12.149. An isosceles trapezoid is circumscribed around a circle of radius $r$. The lateral side of the trapezoid forms an angle $\alpha$ with the smaller base. Find the radius of the circle circumscribed around the trapezoid.
Solution.

Let $B L$ be the height of the given isosceles trapezoid $A B C D$ (Fig. 12.22), then $B L=2 r$. In $\triangle A L B\left(\angle A L B=90^{\circ}\right): A B=\frac{B L}{\sin \angle B A L}=\frac{2 r}{\sin \alpha}$. Since a circle is inscribed in the given trapezoid, then $A B+B C=A B+C D=2 A B$, $A B=C D \Rightarrow \quad \dot{D D}=\frac{A D+B C}{2}=A B=\frac{2 r}{\sin \alpha}$. In $\triangle B L D\left(\angle B L D=90^{\circ}\right):$ $B D=\sqrt{B L^{2}+L D^{2}}=\sqrt{4 r^{2}+\frac{4 r^{2}}{\sin ^{2} \alpha}}=\frac{2 r}{\sin \alpha} \sqrt{1+\sin ^{2} \alpha}$. The radius $R$ of the circle circumscribed around $\triangle A B D: R=\frac{B D}{2 \sin \angle A}=\frac{r \sqrt{1+\sin ^{2} \alpha}}{\sin ^{2} \alpha}$.

Answer: $\frac{r \sqrt{1+\sin ^{2} \alpha}}{\sin ^{2} \alpha}$.