All vertices of an equilateral triangle lie on the sides of a right triangle. One of the sides of the equilateral triangle is parallel to the hypotenuse and its length is three times smaller than the length of the hypotenuse. Find the angles of the right triangle.
Let vertices $K$ and $L$ of the equilateral triangle $KLM$ lie on the legs $BC$ and $AC$ of the right triangle $ABC$, respectively, and vertex $M$ lie on the hypotenuse $AB$. Let $\angle A = \alpha, KL = 1$. Then $\angle KLC = \angle A = \alpha, AB = 3$, and the height $CH = 3 \cos \alpha \sin \alpha$. On the other hand, this height is one and a half times the height of the triangle $KLM$, which is $\frac{3 \sqrt{3}}{4}$. Therefore, $\sin \alpha \cos \alpha = \frac{\sqrt{3}}{4}, \sin 2\alpha = \frac{\sqrt{3}}{2}$, which means $2\alpha = 60^\circ$ or $120^\circ$. Consequently, $\alpha = 30^\circ$ or $\alpha = 60^\circ$.

## Answer

$30^\circ, 60^\circ$.

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