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 ∠BPR.
Join AR and PB
∠PAB=∠PRB=34∘ (Angles in the same segment)
∠APB=∠ARB=90∘ (Angle subtended by the diameter)
⇒ ∠BPQ=90o
In △BPQ
∠PBQ=180−(90+28)=62o
⇒ ∠BPR=62−34=28o (Exterior angle is equal to sum of opposite interior angle)