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:
Answer & explanation
Correct answer: option 2
OA ⊥ BA (Angle between Radius and Tangent)
∠OAP = ∠BAP - ∠OAB = 125° - 90° = 35°
We know that,
OA = OP = OQ (Radius of circle)
So, ∠OAP = ∠OPA = 35°
∠AOQ = ∠OAP + ∠OPA = 35° + 35° = 70°
In Δ AOQ,
∠AOQ = 70°
∠OAQ = ∠OQA (OA = OQ)
∠AOQ + ∠OAQ + ∠OQA = 180°
= 70° + ∠OAQ + ∠OQA = 180°
= ∠OAQ + ∠OQA = 180° - 70° = 110°
= 2∠OQA = 110°/2 = 55°
∠OQA = ∠AQP
So, ∠AQP = 55°