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Question

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


A

50

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B

40

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C

60

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D

30

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Solution

The correct option is C

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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