12.277. One lateral edge of a triangular pyramid is perpendicular to the base plane and equals $l$, the other two form an angle $\alpha$ between them, and with the base plane - the same angle $\beta$. Find the volume of the pyramid.
## Solution.

Let the lateral edge $DA = l$ of the pyramid $DABC$ (Fig. 12.140), perpendicular to the plane of the base, be its height, $AB$ and $AC$ - the projections of the lateral edges $DB$ and $DC$ of the pyramid onto the plane of the base. Then $\angle ABD = \angle ACD = \beta, \angle BDC = \alpha$.
$\triangle DAB = \triangle DAC$ - by the leg and acute angle. Thus, $AB = AC$, $DB = DC$.

If $E$ is the midpoint of $BC$, then $AE \perp BC, DE \perp DC, \angle BDE = \frac{\alpha}{2}$.

In $\triangle DAB (\angle DAB = 90^\circ): AB = l \operatorname{ctg} \beta, DB = \frac{l}{\sin \beta}$.

In $\triangle DEB (\angle DEB = 90^\circ): BE = DB \sin \angle BDE = \frac{l \sin \frac{\alpha}{2}}{\sin \beta}$. In $\triangle AEB (\angle AEB = 90^\circ)$:

\[
\begin{aligned}
& AE = \sqrt{AB^2 - BE^2} = \sqrt{l^2 \operatorname{ctg}^2 \beta - \frac{l^2 \sin^2 \frac{\alpha}{2}}{\sin \beta}} = \frac{l}{\sin \beta} \sqrt{\cos^2 \beta - \sin^2 \frac{\alpha}{2}} = \\
& = \frac{l}{\sin \beta} \sqrt{\frac{1 + \cos 2\beta}{2} - \frac{1 - \cos \alpha}{2}} = \frac{l}{\sin \beta} \sqrt{\cos \left(\beta + \frac{\alpha}{2}\right) \cos \left(\beta - \frac{\alpha}{2}\right)}.
\end{aligned}
\]

The area of the base of the pyramid

\[
S = \frac{1}{2} BC \cdot AE = BE \cdot AE = \frac{l^2 \sin \frac{\alpha}{2} \sqrt{\cos \left(\frac{\alpha}{2} + \beta\right) \cos \left(\frac{\alpha}{2} - \beta\right)}}{\sin^2 \beta}.
\]

The volume of the prism

\[
\begin{aligned}
& V = \frac{1}{3} S \cdot DA = \frac{l^3 \sin \frac{\alpha}{2} \sqrt{\cos \left(\frac{\alpha}{2} + \beta\right) \cos \left(\frac{\alpha}{2} - \beta\right)}}{3 \sin^2 \beta} \\
& \text{Answer: } \frac{l^3 \sin \frac{\alpha}{2} \sqrt{\cos \left(\frac{\alpha}{2} + \beta\right) \cos \left(\frac{\alpha}{2} - \beta\right)}}{3 \sin^2 \beta}
\end{aligned}
\]