$7 \cdot 50$ In isosceles $\triangle A B C$, the altitude on one of the equal sides is $\sqrt{3}$, and the angle between this altitude and the base is $60^{\circ}$. The area of $\triangle A B C$ is
(A) $\sqrt{3}$.
(B) $2 \sqrt{3}$.
(C) 2 .
(D) $\frac{\sqrt{3}}{2}$.
(E) None of the above.
(China Beijing Junior High School Mathematics Competition, 1983)
[Solution] As shown in the figure, given that the angle between the height $BD$ and the base $BC$ is $60^{\circ}$, we know that the base angle $\angle C=30^{\circ}$, and $D$ is on the extension of $CA$, with $BC=2BD=2\sqrt{3}$, and $AB=2$.
Therefore, $S_{\triangle ABC}=\frac{1}{2} AB \cdot BC \sin \angle ABC$
$$
\begin{array}{l}
=\frac{1}{2} \cdot 2 \cdot 2 \sqrt{3} \cdot \sin 30^{\circ} \\
=\sqrt{3} .
\end{array}
$$

Hence, the correct choice is $(A)$.