1. Let the plane angle of the dihedral angle formed by two adjacent lateral faces of a regular tetrahedron be $\theta$. Then the range of $\theta$ is ( ).
(A) $(0, \pi)$
(B) $\left(\frac{\pi}{6}, \pi\right)$
(C) $\left(\frac{\pi}{3}, \frac{2 \pi}{3}\right)$
(D) $\left(\frac{\pi}{3}, \pi\right)$
- 1.D.

As shown in Figure 4, in the regular tetrahedron $V-$ $A B C$, let $A B=B C=C A=$ $a$, and the base angle of the lateral triangular faces be $\alpha$ $\left(0<\alpha<\frac{\pi}{2}\right)$. A plane perpendicular to $V C$ through $A B$ intersects $V C$ at point $D$. Thus, $A D$ and $B D$ are the altitudes on the legs, and $\angle A D B$ is the plane angle of the dihedral angle formed by the adjacent lateral faces. Therefore,
$$
A D=B D=a \sin \alpha, \angle A D B=\theta \text {. }
$$

In the isosceles $\triangle A B D$, by the cosine rule we have
$$
\begin{array}{l}
\cos \theta=\frac{A D^{2}+B D^{2}-A B^{2}}{2 A D \cdot B D}=\frac{2 a^{2} \sin ^{2} \alpha-a^{2}}{2 a^{2} \sin \alpha} \\
=1-\frac{1}{2 \sin ^{2} \alpha}<\frac{1}{2} . \\
\text { Hence } \frac{\pi}{3}<\theta<\pi .
\end{array}
$$