12.102. The volume of a cone is $V$. A pyramid is inscribed in the cone, with an isosceles triangle as its base, where the angle between the lateral sides is $\alpha$. Find the volume of the pyramid.
Solution.

The volume of the pyramid is: $V_{\text {p }}=\frac{1}{3} S H$ (Fig. 12.102). The area of the base of the pyramid $S=\frac{A B^{2} \cdot C B}{4 R}$. From $\triangle A B C \quad C B=2 A B \sin \frac{\alpha}{2}$. Then
$S=\frac{2 A B^{3} \sin \frac{\alpha}{2}}{4 R}=\frac{2 x^{3} \sin \frac{\alpha}{2}}{4 R}$. Hence $R=\frac{2 x^{3} \sin \frac{\alpha}{2}}{4 S}$. The volume of the cone $V=\frac{1}{3} \pi R^{2} H$. Hence $H=\frac{3 V}{\pi R^{2}}$. Substitute into the volume of the pyramid: $V_{\text {p }}=\frac{1}{3} \frac{S \cdot 3 V \cdot 16 S^{2}}{\pi \cdot\left(2 x^{3} \sin \frac{\alpha}{2}\right)^{2}}=\frac{4 V S^{3}}{\pi x^{6} \sin ^{2} \frac{\alpha}{2}}$. On the other hand, the area of the base is: $S=\frac{1}{2} x^{2} \sin \alpha$. Then we finally get:

$$
V_{\mathrm{\Pi}}=\frac{4 V x^{6} \sin ^{3} \alpha}{8 \pi x^{6} \sin ^{2} \frac{\alpha}{2}}=\frac{V}{2 \pi} \frac{\sin \alpha 4 \sin ^{2} \frac{\alpha}{2} \cos ^{2} \frac{\alpha}{2}}{\sin ^{2} \frac{\alpha}{2}}=\frac{2 V}{\pi} \sin \alpha \cos ^{2} \frac{\alpha}{2}
$$

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