12.065. The lateral edge of a regular triangular pyramid is equal to $l$ and forms an angle $\alpha$ with the base plane. Find the volume of the pyramid.
Solution.

Let $SABC$ be a regular triangular pyramid. $SA=l$, $SO \perp (ABC)$, $\angle S A O=\alpha$ (Fig. 12.67). From $\triangle S A O$ we find $S O=l \sin \alpha$, $A O=l \cos \alpha$, hence $A B=A O \sqrt{3}=l \sqrt{3} \cos \alpha$. Therefore,

\[
\begin{aligned}
& S_{\triangle A B C}=\frac{A B^{2} \sqrt{3}}{4}=\frac{3 \sqrt{3} l^{2} \cos ^{2} \alpha}{4} \\
& V_{\text {pyr }}=\frac{1}{3} S_{\triangle A B C} \cdot S O=\frac{1}{3} \cdot \frac{3 \sqrt{3} l^{2} \cos ^{2} \alpha}{4} \cdot l \sin \alpha=\frac{l^{3} \sqrt{3} \sin 2 \alpha \cos \alpha}{8}
\end{aligned}
\]

Answer: $\frac{l^{3} \sqrt{3} \sin 2 \alpha \cos \alpha}{8}$.