12.083. Find the angle between the apothems of two adjacent lateral faces of a regular $n$-sided pyramid if the plane angle at its vertex is $\alpha$.
## Solution.

Let $S_{A B C \ldots K}$ be a regular $n$-sided pyramid, $\angle A S B=\alpha$, $S M$ and $S N$ be the apothems of adjacent lateral faces; we need to find $\angle M S N$ (Fig. 12.85). Draw $S O \perp(A B C)$ and connect $O$ with points $A, B, M$, and

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-776.jpg?height=480&width=406&top_left_y=98&top_left_x=797)

Fig. 12.85 $N$. We have $\angle A O B=\frac{2 \pi}{n}, \angle A B C=\frac{\pi(n-2)}{n}$. Let $A B=a$. From $\triangle M S B$, we find $S M=\frac{a}{2} \operatorname{ctg} \frac{\alpha}{2}$. Next, from $\triangle M N B$ using the cosine rule, we get $M N^{2}=\frac{a^{2}}{2}-\frac{a^{2}}{2} \cos \frac{\pi(n-2)}{n}=\frac{a^{2}}{2}\left(1-\cos \left(\pi-\frac{2 \pi}{n}\right)\right)=$ $=\frac{a^{2}}{2}\left(1+\cos \frac{2 \pi}{n}\right)=a^{2} \cos ^{2} \frac{\pi}{n}$, and from $\triangle M S N$ we obtain $M N^{2}=2 S M^{2}-$ $-2 S M^{2} \cos \angle M S N=2 \frac{a^{2}}{4} \operatorname{ctg}^{2} \frac{\alpha}{2}(1-\cos \angle M S N)=a^{2} \sin ^{2} \frac{\angle M S N}{2} \operatorname{ctg}^{2} \frac{\alpha}{2}$. Therefore, $a^{2} \cos ^{2} \frac{\pi}{n}=a^{2} \sin ^{2} \frac{\angle M S N}{2} \operatorname{ctg}^{2} \frac{\alpha}{2} ; \sin \frac{\angle M S N}{2}=\cos \frac{\pi}{n} \operatorname{tg} \frac{\alpha}{2}$. We get $\angle M S N=2 \arcsin \left(\cos \frac{\pi}{n} \operatorname{tg} \frac{\alpha}{2}\right)$.

Answer: $2 \arcsin \left(\cos \frac{\pi}{n} \operatorname{tg} \frac{\alpha}{2}\right)$.