AHSME 1983 Problem 5

Triangle $ABC$ has a right angle at $C$. If $\sin A = \frac{2}{3}$, then $\tan B$ is $\textbf{(A)}\ \frac{3}{5} \textbf{(B)}\ \frac{\sqrt 5}{3} \textbf{(C)}\ \frac{2}{\sqrt 5} \textbf{(D)}\ \frac{\sqrt{5}}{2} \textbf{(E)}\ \frac{5}{3}$