For a certain hyperbola
\[\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1,\]where $a > b,$ the angle between the asymptotes is $60^\circ.$  Find $\frac{a}{b}.$
I know that the asymptotes of a hyperbola are the lines $y = \pm \frac{b}{a} x,$ so the angle between them is $2 \tan^{-1} \frac{b}{a}.$ If this angle is $60^\circ,$ then I have \[2 \tan^{-1} \frac{b}{a} = 60^\circ,\] which I can simplify by dividing both sides by 2 and applying the inverse tangent function: \[\tan^{-1} \frac{b}{a} = 30^\circ.\] Now I can use the fact that $\tan 30^\circ = \frac{1}{\sqrt{3}}$ to get \[\frac{b}{a} = \frac{1}{\sqrt{3}}.\] Solving for $\frac{a}{b},$ I get \[\frac{a}{b} = \sqrt{3}.\] 