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 ?
Answer & explanation
Correct answer: option 3
∠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°