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:

55°

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°