In a circle with centre O. an arc ABC subtends an angle of `136^(@)` at the centre of the circle. The chord AB is produced to a point P. Then `angle CBP` is equal to :

Options

A. `72^(@)`
B. `44^(@)`
C. `68^(@)`
D. `66^(@)`

Answer

To solve the problem step by step, we will follow the properties of circles and angles subtended by arcs and chords.

### Step 1: Identify the Given Information
- We have a circle with center O.
- An arc ABC subtends an angle of \(136^\circ\) at the center of the circle.
- The chord AB is produced to a point P.
- We need to find the angle \(CBP\).

### Step 2: Draw the Circle and Label Points
- Draw the circle with center O.
- Mark points A, B, and C on the circumference such that arc ABC subtends angle \(AOB = 136^\circ\).
- Extend the chord AB to point P.

### Step 3: Use the Property of Angles in a Circle
According to the property of circles, the angle subtended by an arc at the center (angle AOB) is twice the angle subtended by the same arc at any point on the circumference (angle ACB).

Thus, we can write:
\[ \angle ACB = \frac{1}{2} \times \angle AOB \]
Substituting the given value:
\[ \angle ACB = \frac{1}{2} \times 136^\circ = 68^\circ \]

### Step 4: Identify the Cyclic Quadrilateral
Since points A, B, C, and D (where D is any point on the circumference) form a cyclic quadrilateral, we can use the property that the sum of opposite angles in a cyclic quadrilateral is \(180^\circ\).

### Step 5: Find Angle ABC
Using the cyclic quadrilateral property:
\[ \angle ABC + \angle ACB = 180^\circ \]
Substituting the value of \(\angle ACB\):
\[ \angle ABC + 68^\circ = 180^\circ \]
Thus,
\[ \angle ABC = 180^\circ - 68^\circ = 112^\circ \]

### Step 6: Find Angle CBP
Since AB is extended to point P, angle CBP and angle ABC are supplementary (they form a straight line):
\[ \angle ABC + \angle CBP = 180^\circ \]
Substituting the value of \(\angle ABC\):
\[ 112^\circ + \angle CBP = 180^\circ \]
Thus,
\[ \angle CBP = 180^\circ - 112^\circ = 68^\circ \]

### Conclusion
The angle \(CBP\) is equal to \(68^\circ\).

### Final Answer
\(\angle CBP = 68^\circ\)

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