Let O be the orthocentre of the triangle ABC. If `angleBOC = 150^@`, Then `angleBAC` is

Options

A. `30^@`
B. `60^@`
C. `90^@`
D. `120^@`

Answer

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

### Step-by-Step Solution:

1. **Understanding the Orthocenter**:
The orthocenter \( O \) of triangle \( ABC \) is the point where the altitudes of the triangle intersect. In triangle \( ABC \), the angles at vertices \( A \), \( B \), and \( C \) are related to the angles formed at the orthocenter.

2. **Using the Property of the Orthocenter**:
One important property of the orthocenter is that the angles \( \angle BOC \) and \( \angle BAC \) are related such that:
\[ \angle BOC + \angle BAC = 180^\circ \]
This means that the sum of the angle at the orthocenter and the angle at vertex \( A \) is \( 180^\circ \).

3. **Substituting the Given Angle**:
From the problem, we know that:
\[ \angle BOC = 150^\circ \]
We can substitute this value into the equation from step 2:
\[ 150^\circ + \angle BAC = 180^\circ \]

4. **Solving for \( \angle BAC \)**:
To find \( \angle BAC \), we can rearrange the equation:
\[ \angle BAC = 180^\circ - 150^\circ \]
Simplifying this gives:
\[ \angle BAC = 30^\circ \]

5. **Conclusion**:
Therefore, the measure of angle \( \angle BAC \) is \( 30^\circ \).

### Final Answer:
\[ \angle BAC = 30^\circ \]