If PA and PB are tangents drawn from an external point P to a circle with centre O such that ∠APB = 70°, then ∠OAB is equal to :
Answer & explanation
Correct answer: option 4
We know that,
Sum of the angle of quadrilateral = 360°
Sum of the angle of triangle = 180°
We have,
∠APB = 70°
∠A + ∠B + ∠O + ∠P = 360°
AP and BP are tangent to the circle
∠OAP = ∠OBP = 90°
= 90° + 90° + ∠O + 70° = 360°
= ∠O = 360° – 90° + 90° – 70°
= ∠O = 110°
Now, ∠OBA = ∠OAB
We also know that,
∠AOB + ∠OAB + ∠OBA = 180°
= 110° + 2∠OAB = 180°
⇒ ∠OAB = 35°