Correct option is (c) 27°
We know that the radius and tangent are perpendicular at their point of contact.
∴ OA ⊥ PA and OB ⊥ PB
∴ ∠OAP = ∠OBP = 90°
In quadrilateral OAPB,
⇒ ∠OAP + ∠APB + ∠OBP + ∠AOB = 360°
⇒ 90° + 54° + 90° + ∠AOB = 360°
⇒ 234° + ∠AOB = 360°
⇒ ∠AOB = 360° − 234°
⇒ ∠AOB = 126°
In △AOB,
⇒ OA = OB [ Radii of same circle ]
⇒ ∠OAB = ∠OBA [ Base angles of equal sides are also equal ]
Now,
⇒ ∠OAB + ∠OBA + ∠AOB = 180°
⇒ ∠OAB + ∠OAB + 126° = 180°
⇒ 2∠OAB = 180° − 126°
⇒ 2∠OAB = 54°
∴ ∠OAB = 27°
