Ina circle with centre O, an are ABC subtends an angle of $$110^\circ$$ at the centre of the circle. The chord AB is produced to a point P. Then $$\angle$$CBP is equal to:
Take any point D on the circumference and join AD and DC .
∴ $$\angle$$AOC = 2 × $$\angle$$ADC
⇒ $$\angle$$ADC = 1/2 × $$\angle$$AOC = 1/2 × 110 = 55$$^\circ$$
Now, $$\angle$$PBC = ∠ADC [exterior angle of cyclic quadrilateral]
⇒ $$\angle$$PBC = 55$$^\circ$$
SO , the answer would be option b)55$$^\circ$$.
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