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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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 $\angle {APB}=100^{\circ}$, then $\angle {OAB}$ is equal to:
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:
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