20.36 In a right-angled triangle $ABC$ inscribed in circle $O$, $AB \stackrel{\mathrm{m}}{=} 120^{\circ}$, $\overparen{BC} \stackrel{\mathrm{m}}{=} 72^{\circ}$. A point $E$ is taken on the minor arc $\overparen{A}$ such that $OE \perp AC$. Then the ratio of $\angle OBE$ to $\angle BAC$ is
(A) $\frac{5}{18}$.
(B) $\frac{2}{9}$.
(C) $\frac{1}{4}$.
(D) $\frac{1}{3}$.
(E) $\frac{4}{9}$.
(14th American High School Mathematics Examination, 1963)
[Solution] Given that the minor arc $\overparen{A C} \stackrel{\mathrm{m}}{=} 360^{\circ}-\left(120^{\circ}+72^{\circ}\right)=$ $168^{\circ}$, and from $O E \perp A C$ we know $A E=\overparen{E C}$.
Then $\angle B O E=72^{\circ}+\frac{168^{\circ}}{2}=156^{\circ}$.
Therefore, in the isosceles $\triangle O B E$,
$$
\angle O B E=\frac{1}{2}\left(180^{\circ}-156^{\circ}\right)=12^{\circ} \text {. }
$$

Also, $\angle B A C=\frac{1}{2} \cdot 72^{\circ}=36^{\circ}$. Thus, the required ratio is $1: 3$. Hence, the answer is $(D)$.