12.071. A cone is inscribed in a hemisphere; the vertex of the cone coincides with the center of the circle, which is the base of the hemisphere; the planes of the bases of the cone and the hemisphere are parallel. A line passing through the center of the base of the cone and an arbitrary point on the circumference of the great circle of the hemisphere forms an angle $\alpha$ with the plane of the base of the cone. Find the ratio of the volumes of the hemisphere and the cone.
Solution.

Let the radius of the circle at the base of the hemisphere be denoted by $R$, i.e., $O K=R$, and the radius at the base of the cone by $r$, i.e., $O_{1} L=r$, $\angle O_{1} K O=\alpha$ (Fig. 12.73). Then the height $O_{1} O$ of the cone will be equal to $h=K O \cdot \operatorname{tg} \alpha=R \operatorname{tg} \alpha$. On the other hand, the slant height of the cone $OL$ is the radius of the hemisphere. Then $L O_{1}^{2}=O L^{2}-O O_{1}^{2}$ or $r^{2}=R^{2}-h^{2}$, which means that $r=\sqrt{R^{2}-R^{2} \operatorname{tg}^{2} \alpha}=R \sqrt{\frac{\cos ^{2} \alpha-\sin ^{2} \alpha}{\cos ^{2} \alpha}}=R \frac{\sqrt{\cos 2 \alpha}}{\cos \alpha}$. Then the volume of the cone $V_{\mathrm{k}}=\frac{1}{3} S_{\text {base }} \cdot h=\frac{1}{3} \pi r^{2} h=\frac{1}{3} \pi R^{2} \frac{\cos 2 \alpha}{\cos ^{2} \alpha} \cdot R \operatorname{tg} \alpha$. The volume of the hemisphere $V_{\mathrm{n}}=\frac{2}{3} \pi R^{3}$. The ratio $\frac{V_{\pi}}{V_{\mathrm{k}}}=\frac{2 \cos ^{2} \alpha}{\cos 2 \alpha \operatorname{tg} \alpha}$.

Answer: $\frac{2 \cos ^{2} \alpha}{\cos 2 \alpha \operatorname{tg} \alpha}$.