Trapezoid $A B C D(B C \| A D)$ is inscribed in a circle. It is known that $B C=a, A D=b, \angle C A D=\alpha$. Find the radius of the circle.
The circle circumscribed around trapezoid $ABCD$ is also the circumscribed circle of triangle $ACD$. To find its radius, use the generalized sine theorem.

## Solution

Assume that $a<b$. Let $K$ be the projection of vertex $C$ onto the base $AD$ of trapezoid $ABCD$. Then

$$
\begin{gathered}
DK=\frac{AD-BC}{2}=\frac{b-a}{2}, AK=\frac{AD+BC}{2}=\frac{b+a}{2}, \\
CK=AK \operatorname{tg} \angle CAD=\frac{b+a}{2} \cdot \operatorname{tg} \alpha
\end{gathered}
$$

$$
CD=\sqrt{DK^{2}+CK^{2}}=\sqrt{\left(\frac{b-a}{2}\right)^{2}+\left(\frac{b+a}{2} \cdot \operatorname{tg} \alpha\right)^{2}}
$$

The circle circumscribed around trapezoid $ABCD$ is also the circumscribed circle of triangle $ACD$. Let $R$ be its radius. Then

$$
R=\frac{CD}{2 \sin \angle CAD}=\frac{\sqrt{(b-a)^{2}+(b+a)^{2} \cdot \operatorname{tg}^{2} \alpha}}{4 \sin \alpha}
$$

## Answer

$\frac{\sqrt{(b-a)^{2}+(b+a)^{2} \operatorname{tg}^{2} \alpha}}{4 \sin \alpha}$.