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 :

35°

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°