23.37 In a regular $n$-sided pyramid, the range of the dihedral angle formed by two adjacent lateral faces is
(A) $\left(\frac{n-2}{n} \pi, \pi\right)$.
(B) $\left(\frac{n-1}{n} \pi, \pi\right)$.
(C) $\left(0, \frac{\pi}{2}\right)$.
(D) $\left(\frac{n-2}{n} \pi, \frac{n-1}{n} \pi\right)$.
(China High School Mathematics League, 1994)
[Solution] As shown in the figure, let the regular $n$-sided pyramid be $S-A_{1} A_{2} \cdots A_{n}$. Without loss of generality, assume the base regular $n$-sided polygon is fixed, and the vertex $S$ moves, with the dihedral angle between adjacent sides being $\angle A_{2} H A_{n}$.

When $S$ moves downward and approaches the extreme position (i.e., to the center of the base regular $n$-sided polygon), $\angle A_{2} H A_{n}$ tends to a straight angle;

When $S$ moves upward and approaches infinity, the regular $n$-sided pyramid approaches a regular $n$-sided prism, and $\angle A_{2} H A_{n}$ tends to
$$
\angle A_{2} A_{1} A_{n}=\frac{(n-2) \pi}{n}
$$

Therefore,
$$
\frac{(n-2)}{n} \pi<\angle A_{2} H A_{n}<\pi \text {. }
$$

Thus, the answer is $(A)$.
Note: This problem can also be solved by using the method of special values.