`O` is the centre of a circle and an equilateral `DeltaABC` is inscribed in it. Find the value of `angleBOC`.

Explanation

To solve the problem of finding the value of angle \( BOC \) where \( O \) is the center of the circle and \( \Delta ABC \) is an equilateral triangle inscribed in it, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Properties of an Equilateral Triangle**:
- In an equilateral triangle, all sides are equal and all angles are equal. Therefore, each angle in triangle \( ABC \) is \( 60^\circ \).

2. **Identify the Angles**:
- Let \( \angle BAC \) be one of the angles of the triangle \( ABC \). Since \( ABC \) is equilateral, we have:
\[ \angle BAC = 60^\circ \]

3. **Use the Circle Theorem**:
- According to the theorem of circles, the angle subtended at the center of the circle by an arc is twice the angle subtended at any point on the circumference by the same arc. Therefore, we can express \( \angle BOC \) in terms of \( \angle BAC \):
\[ \angle BOC = 2 \times \angle BAC \]

4. **Substitute the Value of \( \angle BAC \)**:
- Now substituting the value of \( \angle BAC \):
\[ \angle BOC = 2 \times 60^\circ = 120^\circ \]

5. **Final Answer**:
- Thus, the value of \( \angle BOC \) is:
\[ \angle BOC = 120^\circ \]