The tangent at point A on a circle with center O intersects the diameter PQ of the circle when extended, at point B. If ∠BAQ = 105°, then, ∠APQ is equal to:
Show answer & explanation
Correct answer: D
Explanation
The correct answer is Option 4 i.e. 75°.
We know that angle in a semicircle = 90° = ∠PAQ
Now ∠PAB = ∠BAQ - ∠PAQ = 105° - 90° = 15°
Also we know that angle between chord and tangent is equal to angle made by the chord on circle.
Thus ∠PAB = ∠PQA = 15°
⇒ ∠APQ + ∠PQA + ∠PAQ = 180˚ (Angle sum property of triangle)
⇒ ∠APQ + 15˚ + 90˚ = 180˚
⇒ ∠APQ = 75˚