I is the incentre of `Delta ABC, angle ABC=60^(@)` and `angle ACB=50^(@)`. Then find `angle BIC`.

Answer

To find the angle \( \angle BIC \) in triangle \( ABC \) where \( I \) is the incenter, and given that \( \angle ABC = 60^\circ \) and \( \angle ACB = 50^\circ \), we can follow these steps:

### Step-by-Step Solution:

1. **Determine \( \angle BAC \)**:
- We know that the sum of angles in a triangle is \( 180^\circ \).
- Therefore, we can find \( \angle BAC \) as follows:
\[ \angle BAC = 180^\circ - \angle ABC - \angle ACB \]
Substituting the known values:
\[ \angle BAC = 180^\circ - 60^\circ - 50^\circ = 70^\circ \]

2. **Find the angles at the incenter \( I \)**:
- The angles \( \angle IBC \) and \( \angle ICB \) are half of \( \angle ABC \) and \( \angle ACB \) respectively because \( I \) is the incenter.
- Therefore:
\[ \angle IBC = \frac{1}{2} \times \angle ABC = \frac{1}{2} \times 60^\circ = 30^\circ \]
\[ \angle ICB = \frac{1}{2} \times \angle ACB = \frac{1}{2} \times 50^\circ = 25^\circ \]

3. **Use the angle sum property in triangle \( BIC \)**:
- In triangle \( BIC \), the sum of angles is \( 180^\circ \):
\[ \angle IBC + \angle BIC + \angle ICB = 180^\circ \]
- Substituting the known values:
\[ 30^\circ + \angle BIC + 25^\circ = 180^\circ \]

4. **Solve for \( \angle BIC \)**:
- Combine the known angles:
\[ 55^\circ + \angle BIC = 180^\circ \]
- Rearranging gives:
\[ \angle BIC = 180^\circ - 55^\circ = 125^\circ \]

### Final Answer:
\[ \angle BIC = 125^\circ \]