In `triangleABC` the bisectors of `angleB` and `angleC` intersect at I if `angleA=70^(@)` then `angle BIC=…………..`

Options

A. `35^(@)`
B. `75^(@)`
C. `100^(@)`
D. `125^(@)`

Answer

To find the angle \( BIC \) in triangle \( ABC \) where the angle bisectors of \( B \) and \( C \) intersect at point \( I \) and given that \( \angle A = 70^\circ \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Triangle Properties**:
In any triangle, the sum of the interior angles is \( 180^\circ \). Therefore, we can write:
\[ \angle A + \angle B + \angle C = 180^\circ \]

2. **Substitute the Given Angle**:
We know \( \angle A = 70^\circ \). Substituting this into the equation gives:
\[ 70^\circ + \angle B + \angle C = 180^\circ \]

3. **Rearrange to Find \( \angle B + \angle C \)**:
To find the sum of angles \( B \) and \( C \), we rearrange the equation:
\[ \angle B + \angle C = 180^\circ - 70^\circ = 110^\circ \]

4. **Use the Angle Bisector Theorem**:
The angle bisectors divide the angles into two equal parts. Therefore:
\[ \angle IBC = \frac{1}{2} \angle B \quad \text{and} \quad \angle ICB = \frac{1}{2} \angle C \]

5. **Express \( \angle BIC \)**:
In triangle \( BIC \), the sum of the angles is also \( 180^\circ \):
\[ \angle IBC + \angle ICB + \angle BIC = 180^\circ \]
Substituting the bisector angles gives:
\[ \frac{1}{2} \angle B + \frac{1}{2} \angle C + \angle BIC = 180^\circ \]

6. **Substitute the Sum of \( \angle B + \angle C \)**:
We already found that \( \angle B + \angle C = 110^\circ \). Thus:
\[ \frac{1}{2} (110^\circ) + \angle BIC = 180^\circ \]
Simplifying this:
\[ 55^\circ + \angle BIC = 180^\circ \]

7. **Solve for \( \angle BIC \)**:
Rearranging the equation gives:
\[ \angle BIC = 180^\circ - 55^\circ = 125^\circ \]

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