The lateral edge of a regular triangular 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 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 $APC$. The apothem $PK$ can be found from the right triangle $PMK$:

$$
PM = MC = a \sqrt{PM^2 + MK^2} = \sqrt{\left(\frac{a \sqrt{3}}{3}\right)^2 + \left(\frac{a \sqrt{3}}{6}\right)^2} = \frac{a \sqrt{15}}{6}
$$

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 $BG$ is perpendicular to the plane of the face $APC$, so $KD = \frac{1}{2} BG$. From the right triangle $BGN$, we find that

$$
BG = BN \sin \angle BNP = BN \sin \beta = BN \sin \angle PKM =
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_b09eb9152e3973547418g-06.jpg?height=123&width=714&top_left_y=0&top_left_x=672)

Therefore,

![](https://cdn.mathpix.com/cropped/2024_05_06_b09eb9152e3973547418g-06.jpg?height=146&width=411&top_left_y=193&top_left_x=818)

## Answer

$\arcsin \frac{3}{5}$.