$AB$ is the diameter of a circle centered at $O$. $C$ is a point on the circle such that angle $BOC$ is $60^\circ$. If the diameter of the circle is $5$ inches, the length of chord $AC$, expressed in inches, is:
$\text{(A)} \ 3 \qquad \text{(B)} \ \frac {5\sqrt {2}}{2} \qquad \text{(C)} \frac {5\sqrt3}{2} \ \qquad \text{(D)} \ 3\sqrt3 \qquad \text{(E)} \ \text{none of these}$
We see that $\angle A$ is half the measure of $\angle BOC$, so $\angle A = 30^{\circ}$. That makes $ABC$ a $30-60-90$ triangle and sidelength $\overline{AC}$ equal to $\frac{5\sqrt{3}}{2}$.
$\fbox{C}$