11.2. Two circles with radii 1 and 2 have a common center $O$. The area of the shaded region is three times smaller than the area of the larger circle. Find the angle $\angle A O B$.

![](https://cdn.mathpix.com/cropped/2024_05_06_acc874c26ed09747c6f4g-1.jpg?height=439&width=468&top_left_y=1665&top_left_x=811)
Answer: $60^{\circ}$

Solution: Let $\angle A O B=\alpha$. Then the area of the shaded part is $\frac{360^{\circ}-\alpha}{360^{\circ}} \pi+$ $4 \frac{\alpha}{360^{\circ}} \pi-\frac{\alpha}{360^{\circ}} \pi=\left(\frac{360^{\circ}-\alpha}{360^{\circ}}+3 \frac{\alpha}{360^{\circ}}\right) \pi$. The area of the large circle is $4 \pi$. According to the condition, $4 \pi=3\left(\frac{360^{\circ}-\alpha}{360^{\circ}}+3 \frac{\alpha}{360^{\circ}}\right) \pi, 4 \times 360^{\circ}=3\left(360^{\circ}-\alpha\right)+9 \alpha, 1440^{\circ}=1080^{\circ}-6 \alpha, 6 \alpha=$ $360^{\circ}, \alpha=60^{\circ}$.

Criteria: Only the correct answer - 1 point. The formula for calculating the area of a sector is written - 1 point. If it is proven that for an angle of $60^{\circ}$, the given ratio of areas holds - 4 points.