In triangle $BCD$, $\angle C = 90^\circ$, $CD = 3$, and $BD = \sqrt{13}$. What is $\tan B$?
To find $\tan B$, I need to know the ratio of the opposite and adjacent sides of angle $B$. The opposite side is $CD = 3$, and the adjacent side is $BC$. I can use the Pythagorean theorem to find $BC$, since it is a right triangle. $BC^2 + CD^2 = BD^2$. Substituting the given values, I get $BC^2 + 9 = 13$. Solving for $BC$, I get $BC = \sqrt{4} = 2$. Now I have $\tan B = \frac{CD}{BC} = \frac{3}{2}$. 