12.172. A trapezoid is inscribed in a circle. The larger base of the trapezoid forms an angle $\alpha$ with a lateral side, and an angle $\beta$ with a diagonal. Find the ratio of the area of the circle to the area of the trapezoid.
## Solution.

Let $AD$ be the larger base of the given trapezoid $ABCD$, $\angle BAD = \alpha$, $\angle BDA = \beta$ (Fig. 12.42), $BN$ be the height of the trapezoid, and $BD = 1$.

Then, from $\triangle BMD \left(\angle BMD = 90^\circ\right)$:

$BM = BD \sin \angle BDM = \sin \beta$

$DM = BD \cos \angle BDM = \cos \beta$.

The area of the trapezoid

$$
S_{1} = \frac{AD + BC}{2} \cdot BM = DM \cdot BM = \sin \beta \cos \beta = \frac{1}{2} \sin 2\beta
$$

The radius $R$ of the circle circumscribed around $\triangle BAD: R = \frac{BD}{2 \sin \angle A} = \frac{1}{2 \sin \alpha}$.

Then the area of the circle $S_{2} = \pi R^2 = \frac{\pi}{4 \sin^2 \alpha}$.

Thus, $\frac{S_{2}}{S_{1}} = \frac{\pi}{2 \sin^2 \alpha \sin 2\beta}$.

Answer: $\frac{\pi}{2 \sin^2 \alpha \sin 2\beta}$.