In the diagram above, O is the center of the circle, |SQ| = |QR| and ∠PQR = 68°. Calculate ∠PRS

  • A 34°
  • B 45°
  • C 56°
  • D 62°
  • E 68°
waec 1991mathematics

The correct answer is A. 34°

From the figure, < PQR = 68°

\(\therefore\) < QRS = < QSR = \(\frac{180 - 68}{2}\) (base angles of an isos. triangle)

= 56°

\(\therefore\) < PRS = 90° - 56° = 34° (angles in a semi-circle)

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