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

  • A 37o
  • B 53o
  • C 65o
  • D 127o
  • E 147o
waec 1992mathematics

The correct answer is B. 53o

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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