Question : From a point P, two tangents PA and PB are drawn to a circle with centre O. If OP is equal to the diameter of the circle, then $\angle$APB is:
Option 1: 45°
Option 2: 90°
Option 3: 30°
Option 4: 60°
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Correct Answer: 60°
Solution : Given: PA and PB are two tangents to the circle with centre O. Let $r$ be the radius of the circle. $\therefore$ OA $\perp$ PA ⇒ $\angle$OAP = 90° In $\triangle$OPA, $\sin\angle$OPA = $\frac{OA}{OP}=\frac{r}{2r}$ [Given OP is the diameter of the circle] ⇒ $\sin\angle$OPA = $\frac{1}{2}$ = $\sin 30°$ ⇒ $\angle$OPA = 30° Similarly, it can be proved that $\angle$OPB = 30° Now, $\angle$APB = $\angle$OPA + $\angle$OPB = 30° + 30° = 60° Hence, the correct answer is 60°.
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