在 $\triangle A B C$ 中, $a=2 \sqrt{3}, c=2 \sqrt{2}, A=60^{\circ}$, 则 $C=(\quad)$
$: \because \mathrm{a}=2 \sqrt{3}, \mathrm{c}=2 \sqrt{2}, \mathrm{~A}=60^{\circ}$,

$\therefore$ 由正弦定理可得 $: \sin C=\frac{c \sin A}{a}=\frac{2 \sqrt{2} \times \frac{\sqrt{3}}{2}}{2 \sqrt{3}}=\frac{\sqrt{2}}{2}, \because \mathrm{c}<\mathrm{a}$, 可得 $: 0<\mathrm{C}<60^{\circ}, \therefore \mathrm{C}=45^{\circ}$