7.42 In $\triangle A B C$, $\angle A B C=120^{\circ}, \quad A B=3, \quad B C=4$. If a perpendicular line from $A$ to $A B$ intersects with a perpendicular line from $C$ to $B C$ at $D$, then $C D$ equals
(A) 3 .
(B) $\frac{8}{\sqrt{3}}$.
(C) 5 .
(D) $\frac{11}{2}$.
(E) $\frac{10}{\sqrt{3}}$.
(43rd American High School Mathematics Examination, 1992)
[Solution] By the cosine theorem in $\triangle A B C$, we have
$$
A C^{2}=3^{2}+4^{2}-2 \cdot 3 \cdot 4 \cdot \cos 120^{\circ}=37 \text {. }
$$

Thus, $A C=\sqrt{37}$.
If $R$ is the radius of the circumcircle of $\triangle A B C$, then $A C=2 R \sin 120^{\circ}$, i.e., $2 R=\frac{A C}{\sin 120^{\circ}}=\sqrt{37} \cdot \frac{2}{\sqrt{3}}=\frac{2 \sqrt{37}}{\sqrt{3}}$.
Given $\angle D A B=\angle B C D=90^{\circ}$, thus
$B D$ is the diameter of the circumcircle of the triangle, i.e.,
$$
\begin{array}{c}
B D=2 R . \\
\therefore \quad C D^{2}=B D^{2}-B C^{2}=\frac{4 \cdot 37}{3}-4^{2}=\frac{100}{3} . \\
\text { Hence } C D=\frac{10}{\sqrt{3}} .
\end{array}
$$

Therefore, the answer is $(E)$.