在 $\triangle A B C$ 中, $a=2, b=\sqrt{3}, c=1$, 则最小角为 $(\quad)$
$[\because a>b>c, \therefore C$ 最小.

$\because \cos C=\frac{a^{2}+b^{2}-c^{2}}{2 a b}=\frac{2^{2}+(\sqrt{3})^{2}-1^{2}}{2 \times 2 \times \sqrt{3}}=\frac{\sqrt{3}}{2}$,

又 $\because 0<C<\pi, \quad \therefore C=\frac{\pi}{6}$.