8,9

The lateral edge of a regular triangular pyramid forms an angle of $60^{\circ}$ with the base plane. Find the angle between the lateral faces of the pyramid.
Let $ABCD$ be a regular triangular pyramid with vertex $P$, and $F$ be the base of the perpendicular dropped from the midpoint $L$ of edge $BC$ to line $AP$. The line $AL$ is the orthogonal projection of the inclined line $AP$ onto the plane of the base of the pyramid. According to the theorem of three perpendiculars, $AP \perp BC$, so the line $AP$ is perpendicular to two intersecting lines in the plane $BFC$. Therefore, the line $AP$ is perpendicular to the plane $BFC$. Hence, $FL$ is the common perpendicular to the skew lines $AP$ and $BC$. From the right triangle $AFL$, we find that

$$
FL = AL \sin \angle FAL = AL \sin \angle DAM = \frac{a \sqrt{3}}{2} \cdot \sin 60^{\circ} = \frac{3a}{4}
$$

Since the line $AP$ is perpendicular to the plane of triangle $BFC$, the angle between the lateral faces $PAB$ and $PAC$ is the angle $BFC$. Let $\angle BFC = \gamma$. In the isosceles triangle $BFC$, the median $FL$ is also the altitude and the angle bisector, so in the right triangle $BLF$, the angle $BFL$ is $\frac{\gamma}{2}$,

$$
\tan \frac{\gamma}{2} = \frac{BL}{FL} = \frac{\frac{a}{2}}{\frac{3a}{4}} = \frac{2}{3}
$$

Using a known trigonometric formula, we can find the cosine of the angle between the lateral faces of the given pyramid:

$$
\cos \gamma = \frac{1 - \tan^2 \frac{\gamma}{2}}{1 + \tan^2 \frac{\gamma}{2}} = \frac{1 - \frac{4}{9}}{1 + \frac{4}{9}} = \frac{5}{13}
$$

## Answer

$2 \arctan \frac{2}{3} = \arccos \frac{5}{13}$