In the figure, $ P Q $ is tangent at a point $ R $ of the circle with centre $ O $. If $ \angle T R Q=30^{\circ} $, find $ m \angle P R S $.
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Given:
\( P Q \) is tangent at a point \( R \) of the circle with centre \( O \).
\( \angle T R Q=30^{\circ} \).
To do:
We have to find \( m \angle P R S \).
Solution:
In the figure,
$RT$ and $RS$ are joined such that $\angle TRQ = 30^o$
Let $\angle PRS = x^o$
$\angle SRX = 90^o$ (Angle in a semicircle is $90^o$)
$\angle TRQ + \angle SRT + \angle PRS = 180^o$ (Sum of the angles on a straight line is $180^o$)
$30^o + 90^o + x^o = 180^o$
$120^o + x^o = 180^o$
$x^o = 180^o - 120^o$
$x^o= 60^o$
$\angle PRS = 60^o$
Therefore, \( m \angle P R S \) is $60^o$.
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