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Chords $A B$ and $A C$ are equal to each other. The inscribed angle formed by them in the circle is $30^{\circ}$. Find the ratio of the area of the part of the circle enclosed by this angle to the area of the entire circle.
A part of the circle enclosed in the given angle consists of a sector with an angle of $60^{\circ}$ and two equal isosceles triangles.

## Solution

Since triangle ABC is acute, the center of its circumscribed circle is located inside the triangle. Let $O$ be the center of the circle, and $R$ be its radius. The part of the circle enclosed in the given angle consists of sector $BOC$ with an angle $\angle BOC$ equal to $60^{\circ}$, and two equal isosceles triangles $AOB$ and $AOC$ with angles at the vertices equal to $\frac{360^{\circ}-60^{\circ}}{2}=150^{\circ}$.

The area of the sector is one-sixth of the area of the circle, i.e., $\frac{\pi R^{2}}{6}$, and the area of each triangle is

$$
\frac{1}{2} R^{2} \sin 150^{\circ}=\frac{R^{2}}{4}
$$

Therefore, the desired ratio is

$$
\frac{\frac{\pi R^{2}}{6}+\frac{R^{2}}{2}}{\pi R^{2}}=\frac{\pi+3}{6 \pi}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_d1936174aeda27dcc10dg-21.jpg?height=463&width=443&top_left_y=820&top_left_x=813)

## Answer

$\frac{\pi+3}{6 \pi}$