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JAMB Mathematics 1980 Question 46

Question 46

Circle Theorems

In the figure above, PS = SR and RX is a tangent to the circle at R, <QRX is equal to 72°. Find <SPR

Diagram for this question
  1. A.20°
  2. B.36°
  3. C.72°
  4. D.30°
  5. E.15°.
Show the answer and explanation

Answer: B (36°)

Explanation

By the alternate segment theorem <RSQ = <QRX = 72°. This is the exterior angle of isosceles triangle PSR, so 2 × <SPR = 72° and <SPR = 36°.