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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:

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: Correct Answer 55°

Given:

A circle with center O and diameter PQ extended outside the circle to point B

∠BAP = 125°

Concept Used:

Angle formed between radius and tangent of a circle is always 90°

Angles opposite to equal sides are equal.

Sum of two opposite interior angles of a triangle is equal to the opposite exterior angle.

Sum of all angles of triangle is 180°

Calculation:

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OA ⊥ BA  (Angle between Radius and Tangent)

∠OAP = ∠BAP - ∠OAB

⇒ ∠OAP = 125° - 90° = 35°

Since,

OA = OP = OQ (Radius of circle)

So,

∠OAP = ∠OPA = 35°

∠AOQ = ∠OAP + ∠OPA

⇒ ∠AOQ = 35° + 35° = 70°

In Δ AOQ,

∠AOQ = 70°

∠OAQ = ∠OQA (since OA = OQ)

∠AOQ + ∠OAQ + ∠OQA = 180°

⇒ 70° + ∠OAQ + ∠OQA = 180°

⇒ ∠OAQ + ∠OQA = 180° - 70° = 110°

⇒ 2∠OQA = 110°/2 = 55°

Since ∠OQA is the same angle as ∠AQP

∠AQP = 55°

∴ The value of ∠AQP = 55° 

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