The tangent at a point A of a circle with centre O intersects the diameter PQ of the circle (when extended) at the point B. If ∠BAP = 125°, then ∠AQP is equal to:
A. 50°
B. 55°
C. 60°
D. 45°
Answer: Option B
Solution (By Examveda Team)
∵ BAT is a straight line
∠TAP = ∠TAB - ∠BAP
∠TAP = 180° - 125° = 55°
∠AQP = ∠TAP = 55° (by alternate segment theorem)
This Question Belongs to Arithmetic Ability >> Geometry
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