3-3. A pyramid, all lateral edges of which are inclined to the base plane at an angle $\varphi$, has an isosceles triangle with an angle $\alpha$ between the equal sides as its base. Determine the dihedral angle at the edge connecting the vertex of the pyramid with the vertex of the angle $\alpha$.
Solution 3. Let $ABC$ be the base of the given pyramid ($\angle BAC = \alpha$), and $S$ be the vertex of the pyramid. The lateral edges are inclined at equal angles, so they are equal. From the congruence of triangles $ASB$ and $ASC$, it follows that the feet of the perpendiculars dropped from points $B$ and $C$ to the line $AS$ coincide. Let $D$ be the foot of these two perpendiculars, and $E$ be the midpoint of segment $BC$. Then the angle $\theta = \angle CDB$ is the desired angle. Clearly, $DE = AE \sin \varphi$ and $EC = AE \tan \frac{\alpha}{2}$. Therefore, $\tan \frac{\theta}{2} = \frac{EC}{DE} = \frac{\tan \frac{\alpha}{2}}{\sin \varphi}$.

## 2nd Variant