1956 AMC 12 第 49 题

The circle is tangent to one side of triangle $PAB$ and the extensions of the other two, so $O$ is the excenter opposite $P$

The external angle bisectors at $A$ and $B$ form an angle of $90^\circ-\frac12\angle P$

The circle lies opposite $P$ across side $AB,$ so $O$ is the excenter opposite $P.$ Thus $AO$ and $BO$ bisect the exterior angles at $A$ and $B.$ If the interior angles at $A$ and $B$ are $\alpha$ and $\beta,$ then triangle $AOB$ has angles $90^\circ-\frac{\alpha}{2}$ and $90^\circ-\frac{\beta}{2}$ at $A$ and $B.$ Hence $$\begin{aligned} \angle AOB &=180^\circ-\left(90^\circ-\frac\alpha2\right)\\ &\mathrel{\phantom{=}}-\left(90^\circ-\frac\beta2\right)\\ &=\frac{\alpha+\beta}{2} =\frac{180^\circ-40^\circ}{2}\\ &=70^\circ. \end{aligned}$$

Thus, the correct answer is E.