Find the volume of a regular triangular pyramid with the radius $R$ of the circumscribed sphere and the angle $\alpha$ between the lateral edge and the base plane.

#
Let $O$ be the center of the sphere of radius $R$, circumscribed around a regular triangular pyramid $ABCD$ with vertex $D$ (Fig.1). The point $O$ lies on the line $DM$, where $M$ is the center of the base $ABC$. By the problem's condition, $OA=R, \angle DAM=\alpha$. Denote $AB=BC=AC=a$. Then $AM=\frac{a \sqrt{3}}{3}$. Consider the section of the pyramid and the sphere by a plane passing through points $A, D$, and $M$ (Fig.2). We obtain a circle of radius $R$ with center $O$ on the line $MD$. Extend the segment $DM$ beyond point $M$ to intersect the circle at point $P$. From the right triangles $AMD$ and $APD$, we find that

\[
\begin{aligned}
& DM = AM \operatorname{tg} \angle DAM = \frac{a \sqrt{3} \operatorname{tg} \alpha}{3}, \\
& AM^2 = DM \cdot MP, \text{ or } \frac{1}{3} a^2 = \frac{a \sqrt{3} \operatorname{tg} \alpha}{3} \left(2R - \frac{a \sqrt{3} \operatorname{tg} \alpha}{3}\right),
\end{aligned}
\]

from which

\[
a = \frac{2R \sqrt{3} \operatorname{tg} \alpha}{1 + \operatorname{tg}^2 \alpha} = 2R \sqrt{3} \operatorname{tg} \alpha \cos 2\alpha = 2R \sqrt{3} \sin \alpha \cos \alpha = R \sqrt{3} \sin 2\alpha
\]

Therefore,

\[
\begin{gathered}
V_{ABCD} = \frac{1}{3} S_{\triangle ABC} \cdot DM = \frac{1}{3} \cdot \frac{a^2 \sqrt{3}}{4} \cdot \frac{a \sqrt{3} \operatorname{tg} \alpha}{3} = \\
= \frac{1}{12} a^3 \operatorname{tg} \alpha = \frac{1}{12} (R \sqrt{3} \sin 2\alpha)^3 \operatorname{tg} \alpha = \frac{1}{4} R^3 \sqrt{3} \sin^3 2\alpha \operatorname{tg} \alpha.
\end{gathered}
\]

## Answer

\[
\frac{1}{4} R^3 \sqrt{3} \sin^3 2\alpha \operatorname{tg} \alpha = \frac{1}{2} R^3 \sqrt{3} \sin^2 2\alpha \sin 2\alpha.
\]