O is the circumcentre of a triangle `DeltaABC`.The point A and the chord BC are on the opposite side of O.If `angle BOC=150^(@)`.Then the angle `angleBAC` is:

Options

A. `65^(@)`
B. `60^(@)`
C. `70^(@)`
D. `75^(@)`

Answer

To solve the problem, we need to find the angle \( \angle BAC \) given that \( O \) is the circumcenter of triangle \( \Delta ABC \) and \( \angle BOC = 150^\circ \).

### Step-by-Step Solution:

1. **Understanding the Circumcenter**:
- The circumcenter \( O \) is the point where the perpendicular bisectors of the sides of the triangle intersect. It is also the center of the circumcircle that passes through all three vertices \( A, B, \) and \( C \).

2. **Identifying Angles**:
- We are given that \( \angle BOC = 150^\circ \). This angle is formed at the circumcenter \( O \) by the points \( B \) and \( C \).

3. **Using the Property of Angles in a Circle**:
- A key property of circles is that the angle subtended at the center of the circle (in this case, \( \angle BOC \)) is twice the angle subtended at any point on the circumference (in this case, \( \angle BAC \)).
- Mathematically, this can be expressed as:
\[ \angle BOC = 2 \times \angle BAC \]

4. **Setting Up the Equation**:
- Since we know \( \angle BOC = 150^\circ \), we can set up the equation:
\[ 150^\circ = 2 \times \angle BAC \]

5. **Solving for \( \angle BAC \)**:
- To find \( \angle BAC \), we divide both sides of the equation by 2:
\[ \angle BAC = \frac{150^\circ}{2} = 75^\circ \]

6. **Conclusion**:
- Therefore, the angle \( \angle BAC \) is \( 75^\circ \).

### Final Answer:
\[ \angle BAC = 75^\circ \]