12.134. The bases of an isosceles trapezoid are equal to $a$ and $b(a>b)$, and the angle at the larger base is $\alpha$. Find the radius of the circle circumscribed around the trapezoid.
Solution.

Let $BL$ be the height of the given trapezoid $ABCD$ (Fig. 12.7), $BC=b, AD=a$, $\angle A=\alpha, 0^{\circ}<\alpha<90^{\circ}$.

Then $AL=\frac{a-b}{2}$ and from $\triangle ALB\left(\angle ALB=90^{\circ}\right)$:

$$
AB=\frac{a-b}{2 \cos \alpha}
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0896.jpg?height=367&width=553&top_left_y=164&top_left_x=93)

Fig. 12.6

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0896.jpg?height=448&width=488&top_left_y=79&top_left_x=739)

Fig. 12.7

Next, from $\triangle ABD$:

$BD^{2}=AB^{2}+AD^{2}-2AB \cdot AD \cos \angle A=\frac{(a-b)^{2}}{4 \cos ^{2} \alpha}+a^{2}-\frac{2a(a-b)}{2 \cos \alpha} \cdot \cos \alpha=$ $=\frac{a^{2}-2ab+b^{2}}{4 \cos ^{2} \alpha}+a^{2}-a^{2}+ab=\frac{a^{2}-2ab+b^{2}+4ab \cos \alpha}{4 \cos ^{2} \alpha}=$ $=\frac{a^{2}+b^{2}-2ab(1-2 \cos ^{2} \alpha)}{4 \cos ^{2} \alpha}=\frac{a^{2}+b^{2}+2ab \cos 2 \alpha}{4 \cos ^{2} \alpha}$. $BD=\frac{\sqrt{a^{2}+b^{2}+2ab \cos 2 \alpha}}{2 \cos \alpha}$.

The radius of the circle circumscribed around $\triangle BAD$ will be equal to

$$
\begin{aligned}
& R=\frac{BD}{2 \sin \angle A}=\frac{\sqrt{a^{2}+b^{2}+2ab \cos 2 \alpha}}{2 \sin 2 \alpha} \\
& \text { Answer: } \frac{\sqrt{a^{2}+b^{2}+2ab \cos 2 \alpha}}{2 \sin 2 \alpha}
\end{aligned}
$$