In the diagram, PR is a diameter of the circle centre O. RS is a tangent at R and QPR = 58o. Find < QRS

  • A 112o
  • B 116o
  • C 122o
  • D 148o
waec 2006mathematics

The correct answer is C. 122o

PRQ = 90 - 58 = 32o(angle in a semi-circle)

Since PRS = 90o(radius angular to tangent)

QRS = 90 + 32

= 122o

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