[ Linear dependence of vectors $]$ [ Angles between lines and planes $]$

The lateral edge of a regular triangular pyramid forms an angle of $60^{\circ}$ with the base plane. Find the angle between the apothem and the plane of the adjacent lateral face.
Let $a$ be the side of the base $ABC$ of a regular triangular pyramid $ABCP$, $\beta$ be the angle between a lateral face and the base plane, $\phi$ be the angle we are looking for between the apothem $PK$ (where $K$ is the midpoint of $AB$) and the plane of the face $APC$, $M$ be the center of the base $ABC$, $N$ be the midpoint of $AC$, and $D$ be the orthogonal projection of point $K$ onto the plane of the face $APC$. The angle $\phi$ we are looking for is the angle $KPD$, and $\sin \phi = \frac{KD}{PK}$. Therefore, to solve the problem, we need to find the apothem of the pyramid and the distance from point $K$ to the plane of the face $APC$. From the right triangles $PAM$ and $PKM$, we find that

$$
\begin{gathered}
P M = A M \operatorname{tg} \angle P A M = \frac{2}{3} \cdot \frac{a \sqrt{3}}{2} \cdot \operatorname{tg} 60^{\circ} = \frac{a \sqrt{3}}{3} \cdot \sqrt{3} = a, \\
P K = \sqrt{P M^{2} + K M^{2}} = \sqrt{a^{2} + \left(\frac{a \sqrt{3}}{6}\right)^{2}} = \frac{a \sqrt{13}}{2 \sqrt{3}}
\end{gathered}
$$

Since $K$ is the midpoint of the slant $AB$ to the plane of the face $APC$, the distance $KD$ is half the distance from point $B$ to this plane. Let $BG$ be the height of the triangle $PBN$. Then the segment $BG$ is perpendicular to the plane of the face $APC$, so $KD = \frac{1}{2} BG$. From the right triangle $BGN$, we find that

$$
B G = B N \sin \angle B N P = B N \sin \beta = B N \sin \angle P K M =
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_8a44229da56811dd86d6g-25.jpg?height=146&width=637&top_left_y=2434&top_left_x=716)

Therefore,

$$
\sin \phi = \frac{K D}{P K} = \frac{\frac{3 a}{2 \sqrt{13}}}{2 \sqrt{13}} = \frac{3 \sqrt{3}}{13}
$$

Problem