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 = 128°, then ∠OAB is equal to :
Show answer & explanation
Correct answer: A
Explanation
The correct answer is option 1 i.e. 64°
∠APB = 1280
The angle made by the point of contact between the tangent and the center of the circle is 900
The Sum of all angles of a quadrilateral is 3600
In a triangle angles opposite to equal sides are equal.
The sum of all angles of a triangle is 1800
PA and PB are tangents.
∠OAP = 900
∠ OBP = 900
OAPB is a quadrilateral.
∠OAP + ∠APB + ∠PBO + ∠BOA = 3600
900 + 1280 + 900 + ∠BOA = 3600
∠BOA = 3600 - 3080
∠BOA = 520
OA = OB
In ΔOAB -
∠OAB = ∠OBA (In a triangle angles opposite to equal sides are equal)
∠OAB + ∠ OBA + ∠BOA = 1800
2 × ∠OAB + 520 = 1800
∠OAB = (1800 - 520)/2
∠OAB = 640