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Question

In the given figure, PQRS is a cyclic quadrilateral in a circle with centre O. If PSR=130. The value of QPR​ is .

A
40°
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B
50°
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C
60°
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Solution

The correct option is A 40°
Given, PSR=130
We have PSR+PQR=180 [Opposite angles of a cyclic quadrilateral]
130+Q=180
Q=180130=50
Also, PRQ=90 [Angle subtended by diameter on the circle is always 90]
In ΔPRQ,
QPR+PRQ+RQP=180 [Sum of angles of triangle is 180]
QPR+90+50=180
QPR=180140=40

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