jmc

First, from triangle $ABO$, $\angle AOB=180^{\circ }-\angle BAO-\angle ABO$. Note that $AO$ bisects $\angle BAT$ (to see this, draw radii from $O$ to $AB$ and $AT,$ creating two congruent right triangles), so $\angle BAO=\angle BAT/2$. Similarly, $\angle ABO=\angle ABR/2$.

Also, $\angle BAT=180^{\circ }-\angle BAP$, and $\angle ABR=180^{\circ }-\angle ABP$. Hence, $$\begin{array}{rl}\angle AOB& =180^{\circ }-\angle BAO-\angle ABO\\ & =180^{\circ }-\frac{\angle BAT}{2}-\frac{\angle ABR}{2}\\ & =180^{\circ }-\frac{180^{\circ }-\angle BAP}{2}-\frac{180^{\circ }-\angle ABP}{2}\\ & =\frac{\angle BAP+\angle ABP}{2}.\end{array}$$

Finally, from triangle $ABP$, $\angle BAP+\angle ABP=180^{\circ }-\angle APB=180^{\circ }-40^{\circ }=140^{\circ }$, so $$\angle AOB=\frac{\angle BAP+\angle ABP}{2}=\frac{140^{\circ }}{2}=\boxed{70^{\circ }}.$$