If O is the circumcentre of a triangle ABC lying inside the tri angle, then `/_OBC + /_BAC` is equal to

Options

A. `90^@`
B. `60^@`
C. `110^@`
D. `120^@`

Answer

To solve the problem, we need to find the value of the sum of angles \( \angle OBC + \angle BAC \) where \( O \) is the circumcenter of triangle \( ABC \).

### Step-by-Step Solution:

1. **Draw the Circumcircle and Triangle**:
- Draw a circle and inscribe triangle \( ABC \) such that points \( A \), \( B \), and \( C \) lie on the circumference of the circle. Let \( O \) be the circumcenter of triangle \( ABC \).

2. **Identify Angles**:
- Let \( \angle OBC = x \) and \( \angle BAC = y \).

3. **Properties of Isosceles Triangle**:
- Since \( O \) is the circumcenter, the segments \( OB \) and \( OC \) are equal (both are radii of the circumcircle). Therefore, triangle \( OBC \) is isosceles with \( OB = OC \).
- This implies that \( \angle OBC = \angle OCB = x \).

4. **Use Triangle Sum Property**:
- In triangle \( OBC \), the sum of the angles is \( 180^\circ \):
\[ \angle OBC + \angle OCB + \angle BOC = 180^\circ \]
Substituting the known angles:
\[ x + x + \angle BOC = 180^\circ \]
Simplifying gives:
\[ 2x + \angle BOC = 180^\circ \]
Thus,
\[ \angle BOC = 180^\circ - 2x \]

5. **Relate \( \angle BAC \) to \( \angle BOC \)**:
- By the inscribed angle theorem, the angle at the circumference \( \angle BAC \) is half of the angle at the center \( \angle BOC \):
\[ \angle BAC = \frac{1}{2} \angle BOC = \frac{1}{2} (180^\circ - 2x) \]
Simplifying gives:
\[ \angle BAC = 90^\circ - x \]

6. **Calculate \( \angle OBC + \angle BAC \)**:
- Now, we can find the sum:
\[ \angle OBC + \angle BAC = x + (90^\circ - x) \]
Simplifying this expression:
\[ \angle OBC + \angle BAC = 90^\circ \]

### Conclusion:
The final result is:
\[ \angle OBC + \angle BAC = 90^\circ \]