G10.3 $A, P, B$ are three points on a circle with centre $O$.
If $\angle A P B=146^{\circ}$, find $\angle O A B$.
Add a point $Q$ as shown in the diagram.
$$
\begin{array}{l}
\left.\angle A Q B=180^{\circ}-146^{\circ}=34^{\circ} \text { (opp. } \angle \text { s cyclic quad. }\right) \\
\angle A O B=2 \times 34^{\circ}=68^{\circ}\left(\angle \text { at centre twice } \angle \text { at } \odot^{\mathrm{ce}}\right) \\
O A=O B=\text { radii } \\
\begin{aligned}
\angle O A B & =\angle O B A(\text { base } \angle \text { s isos. } \triangle) \\
& =\frac{180^{\circ}-68^{\circ}}{2}=56^{\circ}(\angle \text { s sum of } \triangle)
\end{aligned}
\end{array}
$$