jmc

Because $△BCD$ is a right triangle, we know that $\tan B=\frac{CD}{BC}$.

By the Pythagorean Theorem, $BC=\sqrt{B{D}^2-C{D}^2}=\sqrt{13-9}=\sqrt{4}=2$.

Then $\tan B=\frac{CD}{BC}=\boxed{\frac{3}{2}}$.