$7 \cdot 25$ right $\triangle A B C$, where $\angle C$ is the right angle. If $\sin A=\frac{2}{3}$, then $\operatorname{tg} B$ is
(A) $\frac{3}{5}$.
(B) $\frac{\sqrt{5}}{3}$.
(C) $\frac{2}{\sqrt{5}}$.
(D) $\frac{\sqrt{5}}{2}$.
(E) $\frac{5}{3}$.
(34th American High School Mathematics Examination, 1983)
[Solution 1] In the figure, since $\sin A=\frac{B C}{A B}=\frac{2}{3}$, we can let
$B C=2 x, A B=3 x$, where $x>0$.
And
$$
\begin{aligned}
A C & =\sqrt{A B^{2}-B C^{2}} \\
& =\sqrt{5} x .
\end{aligned}
$$

Thus, $\operatorname{tg} B=\frac{A C}{B C}=\frac{\sqrt{5}}{2}$.
Therefore, the answer is $(D)$.
[Solution 2] $\operatorname{tg} B=\frac{\sin B}{\cos B}=\frac{\cos A}{\sin A}=\frac{\sqrt{1-\sin ^{2} A}}{\sin A}=\frac{\sqrt{5}}{2}$. Therefore, the answer is $(D)$.