# 11.3. (7 points)

The base of the pyramid with vertex $P$ is a quadrilateral $A B C D$, where the sum of angles $A$ and $D$ is five times smaller than the sum of angles $B$ and $C$. Find the angle between the planes of the faces $P A B$ and $P C D$, if both are perpendicular to the base.
Answer: $60^{\circ}$.

Solution: Let the planes $A P B$ and $C P D$ intersect along the line $P X$ (point $X$ lies in the plane $A B C$, see the figure).

![](https://cdn.mathpix.com/cropped/2024_05_06_1fc28fbfec7452221237g-2.jpg?height=417&width=620&top_left_y=865&top_left_x=838)

Since $(A P B) \perp(A B C)$ and $(C P D) \perp(A B C)$, then $P X \perp(A B C)$. Therefore, the angle $\angle B X C$ is the linear angle of the dihedral angle with edge $P X$. From the problem statement, it follows that in the quadrilateral $A B C D$, the sum of angles $A$ and $D$ is $60^{\circ}$, hence $\angle B X C = 120^{\circ}$. The obtained angle is obtuse, so the angle between the specified planes is $60^{\circ}$.