WAEC - Mathematics (1992 - No. 23)

In the diagram above, O is the center of the circle QOR is a diameter and ∠PSR is 37°. Find ∠PRQ

37^o

53^o

65^o

127^o

147^o

Construction: Join PQ.

Then < RSP = 37° = < RQP (angles on the same segment)

But < RPQ = 90° (angle in a semi-circle)

\(\therefore\) < PRQ = 180° - (90° + 37°)

= 53°