Angle-finding problem

Since $\angle OAP=\angle OBP=10^{\circ}$ , quadrilateral $ABPO$ is cyclic . So, we have, $\angle BOP=\angle BAP=\alpha$ Also, $OM=OA=OB=ON=r$ , where $r$ is the radius of the semicircle. So, $\triangle AOM$ and $\triangle BON$ are isosceles. Simple angle chasing leads to $\begin{aligned} \angle MAO&=\frac{180^{\circ}-40^{\circ}}{2}=70^{\circ} \\ \\ \angle BNO&=\frac{180^{\circ}-\alpha}{2}=90^{\circ}-\frac{\alpha}{2} \end{aligned}$ Since quadrilateral $MABN$ is cyclic, $\begin{aligned} \angle MAB&=180^{\circ}-\angle BNM \\ \angle MAO+\angle OAP+\angle BAP&=180^{\circ}-\angle BNO \\ 70^{\circ}+10^{\circ}+\alpha&=180^{\circ}-\left(90^{\circ}-\frac{\alpha}{2}\right) \\ \implies \alpha&=20^{\circ} \end{aligned}$ Therefore, the required angle is $\angle BOP=20^{\circ}$

Rotate $\triangle BOP$ counterclockwise about $O$ so that $B'$ coincides with $A$ :

Then by the exterior angle theorem on $\triangle AOP$ , $\angle APO = \angle AOM - \angle OAP = 40° - 10° = 30°$ .

Since $OP = OP'$ , $\triangle OPP'$ is an isosceles triangle, so $\angle OP'P = \angle OPP = 30°$ .

And by the exterior angle theorem on $\triangle B'OP'$ , $\angle B'OP' = \angle OP'P - \angle OB'P' = 30° - 10° = \boxed{20°}$ .

Since $\angle OAP = \angle OBP$ , $ABPO$ is a cyclic quadrilateral . Then $\angle ABP + \angle AOP = 180^\circ$ . This means that $\angle ABP = \angle AOM = 40^\circ$ . Since $\angle ABO = \angle ABP - \angle OBP = 40^\circ - 10^\circ = 30^\circ$ . Note that $\triangle ABO$ is isosceles, implying that $\angle OAB = \angle ABO = 30^\circ$ and $\angle AOB = 120^\circ$ . Now $\angle BON = 180^\circ - \angle AOB - \angle AOM = 180^\circ - 120^\circ - 40^\circ = \boxed{20}^\circ$ .