The height of a right triangle, drawn from the vertex of the right angle, is equal to $a$ and forms an angle $\alpha$ with the median, drawn from the same vertex. Find the legs of the triangle.

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Let $CH = a$ be the height of the right triangle $ABC$, drawn from the vertex $C$ of the right angle, and $CM$ be the median of this triangle, where $\angle MCH = \alpha$. Then $BM = AM = CM$.

![](https://cdn.mathpix.com/cropped/2024_05_06_3ae27f1f0c68451d31a7g-22.jpg?height=260&width=512&top_left_y=1814&top_left_x=767)

Assume that $BC > AC$. Then the point $M$ lies between $B$ and $H$. By the external angle theorem of a triangle, $\angle MBC = \frac{1}{2} \angle CMH = 45^\circ - \alpha$. Therefore,

$BC = \frac{CH}{\sin \angle HBC} = \frac{a}{\sin \left(45^\circ - \frac{\alpha}{2}\right)}$.

Similarly, $AC = \frac{CH}{\sin \angle HAC} = \frac{a}{\sin \left(45^\circ + \frac{\alpha}{2}\right)}$.

## Answer

$\frac{a \sqrt{2(1 + \sin \alpha)}}{\cos \alpha} = \frac{a}{\sin \left(45^\circ \pm \frac{\alpha}{2}\right)}$.

## Problem