12.219. When a circular sector rotates about one of its extreme radii, a body is formed whose spherical surface area is equal to the area of the conical surface. Find the sine of the central angle of the circular sector.
## Solution.

As a result of the rotation, a body consisting of a cone and a spherical segment with a common base was obtained (Fig. 12.86): the center $O$ of the circular sector is the vertex of the cone, the radius $OB$, around which the rotation was performed, is the axis of the obtained body, the radius $OA$ is the generatrix of the cone, $B$ is the vertex of the spherical segment.

The point $C$ is the center of the base of the cone. Then $\angle ACO=90^{\circ}, CB$ is the height of the spherical segment, $\angle AOB$ is the central angle of the circular sector.

Let $OA=OB=R, \angle AOB=\alpha$, then from $\triangle ACO: OC=R \cos \alpha, CA=R \sin \alpha$. $BC=OB-OC=R(1-\cos \alpha)$.

The area of the conical surface $S_{1}=\pi \cdot CA \cdot OA=\pi R^{2} \sin \alpha$.

The area of the surface of the spherical segment

$S_{2}=2 \pi \cdot OB \cdot CB=2 \pi R^{2}(1-\cos \alpha)$.

Since $S_{1}=S_{2}$, then $\pi R^{2} \sin \alpha=2 \pi R^{2}(1-\cos \alpha) \sin \alpha=2(1-\cos \alpha) \frac{\sin \alpha}{1-\cos \alpha}=$ $=2 ; \operatorname{tg} \frac{\alpha}{2}=\frac{1}{2}$.

Then $\sin \alpha=\frac{2 \operatorname{tg} \frac{\alpha}{2}}{1+\operatorname{tg}^{2} \frac{\alpha}{2}}=\frac{4}{5}$.

Answer: $\frac{4}{5}$.