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: $\boxed{45^\circ \ (\text{option b})}$

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.