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