3.71. Find the ratio of the area of a sector with a given central angle $\alpha$ radians to the area of the circle inscribed in it.
3.71. By the condition, $O A n B$ is a sector, $\angle A O B=\alpha$ radians, $O_{1}$ is the center of the circle inscribed in the sector $O A n B$ (Fig. 3.72). Let $R$ be the radius of the sector, $r$ be the radius of the inscribed circle; then $S_{O A n B}=\frac{1}{2} R^{2} \alpha, S_{O_{1}}=\pi r^{2}$. In $\Delta O_{1} A_{1} O \quad\left(O_{1} A_{1} \perp O A\right) \quad$ we have $r=O_{1} A_{1}=O O_{1} \sin \frac{\alpha}{2}=(R-r) \sin \frac{\alpha}{2}$,

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-544.jpg?height=499&width=456&top_left_y=1233&top_left_x=687)

Fig. 3.72 or $r+r \sin \frac{\alpha}{2}=R \sin \frac{\alpha}{2}$, from which

$$
r=\frac{R \sin \frac{\alpha}{2}}{1+\sin \frac{\alpha}{2}}
$$

Finally, we get

$$
\frac{S_{O A n B}}{S_{O_{1}}}=\frac{R^{2} \alpha\left(1+\sin \frac{\alpha}{2}\right)^{2}}{2 \pi R^{2} \sin ^{2} \frac{\alpha}{2}}=\frac{2 \alpha \cos ^{2}\left(\frac{\pi}{4}-\frac{\alpha}{4}\right)}{\pi \sin ^{2} \frac{\alpha}{2}}
$$

![](https://cdn.mathpix.com/cropped/2024_05_21_1e3d8d2f6605ac23e2a5g-545.jpg?height=520&width=465&top_left_y=661&top_left_x=58)

Fig. 3.73