12.226. At the base of a triangular pyramid lies an isosceles triangle, which has an area of $S$ and the angle at the vertex is $\alpha$. Find the volume of the pyramid if the angle between each lateral edge and the height of the pyramid is $\beta$.
## Solution.

Let $DO$ be the height of the pyramid $DABC$ (Fig. 12.93), $\angle ADO = \angle BDO = \angle CDO = \beta, AB = BC, \angle ABC = \alpha, S_{\triangle ABC} = S$.

$\triangle AOM = \triangle BOM = \triangle COM \Rightarrow OA = OB = OC, O$ is the center of the circumcircle of $\triangle ABC, OA$ is its radius.

If $OA = R$, then $AB = BC = 2R \sin \angle BAC = 2R \sin \left(90^\circ - \frac{\alpha}{2}\right) = 2R \cos \frac{\alpha}{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0975.jpg?height=601&width=593&top_left_y=77&top_left_x=93)

Fig. 12.93

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-0975.jpg?height=613&width=610&top_left_y=67&top_left_x=654)

Fig. 12.94

\[
\begin{aligned}
& S_{\triangle ABC} = \frac{1}{2} AB^2 \sin \angle ABC \Rightarrow \frac{1}{2} \cdot 4R^2 \cos^2 \frac{\alpha}{2} \sin \alpha = S \Rightarrow \\
& \Rightarrow R = \sqrt{\frac{S}{2 \cos^2 \frac{\alpha}{2} \sin \alpha}} = \frac{\sqrt{2S \sin \alpha}}{2 \cos \frac{\alpha}{2} \sin \alpha}
\end{aligned}
\]

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

\[
DO = OA \operatorname{ctg} \angle ADO = \frac{\sqrt{2S \sin \alpha} \operatorname{ctg} \beta}{2 \cos \frac{\alpha}{2} \sin \alpha}
\]

The volume of the pyramid $V = \frac{1}{3} S_{\triangle ABC} \cdot DO = \frac{S \operatorname{ctg} \beta \sqrt{2S \sin \alpha}}{6 \cos \frac{\alpha}{2} \sin \alpha}$.

Answer: $\frac{S \operatorname{ctg} \beta \sqrt{2S \sin \alpha}}{6 \cos \frac{\alpha}{2} \sin \alpha}$.