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Question

In the given figure PQ is a diameter of a circle with centre O and PT is a tangent at P. QT meets the circle at R. If ∠POR = 72, find PTR.


A

44°

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B

54°

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C

64°

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D

34°

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Solution

The correct option is B

54°


OQR = ORQ (OR=OQ (radius))

But ROP = OQR + ORQ (Exterior angle theorem)

72 = 2 OQR

OQR = 36

In QPT

PTQ = 180 - PQT - QPT

PTQ = 180 36 - 90

PTQ = 54

Therefore, PTR = 54


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