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The tangent at a point A on a circle with centre O intersects the diameter PQ of the circle, when extended, at point B. If ∠BAQ = 105°, then ∠APQ is equal to:

The tangent at a point A on a circle with centre O intersects the diameter PQ of the circle, when extended, at point B. If ∠BAQ = 105°, then ∠APQ is equal to: Correct Answer 75°

As we know,

∠PAQ = 90°

∠BAQ = 105° (Given)

⇒ ∠BAP + ∠PAQ = 105°

⇒ ∠BAP = 105° – 90° = 15°

As we know,

∠BAP = ∠AQP = 15°

In △APQ

∠APQ + ∠PAQ + ∠AQP = 180°

⇒ ∠APQ + 90° + 15° = 180°

∴ ∠APQ = 180° – 90° – 15° = 75°

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