A, B and C are the three points on a circle such that the angle subtendsed by the chords AB and AC at the centre O are `90^@` and `110^(@)` respectively. `angle BAC` is equal to

Options

A. `70^(@)`
B. `80^(@)`
C. `90^(@)`
D. `100^(@)`

Answer

To solve the problem, we need to find the angle \( \angle BAC \) given the angles subtended by the chords \( AB \) and \( AC \) at the center \( O \) of the circle.

### Step-by-Step Solution:

1. **Understand the Given Angles**:
- The angle subtended by chord \( AB \) at the center \( O \) is \( 90^\circ \).
- The angle subtended by chord \( AC \) at the center \( O \) is \( 110^\circ \).

2. **Identify the Angles at the Center**:
- Let \( \angle AOB = 90^\circ \) (angle at center for chord \( AB \)).
- Let \( \angle AOC = 110^\circ \) (angle at center for chord \( AC \)).

3. **Use the Properties of Angles in a Circle**:
- The angle subtended at the circumference by a chord is half the angle subtended at the center by the same chord. Therefore:
- \( \angle ACB = \frac{1}{2} \angle AOB = \frac{1}{2} \times 90^\circ = 45^\circ \)
- \( \angle ABC = \frac{1}{2} \angle AOC = \frac{1}{2} \times 110^\circ = 55^\circ \)

4. **Apply the Triangle Angle Sum Property**:
- In triangle \( ABC \), the sum of angles is \( 180^\circ \):
\[ \angle A + \angle B + \angle C = 180^\circ \]
- Substituting the known angles:
\[ \angle BAC + \angle ABC + \angle ACB = 180^\circ \]
\[ \angle BAC + 55^\circ + 45^\circ = 180^\circ \]

5. **Calculate \( \angle BAC \)**:
- Combine the known angles:
\[ \angle BAC + 100^\circ = 180^\circ \]
- Solve for \( \angle BAC \):
\[ \angle BAC = 180^\circ - 100^\circ = 80^\circ \]

### Conclusion:
The angle \( \angle BAC \) is \( 80^\circ \).