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We are given that \( ABCD \) is a square, and angles \( \angle EDC = \angle ECD = 15^\circ \). We are tasked with finding \( \angle EBC \).

### Solution:

1. Since \( ABCD \) is a square, all sides are equal, and all interior angles are \( 90^\circ \). 
   - \( \angle DAB = \angle ABC = \angle BCD = \angle CDA = 90^\circ \).

2. In \( \triangle EDC \), we are told that \( \angle EDC = 15^\circ \) and \( \angle ECD = 15^\circ \). Therefore, the third angle in \( \triangle EDC \), \( \angle DEC \), can be calculated:
   \[
   \angle DEC = 180^\circ - (15^\circ + 15^\circ) = 150^\circ
   \]

3. Now, consider the quadrilateral \( EBCD \). Since \( ABCD \) is a square, we know:
   - \( \angle BCD = 90^\circ \)
   - \( \angle EDC = 15^\circ \)

   We already found that \( \angle DEC = 150^\circ \).

4. We know that \( \angle EBC + \angle DEC = \angle BCD = 90^\circ \) (since the sum of angles around point \( C \) must add up to \( 90^\circ \)).

   Therefore:
   \[
   \angle EBC = 90^\circ - 15^\circ = 45^\circ
   \]

Thus, the value of \( \angle EBC \) is **\( 45^\circ \)**.

The correct answer is:
\[
\boxed{45^\circ \ (\text{option b})}
\]

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Do you have any questions, or would you like more details on any of the steps?

Here are 5 related questions:
1. How do we know that \( ABCD \) is a square from the given information?
2. Why do the angles in \( \triangle EDC \) sum to \( 180^\circ \)?
3. What are the properties of a square that helped in solving this problem?
4. How do we calculate angles in quadrilaterals like \( EBCD \)?
5. Can this problem be solved using trigonometry, and how?

**Tip**: In problems involving geometric figures like squares or triangles, always consider the properties of the shapes (e.g., angle sum, symmetry) for solving angle-related questions.

In the given figure, ABCD is a square in which ∠EDC = ∠ECD = 15°, then find ∠EBC?

This geometry problem involves a square ABCD with given angles ∠EDC = ∠ECD = 15°. The task is to find ∠EBC using properties of squares and the angle sum theorem in triangles. A detailed step-by-step solution is provided, suitable for students in Grades 9-10.

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