In a `Delta ABC`, angle bisector of `angle A, angle B & angle C` cuts circumcircle at P, Q , R respectively . If `angle CRQ=46^(@)& angle A=50^(@)`, then find `angle BQR`.

Answer

To solve the problem, we need to find the angle \( \angle BQR \) in triangle \( ABC \) given that \( \angle CRQ = 46^\circ \) and \( \angle A = 50^\circ \).

### Step-by-Step Solution:

1. **Identify the Angles in Triangle ABC**:
- We know \( \angle A = 50^\circ \).
- Let \( \angle B = b \) and \( \angle C = c \).
- The sum of angles in triangle \( ABC \) is \( 180^\circ \):
\[ \angle A + \angle B + \angle C = 180^\circ \implies 50^\circ + b + c = 180^\circ \implies b + c = 130^\circ \]

2. **Use the Angle Bisector Theorem**:
- The angle bisector of \( \angle C \) intersects the circumcircle at point \( R \).
- According to the angle bisector theorem, the angle \( \angle QNR \) (where \( N \) is the intersection of the angle bisectors) can be expressed as:
\[ \angle QNR = 90^\circ + \frac{\angle A}{2} = 90^\circ + \frac{50^\circ}{2} = 90^\circ + 25^\circ = 115^\circ \]

3. **Calculate \( \angle BQR \)**:
- In triangle \( QNR \), we know:
\[ \angle CRQ = 46^\circ \quad \text{and} \quad \angle QNR = 115^\circ \]
- We can find \( \angle BQR \) using the fact that the sum of angles in triangle \( QNR \) is \( 180^\circ \):
\[ \angle BQR + \angle CRQ + \angle QNR = 180^\circ \]
\[ \angle BQR + 46^\circ + 115^\circ = 180^\circ \]
\[ \angle BQR + 161^\circ = 180^\circ \]
\[ \angle BQR = 180^\circ - 161^\circ = 19^\circ \]

### Final Answer:
\[ \angle BQR = 19^\circ \]