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 $."

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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Updated on: 10-Oct-2022

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