\left.\begin{array}{l}{[\text { Linear dependence of vectors }]} \\ {[\underline{\text { Angles between lines and planes }}]}\end{array}\right]

In a regular quadrilateral pyramid $A B C D P$, the angle between the lateral edge $P A$ and the plane of the base $A B C D$ is equal to the angle between the edge $P A$ and the plane $P B C$. Find this angle.
Let $E$ be the orthogonal projection of point $A$ onto the plane of face $B P C$. Then $\angle A P E$ is the desired angle. Denote it by $\alpha$. Since line $A D$ is parallel to the plane of face $B P C$, all its points are equidistant from this plane. Consider the section of the given pyramid by a plane passing through point $P$, the midpoint $L$ of edge $A D$, and the center $M$ of the base $A B C D$. This plane is perpendicular to lines $A D$ and $B C$, since each of these parallel lines is perpendicular to $P M$ (the height of the pyramid) and $P L$. Let this plane intersect edge $B C$ at point $K$. Then $P K$ is the height and median of the isosceles triangle $B P C$. The height $L F$ of the isosceles triangle $L P K$ is perpendicular to the plane $B P C$, since it is perpendicular to the intersecting lines $P K$ and $B C$ of this plane. Therefore, $L F = A E$. Right triangles $A E P$ and $P M A$ are equal by hypotenuse ($A P$ is common) and acute angle ($\angle A P E = \angle P A M$ by the problem statement), so

$$
P M = A E = L F.
$$

In the isosceles triangle $L P K$, the heights dropped to the base and the lateral side are equal, hence this triangle is equilateral. Let $A D = a$. Then

$$
A M = \frac{a \sqrt{2}}{2}, \quad P M = L M \operatorname{tg} \angle P L M = \frac{a}{2} \cdot \operatorname{tg} 60^{\circ} = \frac{a \sqrt{3}}{2}
$$

$$
\operatorname{tg} \alpha = \operatorname{tg} \angle P A M = \frac{P M}{A M} = \frac{\frac{a \sqrt{3}}{2}}{\frac{a \sqrt{2}}{2}} = \sqrt{\frac{3}{2}}
$$

$$
\cos \alpha = \sqrt{\frac{2}{5}}
$$

## Answer

$\operatorname{arctg} \sqrt{\frac{3}{2}} = \arccos \sqrt{\frac{2}{5}}$