PA and PB are tangents drawn to a circle with centre O from an external point P. If A and B are points on the circle and ∠OBA = 42o, then ∠APB is:
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Correct answer: D
Explanation
The correct answer is option 4 i.e. 84o.
Since tangent AP makes a right angle with radius OA
∴ ∠OAP = 90°
In ΔAOP,
∠AOP + ∠APO + ∠PAO = 180°
⇒ ∠AOP + 15° + 90° = 180°
⇒ ∠AOP = 75°
⇒ ∠AOB = 2 × ∠AOP = 150°
Now, in ΔOAB,
∠AOB = 150°
⇒ ∠OAB + ∠OBA = 30° (∵ OA = OB, ∴ ∠OAB = ∠OBA)
∴ ∠OAB = 15°