$13 \cdot 54$ In $\triangle A B C$, $B C=\sqrt{3}, A C=\sqrt{3}, A B>3$, let $x$ be the maximum number such that the degree of $\angle C$ can exceed, then $x$ equals
(A) 150 .
(B) 120 .
(C) 105 .
(D) 90 .
(E) 60 .
(14th American High School Mathematics Examination, 1963)
[Solution] By the cosine rule $c^{2}=a^{2}+b^{2}-2 a b \cos C$, or
$$
\begin{aligned}
\cos C & =\frac{a^{2}+b^{2}-c^{2}}{2 a b} \\
& =\frac{3+3-c^{2}}{2 \cdot \sqrt{3} \cdot \sqrt{3}}=\frac{6-c^{2}}{6},
\end{aligned}
$$

here $c^{2}>9$, so $\cos C < -\frac{1}{2}$. Hence, $C = 120^{\circ}$. Therefore, the answer is $(B)$.