In the figure, PS = QS = RS and QSR - 100°, find QPR
 

  • A 40°
  • B 50°
  • C 80°
  • D 100°
jamb 1990mathematics

The correct answer is B. 50°

Since PS = QS = RS
S is the centre of circle passing through P, Q, R /PS/ = /RS/ = radius

QPR \(\pm\) \(\frac{100^o}{2}\) = 50°

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