$7 \cdot 2$ If $\sin x=3 \cos x$, then $\sin x \cos x$ equals
(A) $\frac{1}{6}$.
(B) $\frac{1}{5}$.
(C) $\frac{2}{9}$.
(D) $\frac{1}{4}$.
(E) $\frac{3}{10}$.
(39th American High School Mathematics Examination, 1988)
[Solution] Given $\operatorname{tg} x=3$.
$$
\therefore \quad \sin x \cos x=\frac{1}{2} \sin 2 x=\frac{1}{2} \cdot \frac{2 \operatorname{tg} x}{1+\operatorname{tg}^{2} x}=\frac{3}{10} \text {. }
$$

Therefore, the answer is $(E)$.