12.135. 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.
## Solution.

Let $O$ be the center of the given sector $A O B$ (Fig. 12.8), and the center $O_{1}$ of the circle inscribed in it lies on the bisector $O K$ of the angle $A O B$.

Further, let $R$ be the radius of the sector, and $r$ be the radius of the circle.

Then the area of the sector $S_{1}=\frac{\alpha R^{2}}{2}$, and the area of the circle $S_{2}=\pi r^{2}$.

$D$ is the point of tangency of the circle and the radius $O B$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0897.jpg?height=327&width=390&top_left_y=159&top_left_x=128)

Fig. 12.8

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0897.jpg?height=452&width=664&top_left_y=79&top_left_x=578)

Fig. $12.9 \mathrm{a}$

Then $O_{1} D \perp O B, O_{1} D=r$.

$$
\begin{aligned}
& \text { In } \triangle O_{1} D O\left(\angle O_{1} D O=\frac{\pi}{2}\right): O O_{1}=\frac{r}{\sin \frac{\alpha}{2}} \\
& R=O K=O O_{1}+O_{1} K=\frac{r}{\sin \frac{\alpha}{2}}+r=\frac{r\left(1+\sin \frac{\alpha}{2}\right)}{\sin \frac{\alpha}{2}}=\frac{r\left(1+\cos \left(\frac{\pi}{2}-\frac{\alpha}{2}\right)\right)}{\sin \frac{\alpha}{2}}=
\end{aligned}
$$

$$
=\frac{2 r \cos ^{2}\left(\frac{\pi}{4}-\frac{\alpha}{4}\right)}{\sin \frac{\alpha}{2}}, \frac{S_{1}}{S_{2}}=\frac{\alpha}{2 \pi} \cdot \frac{R^{2}}{r^{2}}=\frac{4 \alpha \cos ^{4}\left(\frac{\pi}{4}-\frac{\alpha}{4}\right)}{2 \pi \sin ^{2} \frac{\alpha}{2}}=\frac{2 \alpha \cos ^{4}\left(\frac{\pi}{4}-\frac{\alpha}{4}\right)}{\pi \sin ^{2} \frac{\alpha}{2}} .
$$

Answer: $\frac{2 \alpha \cos ^{4}\left(\frac{\pi}{4}-\frac{\alpha}{4}\right)}{\pi \sin ^{2} \frac{\alpha}{2}}$.