The tangent at point A on the circle with center O intersects the diameter PQ of the circle, when extended at point B. If ∠BAQ = 110°, then ∠APQ is equal to ?

70°

∠BAQ = ∠BAP + ∠PAQ [∠PAQ = 90°,angle made by diameter]

∠BAP = 110° - 90° = 20°

OA is radius of the circle and make 90° angle with tangent BA i.e. ∠BAO = 90°

∠BAQ = ∠BAO + ∠OAQ

110° = 90° + ∠OAQ

∠OAQ = 20°

∴ ∠OAQ = ∠AQO [angle by radius]

⇒ OA = OQ

⇒ ∠OAQ = ∠AQO = 20°

Now, In ΔPAQ

∠APQ + ∠PAQ + ∠AQP = 180°

∠APQ + 90° + 20° = 180°

∠APQ = 70°