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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