Internal bisectors of angles `/_B and /_C` of a triangle ABC meet at O. If `/_BAC = 80^@`, then the value of `/_BOC ` is

Options

A. `120^@`
B. `140^@`
C. `110^@`
D. `130^@`

Answer

To find the value of angle \( \angle BOC \) in triangle \( ABC \) where the internal bisectors of angles \( B \) and \( C \) meet at point \( O \) and \( \angle BAC = 80^\circ \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Information**:
- We know that \( \angle BAC = 80^\circ \).
- Let \( \angle ABC = B \) and \( \angle ACB = C \).

2. **Use the Triangle Sum Property**:
- The sum of the angles in triangle \( ABC \) is \( 180^\circ \):
\[ \angle A + \angle B + \angle C = 180^\circ \]
- Substituting the known value:
\[ 80^\circ + B + C = 180^\circ \]
- Rearranging gives:
\[ B + C = 180^\circ - 80^\circ = 100^\circ \]

3. **Express \( \angle BOC \)**:
- The internal bisectors \( OB \) and \( OC \) divide angles \( B \) and \( C \) respectively:
\[ \angle BOC = 180^\circ - \frac{1}{2}(\angle B + \angle C) \]
- Since \( B + C = 100^\circ \):
\[ \angle BOC = 180^\circ - \frac{1}{2}(100^\circ) \]

4. **Calculate \( \angle BOC \)**:
- Calculate \( \frac{1}{2}(100^\circ) = 50^\circ \):
\[ \angle BOC = 180^\circ - 50^\circ = 130^\circ \]

### Final Answer:
\[ \angle BOC = 130^\circ \]