Acute-angled $\triangle ABC$ is inscribed in a circle with center at $O$; $\stackrel \frown {AB} = 120^\circ$ and $\stackrel \frown {BC} = 72^\circ$.
A point $E$ is taken in minor arc $AC$ such that $OE$ is perpendicular to $AC$. Then the ratio of the magnitudes of $\angle OBE$ and $\angle BAC$ is:
$\textbf{(A)}\ \frac{5}{18}\qquad \textbf{(B)}\ \frac{2}{9}\qquad \textbf{(C)}\ \frac{1}{4}\qquad \textbf{(D)}\ \frac{1}{3}\qquad \textbf{(E)}\ \frac{4}{9}$
Because $\stackrel \frown {AB} = 120^\circ$ and $\stackrel \frown {BC} = 72^\circ$, $\stackrel \frown {AC} = 168^\circ$.  Also, $OA = OC$ and $OE \perp AC$, so $\angle AOE = \angle COE = 84^\circ$.  Since $\angle BOC = 72^\circ$, $\angle BOE = 156^\circ$.  Finally, $\triangle BOE$ is an [isosceles triangle](https://artofproblemsolving.com/wiki/index.php/Isosceles_triangle), so $\angle OBE = 12^\circ$.  Because $\angle BAC = \frac{1}{2} \cdot 72 = 36^\circ$, the ratio of the magnitudes of $\angle OBE$ and $\angle BAC$ is $\frac{12}{36} = \frac{1}{3}$, which is answer choice $\boxed{\textbf{(D)}}$.