In the following diagram, $\overline{AO}= 2$, $\overline{BO} = 5$, and $\angle{AOB} = 60^\circ$. Point $C$ is selected on $\overline{BO}$ randomly. Find the probability that $\triangle{AOC}$ is an acute triangle.
Let's start with the boundary situation: when $\triangle{AOC}$ is a right triangle.
Draw line $\overline{AC_1}\perp \overline{OB}$ and line $\overline{AC_2} \perp \overline{AO}$ where both $C_1$ and $C_2$ are points on $OB$. Based on the $30^{\circ} - 60^{\circ} - 90^{\circ}$ right triangle property, $\overline{OC_1}=1$ and $\overline{OC_2}=4$.
Clearly if point $C$ falls to the left of $C_1$, $\angle{ACO}$ will be an obtuse angle. If $C$ falls to the right of $C_2$, $\angle{OAC}$ will be an obtuse angle. If and only if $C$ falls between $C_1$ and $C_2$ will $\triangle{AOC}$ be an acute triangle.
Therefore the answer is $\frac{\overline{C_1 C_2}}{\overline{OB}} = \frac{3}{5}$.