PA and PB are two tangents from a point P outside the circle with centre O. If A and B are points on the circle such that ∠APB = 142°, then ∠OAB is equal to:
Show answer & explanation
Correct answer: C
Explanation
The correct answer is Option 3 i.e 71°.
Given:
∠APB = 142°
Concept used:
Radius makes 90° to the point of contact of tangent.
The line joining the center and meeting point of two tangents bisects the angle.
The line joining center makes 90° to the chord.
∠APO = 1/2 × APB
⇒ ∠APO = 1/2 × 142°
⇒ ∠APO = 71°
∠OAP = 90°
In ΔΑΜΡ,
∠OAP + ∠APO + ∠POA = 180° [Angle Sum Property]
⇒ 90° + 71° + ∠PAM = 180°
⇒ ∠POA = 180° -161°
⇒ ∠POA = 19°
∠OAP = 90°
⇒ ∠OAB + AMO + MOA = 180°
⇒ ∠OAB = 71°
Thus, ∠OAB is equal to 71°.