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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