Question : Two circles touch externally at P. QR is a common tangent of the circle touching the circles at Q and R. Then, the measure of $\angle QPR$ is:
Option 1: $60^{\circ}$
Option 2: $30^{\circ}$
Option 3: $90^{\circ}$
Option 4: $45^{\circ}$
Correct Answer: $90^{\circ}$
Solution :
$\angle$POQ = $\angle$POR = $90^\circ$
OQ = OP = OR (Tangents drawn from the same external point)
$\therefore$ $\angle$OQP = $\angle$OPQ = $\angle$ORP = $\angle$OPR
In $\triangle$POQ,
$\angle$POQ + $\angle$OPQ + $\angle$OQP = $180^\circ$
⇒ $90^\circ$ + $\angle$OPQ + $\angle$OPQ = $180^\circ$
⇒ 2$\angle$OPQ = $180^\circ-90^\circ$
$\therefore \angle$OPQ = $45^\circ$
Similarly,
$\angle$OPR = $45^\circ$
$\therefore$ $\angle$QPR $=\angle$OPQ + $\angle$OPR $=45^\circ + 45^\circ = 90^\circ$
Hence, the correct answer is $90^\circ$.
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