The internal bisectors of `/_ABC and /_ACB `of `DeltaABC `meet each other at O. If `/_BOC =110^@`, then `/_BAC` is equal to

Options

A. `40^@`
B. `55^@`
C. `90^@`
D. `110^@`

Answer

To solve the problem, we need to find the measure of angle \( \angle BAC \) given that the internal bisectors of angles \( \angle ABC \) and \( \angle ACB \) meet at point \( O \) and that \( \angle BOC = 110^\circ \).

### Step-by-Step Solution:

1. **Understanding the Relationship**:
The internal angle bisectors of angles \( \angle ABC \) and \( \angle ACB \) intersect at point \( O \). The angle formed at point \( O \) between these two bisectors is \( \angle BOC \).

2. **Using the Angle Bisector Theorem**:
We know from geometry that the angle formed at the intersection of the angle bisectors can be expressed as:
\[ \angle BOC = 90^\circ + \frac{1}{2} \angle BAC \]

3. **Substituting the Given Value**:
We are given that \( \angle BOC = 110^\circ \). We can substitute this value into the equation:
\[ 110^\circ = 90^\circ + \frac{1}{2} \angle BAC \]

4. **Isolating \( \angle BAC \)**:
To find \( \angle BAC \), we first subtract \( 90^\circ \) from both sides of the equation:
\[ 110^\circ - 90^\circ = \frac{1}{2} \angle BAC \]
This simplifies to:
\[ 20^\circ = \frac{1}{2} \angle BAC \]

5. **Solving for \( \angle BAC \)**:
Now, we multiply both sides by 2 to solve for \( \angle BAC \):
\[ \angle BAC = 2 \times 20^\circ = 40^\circ \]

### Final Answer:
Thus, the measure of \( \angle BAC \) is \( 40^\circ \).

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