A right triangle with an acute angle $\alpha$ is located inside a circle of radius $r$ such that the hypotenuse is a chord of the circle, and the vertex of the right angle lies on the diameter parallel to the hypotenuse. Find the area of the triangle.
Let the vertex of the right angle $C$ of triangle $ABC$ lie on the diameter $MN$, which is parallel to the hypotenuse $AB$, and let $P$ and $Q$ be the projections of points $A$ and $B$ on $MN$ respectively.

$\angle A = \alpha$, $O$ is the center of the circle.

Note that $CH = AC \sin \alpha = AB \cos \alpha \sin \alpha = c \sin 2 \alpha$. Moreover, $c^2 + CH^2 = r^2$, which means $c^2 = \frac{r^2}{1 + \sin^2 2 \alpha}$.

Therefore, $S_{ABC} = \frac{1}{2} AB \cdot CH = c^2 \sin 2 \alpha = \frac{r^2 \sin 2 \alpha}{1 + \sin^2 2 \alpha}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_3ae27f1f0c68451d31a7g-13.jpg?height=429&width=497&top_left_y=0&top_left_x=789)

## Answer

$\frac{r^2 \sin 2 \alpha}{1 + \sin^2 2 \alpha}$