12.117. The base of a pyramid inscribed in a cone is a quadrilateral with adjacent sides equal in pairs, and the angle between one pair of adjacent sides is $\alpha$. Find the ratio of the volume of the pyramid to the volume of the cone.
Solution.

The volume of the pyramid $V_{\mathrm{n}}=\frac{1}{3} S_{\text {osn }} \cdot H$ (Fig. 12.117). The area of the base of the pyramid $S_{\text {osn }}=2 S_{\triangle A B C}=2 \cdot \frac{1}{2} A B \cdot B C\left(\angle A B C=90^{\circ}\right)$. From $\triangle A B C$ $A B=2 O C \cdot \sin \frac{\alpha}{2} ;$ and $B C=2 R \cos \frac{\alpha}{2}$. Then $S_{\text {osn }}=4 R^{2} \cos \frac{\alpha}{2} \sin \frac{\alpha}{2}=2 R^{2} \sin \alpha$. The volume of the cone $V_{\mathrm{k}}=\frac{1}{3} \pi R^{2} H$. Then the required ratio:

$$
\frac{V_{\mathrm{I}}}{V_{\mathrm{k}}}=\frac{S_{\mathrm{osn}} \cdot H}{\pi R^{2} \cdot H}=\frac{2 R^{2} \sin \alpha}{\pi R^{2}}=\frac{2 \sin \alpha}{\pi} \text {. }
$$

Answer: $\frac{2 \sin \alpha}{\pi}$.

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-803.jpg?height=672&width=532&top_left_y=50&top_left_x=124)

Fig. 12.118

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-803.jpg?height=482&width=442&top_left_y=221&top_left_x=729)

Fig. 12.119