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Question

C is the centre of a circle passing through the points P,Q,RandS taken in order. If PSR=120 and PQ is a diameter, then QPR=

A
30
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B
40
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C
45
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D
50
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Solution

The correct option is A 30
Refer figure:
Given that: PSR=120°
PRQ=90° (Diameter of the circle subtends an angle of 90°)
Consider cyclic quadrilateral PQRS:
PSR+PQR=180° (sum of opposite angles of a cyclic quadrilateral is 180°)
Thus, PQR=180°120°=60°
In PQR:
QPR=180°PRQPQR .....(Sum of three angles of a triangle is 180°)
QPR=180°90°60°=30°

Answer: QPR=30°

396722_327980_ans.png

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