If PA and PB are tangents drawn to a circle with centre O at A and B from external point P such that ∠ APB = 78°, then ∠ OAB is equal to:
Show answer & explanation
Correct answer: A
Explanation
The correct answer is Option 1 i.e. 39°.
In the given figure,
∠APB + ∠OAP + ∠ OBP + ∠ AOB = 360°
Since, PA and PB are tangents, ∠OAP = ∠OBP = 90°
⇒ ∠AOB = 360° - 258° = 102°
Now, in triangle AOB
∠AOB + ∠OAB + ∠OBA = 180°
⇒ 102° + 2∠OAB = 180°
⇒ 2∠OAB = 180° - 102°
⇒ ∠OAB = 78°/2 = 39°
Hence the reqired answer is 39°.