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 $\angle \mathrm{APB}=142^{\circ}$, then $\angle \mathrm{OAB}$ is equal to:
Option 1: 31°
Option 2: 58°
Option 3: 71°
Option 4: 64°
Correct Answer: 71°
Solution :
Given: PA and PB are tangents
$\angle OAP = 90^\circ $
$\angle OBP = 90^\circ $
As, OAPB is a quadrilateral
$\angle OAP + \angle APB + \angle PBO + \angle BOA = 360^\circ $
⇒ $90^\circ + 142^\circ + 90^\circ + \angle BOA = 360^\circ $
⇒ $\angle BOA = 360^\circ - 322^\circ $
⇒ $\angle BOA = 38^\circ $
As, OA = OB (Radius)
In ΔOAB, $\angle OAB = \angle OBA$ (In an isosceles triangle angles opposite to equal sides are equal)
$\angle OAB + \angle OBA + \angle BOA = 180^\circ $
⇒ $2 × \angle OAB + 38^\circ = 180^\circ $
⇒ $\angle OAB = \frac{(180^\circ - 38^\circ)}{2}$
∴ $\angle OAB = 71^\circ$
Hence, the correct answer is 71°
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