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

The lateral face of a regular quadrilateral pyramid forms an angle of $45^{\circ}$ with the plane of the base. Find the angle between the apothem of the pyramid and the plane of the adjacent face.
Let $A B C D P$ be a given regular quadrilateral pyramid with vertex $P ; A B=B C=C D=A D=a ; M-$ the center of the square $A B C D ; K, L, N$ and $E$ - the midpoints of segments $A B, B C, C D$ and $A D$ respectively. Since $P K \perp A B$ and $M K \perp A B$, the angle $P K M$ is the linear angle of the dihedral angle between the plane of the lateral face $A B P$ and the plane of the base of the pyramid. By the condition $\angle P K M=45^{\circ}$. The apothem $P K$ of the pyramid is found from the right triangle $Р K M:$

$$
P K=\frac{K M}{\cos \angle I^{\prime} K M}=\frac{\frac{a}{2}}{\cos 45^{a}}=\frac{\sqrt{2}}{2} .
$$

Let's find the angle between the apothem $P K$ and the plane of the face $A P D$. Let $H$ be the orthogonal projection of point $K$ onto the plane of the face $A P D, \phi$ - the desired angle. Then $\phi=\angle K P H$ and $\sin \phi=\frac{\kappa H}{I^{\prime} K}$. Since $K-$ is the midpoint of segment $A B$, the distance $К H$ from point $K$ to the plane of the face $A P D$ is half the distance from point $B$ to this plane, and since the line $B C$ is parallel to the plane $A P D$, all its points are equidistant from this plane. Since $E P L$ is an isosceles right triangle with hypotenuse $E L=a$, the distance from point $L$ to the plane of the face $A P D$ is $\frac{a \sqrt{2}}{2}$. Therefore,

$$
K H=\frac{1}{2} L P=\frac{a \sqrt{2}}{4}
$$

Thus,

$$
\sin \phi=\frac{K H}{1 \Gamma K}=\frac{\frac{a \sqrt{2}}{\frac{1}{\alpha}}}{2}=\frac{1}{2}
$$

Therefore, $\phi=30^{\circ}$.

## Answer

$30^{\circ}$.