2}{\sin \beta}=(\operatorname{ctg} \beta+3) \cos \beta \Leftrightarrow 2=(\operatorname{ctg} \beta+3) \sin \beta \cos \beta \Leftrightarrow \\
\Leftrightarrow 2=3 \sin \beta \cos \beta+\cos 2 \beta \Leftrightarrow 2 \sin 2 \beta-3 \sin \beta \cos \beta+\cos 2 \beta=0 \Leftrightarrow \\
\Leftrightarrow 2 \operatorname{tg} 2 \beta-3 \operatorname{tg} \beta+1=0
\end{gathered}
$$

Thus, $\operatorname{tg} \beta=1$ or $\operatorname{tg} \beta=\frac{1}{2}$. Since $B C>A C$, only the second root satisfies the problem's condition. Therefore,

$$
\beta=\operatorname{arctg} \frac{1}{2}=\operatorname{arcctg} 2
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_808033c11c6fe49d1738g-10.jpg?height=388&width=698&top_left_y=2227&top_left_x=697)

Second solution method.

Let $E L=x, B E=C E=2 y$. From the right triangles $E L A, B L E$ and $A B C$, we find that

$$
A L=\sqrt{A E^{2}-E L^{2}}=\sqrt{10 r^{2}-r^{2}}=3 x
$$

$$
\begin{gathered}
B L=\sqrt{B E^{2}-L E^{2}}=\sqrt{y^{2}-r^{2}}, \\
\left(3 x+\sqrt{y^{2}-r^{2}}\right) 2=(10 x 2-y 2)+4 y 2 \Leftrightarrow \\
\Leftrightarrow 3 x \sqrt{y^{2}-r^{2}}=x 2+y 2 \Leftrightarrow y 4-7 x 2 y 2+10 x 4=0,(y 2-5 x 2)(y 2-2 x 2)=0 .
\end{gathered}
$$

By the problem's condition $B C>A C$, so $y \neq x \sqrt{2}$ (otherwise, triangle $B L E$, and thus $A B C$, would be isosceles. Therefore, $y=x \sqrt{5}$. Then

$$
B L=\sqrt{5 r^{2}-r^{2}}=2 x, \operatorname{ctg} \angle A B C=\frac{B L}{L E}=2 .
$$

## Answer

$\operatorname{arctg} 2, \operatorname{arcctg} 2$.