In the following figure, PQ is a chord of a circle with center O and PT is a tangent. If $\angle QPT\ =\ 60^{o}$, find $\angle PRQ$.
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Given: A chord PQ of the circle O and tangent PT. Where P is the point of contact.$\angle $QPT=60^{o}$
To do: To find $\angle $PRQ=?
Solution:
$\angle OPT=90^{o} \ \ \ \left( \because radius\ is\ always\ perpendicular\ to\ the\ tangent\ at\ the\ point\ of\ contact\right)$
So, $\angle OPQ=\angle OPT-\angle QPT$
$=90^{o}-60^{o}$
$=30^{o}$
Now in $\vartriangle OPQ$,
$\angle OPQ=\angle OQP=30^{o} \ \ \ \ \ \ \left( \because OP=OQ=radius\ of\ the\ circle\right)$
$\therefore \ \angle POQ=180^{o} -\left( 30^{o} \ \ +30^{o} \right)$
$=120^{o}$
As we know $\vartriangle OPQ$ has a reflection $\vartriangle PEQ$.
$\therefore \ \angle POQ=\angle PRQ=120^{o}$
Therefore $\angle PRQ=120^{o}$.
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