Question

AB is a diameter of the circle and A, P, B, R are four points on the circle. The lines AP, RB intersect at Q. Find $\angle BPR.$

Answer 1

Join AR and PB

$\angle PAB=\angle PRB={34}^{\circ }$ (Angles in the same segment)

$\angle APB=\angle ARB={90}^{\circ }$ (Angle subtended by the diameter)

$\Rightarrow \angle BPQ={90}^o$

In $△BPQ$

$\angle PBQ=180-(90+28)={62}^o$

$\Rightarrow \angle BPR=62-34={28}^o$ (Exterior angle is equal to sum of opposite interior angle)