[ [ correct pyramid ]

The dihedral angle at the base of a regular $n$-sided pyramid is $\beta$. Find the dihedral angle between adjacent lateral faces.
Let the side of the base $A_1 A_2 A_n$ of the regular pyramid $P A_1 A_2 A_n$ with vertex $P$ be $a$, the height $PO$ of the pyramid be $h$, the area of a lateral face be $S$, and the angle between adjacent lateral faces be $\gamma$. Drop a perpendicular $A_1 M$ from vertex $A_1$ to the line $O A_2$. Then $A_1 M$ is perpendicular to the plane $A_2 O P$. Since $A_2 O P$ is the bisector plane of the dihedral angle between the lateral faces $A_1 P A_2$ and $A_2 P A_3$, the angle between the planes $A_1 P A_2$ and $A_2 O P$ is $\frac{\gamma}{2}$. If $S'$ is the area of the orthogonal projection of the lateral face $A_1 P A_2$ onto the plane $A_2 O P$, then $S' = S \cos \frac{\gamma}{2}$. On the other hand, if $L$ is the midpoint of $A_1 A_2$, then

\[
\begin{gathered}
P L = \frac{25}{A_1 A_2} = \frac{2 \pi}{a}, \quad O P = P L \sin \beta = \frac{25 d n / P}{\Delta}, \\
\angle A_1 A_2 M = \frac{1}{2} \cdot \frac{18 \operatorname{sp}[(n-2)]}{n} = 90^\circ - \frac{18 \pi p}{n},
\end{gathered}
\]

\[
\angle M A_1 A_2 = 90^\circ - \angle A_1 A_2 M = 90^\circ - \left(90^\circ - \frac{18 \pi r}{n}\right) = \frac{18 r^r}{n},
\]

\[
A_2 M = A_1 A_2 \sin \angle M A_1 A_2 = a \sin \frac{18 \pi}{n}
\]

Therefore,

\[
S' = S_{\triangle A_2 M P} = \frac{1}{2} A_2 M \cdot O P = \frac{1}{2} a \sin \frac{18 \pi P}{n} \cdot \frac{2 S \Delta \operatorname{dn} \beta}{4} = S \sin \frac{18 \pi T}{n} \sin \beta.
\]

Therefore,

\[
S \cos \frac{\gamma}{2} = S \sin \frac{18 \pi \Gamma}{n} \sin \beta
\]

Thus,

\[
\cos \frac{\gamma}{2} = \sin \beta \sin \frac{18 \pi \Gamma}{n}
\]

## Answer

$2 \arccos \left(\sin \beta \sin \frac{18 \pi}{n}\right)$.