12.390. A sphere is inscribed in a spherical sector of radius $R$ (Fig. 12.239). Find the radius of the circle of contact between the surfaces of the sphere and the sector, if the central angle in the axial section of the spherical sector is $\alpha$.
Solution.

Consider the axial section of the given set of bodies.

$$
\begin{array}{ll}
\text { Fig. } 12.239 & \text { In } \triangle O C A\left(\angle O C A=\frac{\pi}{2}\right):
\end{array}
$$

$O A=\frac{r}{\sin \frac{\alpha}{2}}$, where $r$ is the radius of the given sphere.

$$
R=A B=A O+O B=\frac{\sin \frac{u}{2}}{\sin \frac{\alpha}{2}}+r=r \cdot \frac{1+\sin \frac{\alpha}{2}}{\sin \frac{\alpha}{2}}
$$

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

$\triangle C D O$ and $\triangle A C O$ are right triangles with a common acute angle at vertex $O$. Then $\angle O C D=\angle O A C=\frac{\alpha}{2}$. In $\triangle C D O\left(\angle C D O=\frac{\pi}{2}\right)$, the sought radius of the circle of tangency $D C=C O \cos \angle O C D=r \cos \frac{\alpha}{2}=\frac{R \sin \frac{\alpha}{2} \cos \frac{\alpha}{2}}{2 \cos ^{2}\left(\frac{\pi}{4}-\frac{\alpha}{4}\right)}$.
Answer: $\frac{R \sin \alpha}{\cos ^{2}\left(\frac{\pi \alpha}{4}\right)}$.

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

## Solutions to Chapter 13

## APPLICATION OF EQUATIONS TO PROBLEM SOLVING