Question : PA and PB are two tangents from a point P outside the circle with centre O at the points A and B on it. If $\angle A P B=130^{\circ}$, then $\angle O A B$ is equal to:
Option 1: 45°
Option 2: 50°
Option 3: 35°
Option 4: 65°
Correct Answer: 65°
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 + 130^\circ + 90^\circ + \angle BOA = 360^\circ $
⇒ $\angle BOA = 360^\circ - 308^\circ $
⇒ $\angle BOA = 50^\circ $
As, OA = OB (Radius)
In ΔOAB, $\angle OAB = \angle OBA$ (In a triangle angles opposite to equal sides are equal)
$\angle OAB + \angle OBA + \angle BOA = 180^\circ $
⇒ $2 × \angle OAB + 50^\circ = 180^\circ $
⇒ $\angle OAB = \frac{(180^\circ - 50^\circ)}{2}$
∴ $\angle OAB = 65^\circ$
Hence, the correct answer is $65^\circ$.
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