12.060. The base of the pyramid is an isosceles triangle, with the lateral side equal to ' $a$, and the angle at the vertex equal to $\alpha$. All lateral edges are inclined to the plane of the base at an angle $\beta$. Find the volume of the pyramid.
Solution.

Given that $SABC$ is a triangular pyramid, $AB=BC=a$, $\angle ABC=\alpha, SO \perp (ABC), \angle SAO=\angle SBO=\angle SCO=\beta$ (Fig. 12.62). Since $\triangle SOA = \triangle SOB = \triangle SOC$ (by the leg and adjacent angle), then $OA=OB=OC$, i.e., $O$ is the center of the circle circumscribed around $\triangle ABC$. Let $OA=R$; then $AB=2R \sin \angle ACB$ or

\[
\begin{aligned}
& a=2R \sin \left(\frac{\pi}{2}-\frac{\alpha}{2}\right)=2R \cos \frac{\alpha}{2}, \text{ so } R=\frac{a}{2 \cos \frac{\alpha}{2}}. \text{ From } \triangle SOA \text{ we find } \\
& SO=\frac{a \tan \beta}{2 \cos \frac{\alpha}{2}}, \text{ hence } \\
& V_{\text{pyr}}=\frac{1}{3} S_{\triangle ABC} \cdot SO=\frac{1}{6} a^2 \sin \alpha \frac{a \tan \beta}{2 \cos \frac{\alpha}{2}}=\frac{1}{6} a^3 \sin \frac{\alpha}{2} \tan \beta
\end{aligned}
\]