在 $\triangle A B C$ 中, $\tan C=\frac{\sqrt{3}}{3}, \cos A=\frac{\sqrt{3}}{2}$, 则 $\angle B=(\quad)$
$\because \tan C=\frac{\sqrt{3}}{3}, \cos A=\frac{\sqrt{3}}{2}$,

$\therefore \angle C=30^{\circ}, \angle A=30^{\circ}$,

$\therefore \angle B=120^{\circ}$.

故选 $C$.