7. 在 $\triangle A B C$ 中, $A C=\sqrt{6}, B C=2, B=60^{\circ}$, 则 $C=$
由正弦定理得 $\frac{2}{\sin A}=\frac{\sqrt{6}}{\sin 60^{\circ}}, \therefore \sin A=\frac{\sqrt{2}}{2}$.

$\because B C=2<A C=\sqrt{6}, \therefore A$ 为锐角. $\therefore A=45^{\circ}$.

$\therefore C=75^{\circ}$.