4. Given three planes $\alpha, \beta, \gamma$, the angle between each pair of planes is $\theta$, and $\alpha \cap \beta=a, \beta \cap \gamma=$ $b, \gamma \cap \alpha=c$. If proposition 甲: $\theta>\frac{\pi}{3}$; proposition 乙: $a, b, c$ intersect at one point. Then
A. 甲 is a sufficient condition for 乙 but not necessary
B. 甲 is a necessary condition for 乙 but not sufficient
C. 甲 is a necessary and sufficient condition for 乙
D. None of A, B, C is correct
Given that three planes intersect pairwise, the three lines of intersection either intersect at one point or are parallel. When $a / / b / / c$, $\theta=\frac{\pi}{3} \Rightarrow \theta \neq \frac{\pi}{3}$, then $a, b, c$ intersect at one point. When $a, b, c$ intersect at one point, as shown in the figure, in the regular triangular pyramid $S-A B C$, $\alpha=$ side face $S A C, \beta=$ side face $S A B, \gamma=$ side face $S B C$, thus $a=S A, b=S B, c=S C, \angle B E C=\varphi$. Let $A B=2, S A=m$, then from $\triangle S A D \sim \triangle B A E$ $\Rightarrow \frac{S A}{A B}=\frac{S D}{B E} \Rightarrow B E=\frac{A B \cdot S D}{S A}=\frac{2 \sqrt{m^{2}-1}}{m}>1 \Rightarrow m^{2}>\frac{4}{3}$. Therefore, in $\triangle B E C$, $\cos \varphi=\frac{2 \cdot \frac{4\left(m^{2}-1\right)}{m^{2}}-4}{2 \cdot \frac{4\left(m^{2}-1\right)}{m^{2}}}=\frac{m^{2}-2}{2\left(m^{2}-1\right)}$ $=\frac{1}{2}-\frac{1}{2\left(m^{2}-1\right)} \in\left(-1, \frac{1}{2}\right) \Rightarrow \varphi>\frac{\pi}{3}$. Note that when $\varphi>\frac{\pi}{2}$, $\theta=\pi-\varphi$, meaning that condition A is sufficient but not necessary for condition B, so the answer is $A$.