7.31 If the area of $\triangle A B C$ is 64 square inches, and the geometric mean (mean proportional) of sides $A B$ and $A C$ is 12 inches, then $\sin A$ equals
(A) $\frac{\sqrt{3}}{2}$.
(B) $\frac{3}{5}$.
(C) $\frac{4}{5}$.
(D) $\frac{8}{9}$.
(E) $\frac{15}{17}$.
(23rd American High School Mathematics Examination, 1972)
[Solution] As shown in the figure, $S_{\triangle A B C}=64=\frac{1}{2} A B \cdot A C \cdot \sin A$, let $A B \cdot A C=144$,
thus $\sin A=\frac{2 \cdot 64}{144}=\frac{8}{9}$. Therefore, the answer is $(D)$.