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:
Option 1: $45^{\circ}$
Option 2: $35^{\circ}$
Option 3: $70^{\circ}$
Option 4: $50^{\circ}$
Correct Answer: $50^{\circ}$
Solution : According to the question, $\angle$APB = 100º Also, $\angle$ OAP = $\angle$ OBP = 90º (the angle between the tangent and radius at the point of contact is 90º) Since the sum of all angles of a quadrilateral is 360º. $\angle$ OAP + $\angle$ OBP + $\angle$ ABP + $\angle$ AOB = 360º ⇒ 90º + 90º + 100º + $\angle$ AOB = 360º ⇒ $\angle$ AOB = 360º – 280º = 80º Now, in triangle AOB, ⇒ $\angle$ AOB + $\angle$ OAB + $\angle$ OBA = 180º Let $\angle$ OAB = $\angle$ OAB = $x$ (both angles are equal since angles opposite to equal sides are equal) ⇒ 80º + $x$ + $x$ = 180º ⇒ 2$x$ = 100º ⇒ $x$ = 50º Hence, the correct answer is $50^{\circ}$.
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