The arc ABC of a circle with centre O subtends 132 at the centre. The chord AB is extended to the point P. The angle $$\angle$$CBP is equal to

As per the given question,

Angle subtended by the arc =132

Now,

As we know, from the property of circle,

$$2 \times \angle ADC=\angle AOC$$

$$\Rightarrow 2 \times \angle ADC=132^\circ$$

$$\Rightarrow \angle ADC=\dfrac{132^\circ}{2}$$

$$\Rightarrow \angle ADC=66^\circ----------------(i)$$

Now, ABCD is a cyclic quadrilateral,

So as the property of the cyclic quadrilateral

$$\angle ADC +\angle ABC=180$$

$$\Rightarrow \angle ABC=180-\angle ADC$$

$$\Rightarrow \angle ABC=180-66=114^\circ -------------------(ii)$$

We know that angle subtended on the straight line$$ =180^\circ$$

Now, $$\angle ABC +\angle CBP=180$$

$$\angle CBP=180-\angle ABC =180-114=66^\circ$$

$$\Rightarrow \angle CBP=66^\circ$$