[ Law of Sines

In a cyclic quadrilateral, two opposite sides are perpendicular to each other, one of them is equal to $a$, and the angle adjacent to it is divided by the diagonal into parts $\alpha$ and $\beta$ (angle $\alpha$ is adjacent to the given side). Find the diagonals of the quadrilateral.

#
The radius of the circle circumscribed around a triangle is equal to the side of the triangle divided by twice the sine of the opposite angle.

## Solution

Let angle $D$ of the quadrilateral $ABCD$ be acute,

$$
CD \perp AB, CD=a, \angle CDB=\alpha, \angle BDA=\beta
$$

$R$ - radius of the circle. Then

$$
\angle BAD=90^{\circ}-\alpha-\beta, \angle BAC=\alpha, \angle CAD=90^{\circ}-2\alpha-\beta
$$

Therefore,

$$
\begin{gathered}
R=\frac{CD}{2 \sin \left(90^{\circ}-2\alpha-\beta\right)}=\frac{a}{2 \cos (2\alpha+\beta)} \\
AC=2R \sin (\alpha+\beta)=\frac{a \sin (\alpha+\beta)}{\cos (2\alpha+\beta)}
\end{gathered}
$$

$$
BD=2R \sin \left(90^{\circ}-\alpha-\beta\right)=\frac{a \cos (\alpha+\beta)}{\cos (2\alpha+\beta)}
$$

If $D$ is an obtuse angle, the solution is similar.

![](https://cdn.mathpix.com/cropped/2024_05_06_96a7587f2ba0a7b3aee2g-47.jpg?height=378&width=380&top_left_y=217&top_left_x=839)

## Answer

$\frac{a \sin (\alpha+\beta)}{\cos (2\alpha+\beta)} ; \frac{a \cos (\alpha+\beta)}{\cos (2\alpha+\beta)}$