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 $\angle \mathrm{APB}=128^{\circ}$, then $\angle \mathrm{OAB}$ is equal to:
Option 1: 72°
Option 2: 52°
Option 3: 38°
Option 4: 64°
Correct Answer: 64°
Solution :
$\angle APB =128^{\circ}$
Let $\angle OAB = x$
$\angle PAB = 90^{\circ} - x$
Since PA and PB are tangents from the same external point P,
⇒ PA = PB
$\therefore \triangle APB$ is an isosceles triangle.
⇒ $\angle PAB = \angle PBA = 90^{\circ} - x$
In $ \triangle APB$,
$\angle APB + \angle PAB + \angle PBA = 180^{\circ}$
⇒ $128^{\circ} + (90^{\circ} - x) + (90^{\circ} - x) = 180^{\circ}$
⇒ $128^{\circ} = 2x$
⇒ $x = 64^{\circ}$
Hence, the correct answer is $64^{\circ}$.
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