3.36. The base of the pyramid is an isosceles triangle with a lateral side equal to \(a\), and the angle at the vertex is \(\alpha\). All lateral edges are inclined to the base plane at an angle \(\beta\). Find the volume of the pyramid.
3.36. Given that $S A B C-$ is a triangular pyramid, $A B=B C=$ $=a, \angle A B C=\alpha, S O \perp(A B C), \angle S A O=$ $=\angle S B O=\angle S C O=\beta$ (Fig. 3.38). Since $\triangle S O A=\triangle S O B=\triangle S O C$ (by the leg and adjacent angle), then $O A=O B=O C$, i.e., $O$ is the center of the circle circumscribed around $\triangle A B C$. Let $O A=R$; then $A B=2 R \sin \angle A C B$, or

$$
a=2 R \sin \left(\frac{\pi}{2}-\frac{\alpha}{2}\right)=2 R \cos \frac{\alpha}{2} \text {. }
$$

Therefore, $R=\frac{a}{2 \cos \frac{\alpha}{2}}$. Next, from $\triangle S O A$ we find $S O=\frac{a \operatorname{tg} \beta}{2 \cos \frac{\alpha}{2}}$.

Finally, we obtain

$$
V_{\text {map }}=\frac{1}{3} S_{\triangle A B C} \cdot S O=\frac{1}{6} a^{2} \sin \alpha \frac{a \operatorname{tg} \beta}{2 \cos \frac{\alpha}{2}}=\frac{1}{6} a^{3} \sin \frac{\alpha}{2} \operatorname{tg} \beta
$$