Inside a square ABCD, `Delta BEC` is an equilateral triangle. If CE and BD interesect at O, then `angle BOC` is equal to

Options

A. `60^(@)`
B. `75^(@)`
C. `90^(@)`
D. `120^(@)`

Answer

To solve the problem, we need to find the angle \( \angle BOC \) where \( O \) is the intersection of the diagonals \( CE \) and \( BD \) in the square \( ABCD \) with the equilateral triangle \( \Delta BEC \).

### Step-by-Step Solution:

1. **Identify the Angles in the Square**:
- In square \( ABCD \), each angle is \( 90^\circ \).
- Therefore, \( \angle ABC = 90^\circ \).

**Hint**: Remember that all angles in a square are right angles.

2. **Identify the Angles in the Equilateral Triangle**:
- In triangle \( BEC \), since it is an equilateral triangle, all angles are equal to \( 60^\circ \).
- Thus, \( \angle EBC = 60^\circ \) and \( \angle ECB = 60^\circ \).

**Hint**: In an equilateral triangle, all angles are equal and sum up to \( 180^\circ \).

3. **Determine the Angle \( \angle DBC \)**:
- Since \( BD \) is a diagonal of the square, it bisects \( \angle ABC \).
- Therefore, \( \angle DBC = \frac{1}{2} \times 90^\circ = 45^\circ \).

**Hint**: The diagonal of a square bisects the angles at the vertices.

4. **Apply the Triangle Angle Sum Property**:
- In triangle \( BOC \), the sum of the angles must equal \( 180^\circ \).
- We know \( \angle DBC = 45^\circ \) and \( \angle EBC = 60^\circ \).
- Let \( \angle BOC = x \).

**Hint**: The sum of angles in any triangle is \( 180^\circ \).

5. **Set Up the Equation**:
\[ \angle BOC + \angle DBC + \angle EBC = 180^\circ \]
\[ x + 45^\circ + 60^\circ = 180^\circ \]

6. **Solve for \( x \)**:
\[ x + 105^\circ = 180^\circ \]
\[ x = 180^\circ - 105^\circ = 75^\circ \]

**Hint**: Isolate the variable to find its value.

7. **Conclusion**:
- Therefore, \( \angle BOC = 75^\circ \).

### Final Answer:
The angle \( \angle BOC \) is equal to \( 75^\circ \).