12.270. Find the dihedral angle at the vertex of a regular quadrilateral pyramid if it is equal to the angle between a lateral edge and the plane of the pyramid's base.
Solution.

Let $S O$ be the height of the regular pyramid $S A B C D$ (Fig. 12.134), $\angle S D O = \angle C S D, \angle S D O = \angle C S D = \alpha$.

Then $\angle S D C = 90^{\circ} - \frac{\alpha}{2}$:

$S O$ is the oblique to the plane of the base of the pyramid, $O D$ is its projection onto this plane, $C D$ is a line lying in this plane.

We get

$\cos \angle S D C = \cos \angle S D O \cdot \cos \angle O D C \Rightarrow$

$\Rightarrow \cos \left(90^{\circ} - \frac{\alpha}{2}\right) = \cos \alpha \cos 45^{\circ} \Rightarrow \sqrt{2} \sin \frac{\alpha}{2} \cos \alpha$.

Since the angle $\alpha$ is acute, then $\cos \alpha > 0, \sin \frac{\alpha}{2} > 0$ and

$$
\begin{aligned}
& \cos ^{2} \alpha = 2 \sin ^{2} \frac{\alpha}{2} ; \cos ^{2} \alpha = 1 - \cos \alpha ; \cos ^{2} \alpha + \cos \alpha - 1 = 0 \\
& \cos \alpha = \frac{\sqrt{5} - 1}{2} ; \alpha = \arccos \frac{\sqrt{5} - 1}{2}
\end{aligned}
$$

Answer: $\arccos \frac{\sqrt{5} - 1}{2}$.