ABC is a cyclic triangle and the bisectors of `angle BAC, angle ABC and angle BCA` meet the circle at P,Q and R respectively. Then the angle `angle RQP` is:

Options

A. `90^(@)-(angleB)/(2)`
B. `90^(@)+(angleB)/(2)`
C. `90^(@)+(angleC)/(2)`
D. `90^(@)+(angleB)/(2)`

Answer

To solve the problem, we need to find the angle \( \angle RQP \) in the cyclic triangle \( ABC \) where the angle bisectors of \( \angle BAC, \angle ABC, \) and \( \angle BCA \) meet the circumcircle at points \( P, Q, \) and \( R \) respectively.

### Step-by-Step Solution:

1. **Understanding the Configuration**:
- Since \( ABC \) is a cyclic triangle, all vertices \( A, B, C \) lie on a circle. The angle bisectors of the angles \( A, B, \) and \( C \) will intersect the circumcircle at points \( P, Q, R \).

2. **Using the Properties of Cyclic Triangles**:
- In a cyclic triangle, the opposite angles sum up to \( 180^\circ \). This property will be useful in our calculations.

3. **Identifying Angles**:
- Let \( \angle BAC = A \), \( \angle ABC = B \), and \( \angle BCA = C \).
- The angles at points \( P, Q, R \) can be expressed in terms of \( A, B, \) and \( C \):
- \( \angle BAP = \frac{A}{2} \) (since \( P \) is on the angle bisector of \( A \))
- \( \angle ABQ = \frac{B}{2} \)
- \( \angle ACR = \frac{C}{2} \)

4. **Finding \( \angle RQP \)**:
- To find \( \angle RQP \), we can use the fact that \( \angle RQP \) is the exterior angle for triangle \( AQR \):
\[ \angle RQP = \angle AQR + \angle ARQ \]
- Using the inscribed angle theorem, we know:
\[ \angle AQR = \frac{1}{2} \angle A \quad \text{and} \quad \angle ARQ = \frac{1}{2} \angle B \]
- Therefore:
\[ \angle RQP = \frac{1}{2} A + \frac{1}{2} B \]

5. **Relating to Angle \( C \)**:
- Since \( A + B + C = 180^\circ \), we can express \( C \) as:
\[ C = 180^\circ - (A + B) \]
- Thus:
\[ \angle RQP = \frac{1}{2} (A + B) = \frac{1}{2} (180^\circ - C) = 90^\circ - \frac{C}{2} \]

6. **Final Result**:
- Therefore, we conclude that:
\[ \angle RQP = 90^\circ - \frac{C}{2} \]

### Conclusion:
The angle \( \angle RQP \) in the cyclic triangle \( ABC \) where the angle bisectors meet the circumcircle at points \( P, Q, R \) is given by \( 90^\circ - \frac{C}{2} \).