12.049. The plane angle at the vertex of a regular hexagonal pyramid is equal to the angle between a lateral edge and the plane of the base. Find this angle.
## Solution.

Let OABCDEF be a regular hexagonal pyramid, $O O_{1} \perp (A B C)$, $\angle F O E = \angle O F O_{1}$ (Fig. 12.51). Let $O F = l$ and $\angle F O E = \alpha$; then $F E = \sqrt{2 l^{2} - 2 l^{2} \cos \alpha} = 2 l \sin \frac{\alpha}{2}$. Considering that $O_{1} F = F E$, in $\triangle O O_{1} F$ we have $\cos \alpha = \frac{2 l \sin \frac{\alpha}{2}}{l} ; -2 \sin^{2} \frac{\alpha}{2} + 1 = 2 \sin \frac{\alpha}{2} ; 2 \sin^{2} \frac{\alpha}{2} + 2 \sin \frac{\alpha}{2} - 1 = 0$; $\sin \frac{\alpha}{2} = \frac{-1 + \sqrt{3}}{2} ; \frac{\alpha}{2} = \arcsin \frac{\sqrt{3} - 1}{2} ; \alpha = 2 \arcsin \frac{\sqrt{3} - 1}{2}$.

Answer: $2 \arcsin \frac{\sqrt{3} - 1}{2}$.