In the given figure, O is the centre of the circle and arc ABC subtends an angle of 130° at the centre. If AB is extended to P, find ∠PBC.

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Solution

Take any point D on the major arc CA and then join AD and DC.

Since the angle subtended by an arc on the centre is twice the angle subtended by it on the circumference, we have:
∠AOC = 2∠ADC
⇒ 130° = 2∠ADC [∵ ∠AOC = 130°]
∴ ∠ADC = 65° ...(i)
We know that the exterior angle of a cyclic quadrilateral is equal to the interior opposite angle.
Hence, ∠PBC = ∠ADC = 65°

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