Given that \(\cos(w^\circ) = \sin(z^\circ)\), we can use the complementary angle identity for sine and cosine:
\[ \sin(z^\circ) = \cos(90^\circ - z^\circ) \]
Thus, we can say:
\[ \cos(w^\circ) = \cos(90^\circ - z^\circ) \]
This implies that:
- \(w^\circ = 90^\circ - z^\circ\) (if both angles are in the same quadrant)
- or \(w^\circ = 90^\circ + z^\circ\) (if considering the periodic nature of cosine)
Since we are considering angles in triangles, we will stick with the first case:
\[ w + z = 90^\circ \]
This means that:
\[ z = 90^\circ - w \]
In a triangle, if \( \angle A = w^\circ \) and \(\angle C = z^\circ\), then \(\angle A + \angle C = 90^\circ\). Therefore, \(\angle A\) and \(\angle C\) are complementary angles.
Now considering the triangles \( \Delta ABD \) and \( \Delta ECD \):
- \( \angle D = 90^\circ \)
- \( \angle A = w^\circ \)
- \( \angle E = 90^\circ - w^\circ = z^\circ \)
- And since both angles in respective triangles are summing to 90 when added to angle D (for triangles ABD and ECD), we can infer similarity.
Thus, we can conclude:
Statement: \( w = z \) and \( \Delta ABD \sim \Delta ECD \) are more proper with the first conditions.
The options can be interpreted again in terms of relations:
- The similarity of the triangles implies \( \Delta ABD \sim \Delta ECD \).
The correct statement is:
Answer: \( w = z \) and \( \Delta ABD \sim \Delta ECD \)