3.95. The base of the pyramid is an isosceles triangle with an angle $\alpha$ between the lateral sides. The pyramid is placed in a certain cylinder such that its base is inscribed in the base of this cylinder, and the vertex coincides with the midpoint of one of the cylinder's generators. The volume of the cylinder is $V$. Find the volume of the pyramid.
3.95. By condition, $S A B C$ is a pyramid, $C A=C B, \angle A C B=a$, $\triangle A B C$ is inscribed in the base of the cylinder, $S \in K M, K M$ is the generatrix of the cylinder, $K S=S M, V_{\text {cyl }}=V$ (Fig. 3.95). Let the radius of the base of the cylinder be $R$, and its height be $H$; then $V=\pi R^{2} H$, from which $R^{2} H=\frac{V}{\pi}$. We have $A B=2 R \sin \alpha$ and, therefore,

$$
S_{\triangle A B C}=\frac{4 R^{2} \sin ^{2} \alpha \sin ^{2}\left(\frac{\pi}{2}-\frac{\alpha}{2}\right)}{2 \sin \alpha}=2 R^{2} \sin \alpha \cos ^{2} \frac{\alpha}{2}
$$

Finally, we get

$V_{\text {pyr }}=\frac{1}{3} \cdot 2 R^{2} \sin \alpha \cos ^{2} \frac{\alpha}{2} \cdot \frac{H}{2}=\frac{1}{3} R^{2} H \sin \alpha \cos ^{2} \frac{\alpha}{2}=\frac{V}{3 \pi} \sin \alpha \cos ^{2} \frac{\alpha}{2}$