In a circle with centre O, an arc ABC subtends an angle of 132° at the center of the circle. Chord AB is produced to point P. Then ∠CBP is equal to∶
66°
As we know
The angle subtended by an arc at the centre of a circle is double the size of the angle subtended by the same arc at the circle circumference
Take any point Q on the circumference and join AQ and QC.
∴ ∠AOC = 2 × ∠AQC
⇒ ∠AQC = 1/2 × ∠AOC = 1/2 × 132° = 66°
Now, ∠CBP = ∠AQC [Exterior angle of cyclic quadrilateral]
⇒ ∠CBP = 66°
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