$\begin{array}{l}\angle A B R=35^{\circ}+y^{\circ}=50^{\circ}(\text { ext. } \angle \text { of } \triangle) \\ \angle A P R=\angle A B R=50^{\circ}(\angle \text { s in the same segment }) \\ p+50=90(\angle \text { in semi-circle }) \\ p=40\end{array}$
I2.4 In the figure, $A B$ is a diameter of the circle. $A P Q$ and $R B Q$ are straight lines.
If $\angle P A B=35^{\circ}, \angle P Q B=y^{\circ}$ and $\angle R P B=p^{\circ}$, find $p$.