Question

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:

• A. 38°
• B. 64°
• C. 60
• D. 72°

Answer 1

Answer:

B. 64°

PA and PB are tangents

∠OAP = 90°

∠ OBP = 90° 

OAPB is a quadrilateral ,

∠OAP + ∠APB + ∠PBO + ∠BOA = 360° 

= 90° + 128° + 90° + ∠BOA = 360°

∠BOA = 360° – 308° 

∠BOA = 52° 

OA = OB (Radii)

In ΔOAB, ∠OAB = ∠OBA (In a triangle angles opposite to equal sides are equal)

∠OAB + ∠ OBA + ∠BOA = 180°

2 × ∠OAB + 52° = 180° 

∠OAB = (180° – 52°)/2 

∠OAB is 64°