Question : In the given figure, $P$ is the centre of the circle. If $QS=PR$, then what is the ratio of $\angle RSP$ to the $\angle TPR$?
Option 1: 1 : 4
Option 2: 2 : 5
Option 3: 1 : 3
Option 4: 2 : 7
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Correct Answer: 1 : 3
Solution :
In the given figure,
We have, $SQ = PR$.
$PQ$ and $PR$ are the radii of the circle.
So, $PQ=PR$
This is the property of an isosceles triangle.
$\triangle PSQ$ is an isosceles triangle.
Let $\angle PSQ = \theta$, then $\angle SPQ = \theta$.
From the Exterior Angle Theorem,
In $\triangle PQS$,
$\angle PQR = \angle PSQ + \angle SPQ = \theta + \theta = 2\theta$
$\angle PQR =\angle PRQ = 2\theta$
Again, the Exterior Angle Theorem,
In $\triangle PSR$,
$\angle TPR = \angle PSR + \angle PRS = \theta + 2\theta = 3\theta$
$\angle RSP : \angle TPR = 1:3$.
Hence, the correct answer is 1 : 3.
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