Triangle $ABC$ has a right angle at $C$. If $\sin A = \frac{2}{3}$, then $\tan B$ is
$\textbf{(A)}\ \frac{3}{5}\qquad \textbf{(B)}\ \frac{\sqrt 5}{3}\qquad \textbf{(C)}\ \frac{2}{\sqrt 5}\qquad \textbf{(D)}\ \frac{\sqrt{5}}{2}\qquad \textbf{(E)}\ \frac{5}{3}$
Since $\sin$ can be seen as "opposite over hypotenuse" in a right triangle, we can label the diagram as shown.

By the Pythagorean Theorem, we have:
\[y^2+(2x)^2=(3x)^2\]
\[y=\sqrt{9x^2-4x^2}\]
\[y=\sqrt{5x^2}\]
\[y=x\sqrt{5}\]
so $\tan{B} = \frac{x \sqrt{5}}{2x} = \frac{\sqrt{5}}{2}$, which is choice $\boxed{\textbf{(D)}}$.