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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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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Question : In a $\triangle \mathrm{ABC}$, the bisectors of $\angle \mathrm{B}$ and $\angle \mathrm{C}$ meet at $\mathrm{O}$. If $\angle \mathrm{BOC}=142^{\circ}$, then the measure of $\angle \mathrm{A}$ is:
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