In the figure given below 'O' is the centre of the circle. If QR = OP and ∠ORP = 20°. Find the value of 'x' giving reasons

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#### Solution

Now,

OP = QR....given

So,OP = OT = OQ = QR

In ΔRQP

RQ = QO

So `angleQRO = angleQOR = 20^@`

So by sum of angles in ΔRQP

`angleRQO = 140^@`

Now

`angleRQO + angleOQP = 180^@` .....linear pair

`angle OQP = 40^@`

in ΔPOQ

OQ = PO...radii

So `angleQPO - angle OQP = 40^@`

So by sumof angle in ΔOQP

`anglePOQ - 100^@`

Now

`anglePOT + anglePOQ +angle QOR = 180^@` ......angles in straight line

`x= 60^@`

Concept: Circumscribing and Inscribing a Circle on a Triangle

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