12.213. A perpendicular, equal to $a$, is dropped from the base of the height of a regular triangular pyramid onto a lateral face. Find the volume of the pyramid if the angle of inclination of the lateral edge to the plane of the base is $\alpha$.
Solution.

Let $SO$ be the height of the regular pyramid $SABC$ (Fig. 12.80), $\angle SCO = \alpha$, and $D$ the midpoint of $BC$. Then $AD \perp BC$, $SD \perp BC$, and the edge $BC$ is perpendicular to the plane $ADS$. The planes $ADS$ and $BSC$ are perpendicular.

In the plane $ADS$, draw a perpendicular $OM$ from point $O$ to $SD$. Since the planes $ADS$ and $BSC$ are perpendicular, $OM$ is perpendicular to the plane $BSC$ and equals $a$.

Let $BC = x$, $\angle ODS = \varphi \Rightarrow OC = \frac{x \sqrt{3}}{3}$, $OD = \frac{x \sqrt{3}}{6}$, $S_{\triangle ABC} = \frac{x^2 \sqrt{3}}{4}$.

In $\triangle SOC (\angle SOC = 90^\circ)$: $SO = OC \tan \angle SCO = \frac{x \sqrt{3}}{3} \tan \alpha$.

In $\triangle SOD (\angle SOD = 90^\circ)$: $SO = OD \tan \angle ODS = \frac{x \sqrt{3}}{6} \tan \varphi$.

Then $\frac{x \sqrt{3}}{3} \tan \alpha = \frac{x \sqrt{3}}{6} \tan \varphi \Leftrightarrow \tan \varphi = 2 \tan \alpha$.

In $\triangle OMD (\angle OMD = 90^\circ)$:

$$
\begin{aligned}
& OD = \frac{OM}{\sin \angle ODM} = \frac{a}{\sin \varphi} = \frac{a}{\tan \varphi \cos \varphi} = \frac{a \sqrt{1 + \tan^2 \varphi}}{\tan \varphi} = \frac{a \sqrt{1 + 4 \tan^2 \alpha}}{2 \tan \alpha} \Rightarrow \\
& \Rightarrow \frac{x \sqrt{3}}{6} = \frac{a \sqrt{1 + 4 \tan^2 \alpha}}{2 \tan \alpha} \Leftrightarrow x = \frac{\sqrt{3} a \sqrt{1 + 4 \tan^2 \alpha}}{\tan \alpha}
\end{aligned}
$$

Volume of the pyramid:

$$
\begin{aligned}
& V = \frac{1}{3} S_{\triangle ABC} \cdot SO = \frac{1}{3} \cdot \frac{x^2 \sqrt{3}}{4} \cdot \frac{x \sqrt{3}}{3} \tan \alpha = \frac{1}{12} x^3 \tan \alpha = \\
& = \frac{1}{12} \cdot \frac{3 \sqrt{3} a^3 \sqrt{1 + 4 \tan^2 \alpha}}{\tan^3 \alpha} \cdot \tan \alpha = \frac{\sqrt{3} a^3 \sqrt{(1 + 4 \tan^2 \alpha)^3}}{4 \tan^2 \alpha} \\
& \text{Answer: } \frac{\sqrt{3} a^3 \sqrt{(1 + 4 \tan^2 \alpha)^3}}{4 \tan^2 \alpha}
\end{aligned}
$$