$20.5 \triangle P A B$ is formed by three tangents to circle $O$, and $\angle A P B=40^{\circ}$, then $\angle A O B$ is
(A) $45^{\circ}$.
(B) $50^{\circ}$.
(C) $55^{\circ}$.
(D) $60^{\circ}$.
(E) $70^{\circ}$.
(7th American High School Mathematics Examination, 1956)
$$
\begin{array}{l}
\text { [Solution] It is easy to know that } \angle P A B + \angle P B A = 180^{\circ} \\
- \angle B P A = 140^{\circ}, \\
\text { Also, } \angle T A S = 180^{\circ} - \angle P A B, \angle R S S = 180^{\circ} - \angle P B A, \\
\text { Therefore, } \angle T A S + \angle R S S = 360^{\circ} - 140^{\circ} = 220^{\circ} .
\end{array}
$$

Since $O A$ and $O B$ bisect $\angle T A S$ and $\angle R S S$ respectively, we have
$$
\begin{array}{l}
\angle O A S + \angle O S S = \frac{1}{2} \cdot 220^{\circ} = 110^{\circ}. \\
\therefore \quad \angle A O B = 180^{\circ} - 110^{\circ} = 70^{\circ}.
\end{array}
$$

Therefore, the answer is $(E)$.