12.337. The base of the pyramid is an isosceles triangle, with an area of $S$, and the angle between the lateral sides is $\alpha$. All lateral edges of the pyramid form the same angle with the base plane. Find this angle if the volume of the pyramid is $V$.
## Solution.

Let $DO$ be the height of the pyramid $DABC$, with $AB = AC$ and $\angle BAC = \alpha$ (Fig. 12.198). Since $\angle DAO = \angle DBO = \angle DCO$, $O$ is the center of the circumcircle of $\triangle ABC$.

If the radius of this circle is $R$, then $AB = 2R \sin \angle ACB = 2R \sin \left(90^\circ - \frac{\alpha}{2}\right) = 2R \cos \frac{\alpha}{2}$.

$$
S = \frac{1}{2} AB^2 \sin \angle BAC = \frac{1}{2} \cdot 4R^2 \cos^2 \frac{\alpha}{2} \sin \alpha; \quad R = \frac{\sqrt{2S}}{2 \cos \frac{\alpha}{2} \sqrt{\sin \alpha}}
$$

Let $H$ be the height of the pyramid. Since $V = \frac{1}{3} S H$, then $H = \frac{3V}{S}$.

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

$\operatorname{tg} \angle EAO = \frac{DO}{OA} = \frac{H}{R} = \frac{3V}{S} \cdot \frac{2 \cos \frac{\alpha}{2} \sqrt{\sin \alpha}}{\sqrt{2S}} = \frac{3V \cos \frac{\alpha}{2}}{S} \sqrt{\frac{2 \sin \alpha}{S}}$.

Answer: $\operatorname{arctg} \frac{3V \cos \frac{\alpha}{2}}{S} \sqrt{\frac{2 \sin \alpha}{S}}$.

![](https://cdn.mathpix.com/cropped/2024_05_21_3edb7be591bff54ce350g-1097.jpg?height=720&width=1167&top_left_y=76&top_left_x=93)

Fig. 12.200