In the diagram above, PQ is a tangent at T to the circle ABT. ABC is a straight line and TC bisects ∠BTO. Find x.

  • A 20o
  • B 30o
  • C 35o
  • D 40o
  • E 50o
waec 1990mathematics

The correct answer is E. 50o

From the figure < TAB = < BTQ = 40° (alternate segment)

\(\therefore\)< ATB = 180° - (70° + 40°) = 70° (angle on a straight line)

< BTC = \(\frac{40°}{2} = \frac{< BTQ}{2}\)

\(\therefore < BTQ = 40°\)

x° = 180° - (40° + 70° + 20°) 

= 50°

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