Let $M$ and $N$ be points on the sides $BC$ and $CD$, respectively, of a square $ABCD$. If $|BM|=21$, $|DN|=4$, and $|NC|=24$, what is $m(\widehat{MAN})$?

$ 
\textbf{(A)}\ 15^\circ
\qquad\textbf{(B)}\ 30^\circ
\qquad\textbf{(C)}\ 37^\circ
\qquad\textbf{(D)}\ 45^\circ
\qquad\textbf{(E)}\ 60^\circ
$
1. **Rotate the square around point \( A \) by \( 90^\circ \) counterclockwise:**
   - When we rotate the square \( ABCD \) by \( 90^\circ \) counterclockwise around point \( A \), point \( D \) maps to point \( B \). 
   - Consequently, point \( N \) on side \( CD \) maps to a new point \( N' \) on side \( BC \).

2. **Determine the coordinates of points \( M \) and \( N \):**
   - Let the side length of the square \( ABCD \) be \( s \).
   - Given \( |BM| = 21 \), point \( M \) is 21 units from \( B \) along \( BC \).
   - Given \( |DN| = 4 \) and \( |NC| = 24 \), point \( N \) is 4 units from \( D \) along \( CD \).

3. **Calculate the side length \( s \) of the square:**
   - Since \( |DN| + |NC| = s \), we have \( 4 + 24 = s \).
   - Therefore, \( s = 28 \).

4. **Find the coordinates of points \( M \) and \( N \):**
   - Point \( B \) is at \( (0, 0) \).
   - Point \( C \) is at \( (28, 0) \).
   - Point \( D \) is at \( (28, 28) \).
   - Point \( M \) is at \( (21, 0) \) since it is 21 units from \( B \) along \( BC \).
   - Point \( N \) is at \( (28, 4) \) since it is 4 units from \( D \) along \( CD \).

5. **Rotate point \( N \) to find \( N' \):**
   - Rotating point \( N \) by \( 90^\circ \) counterclockwise around \( A \) (which is at \( (0, 28) \)) gives:
     \[
     N' = (0 + (4 - 28), 28 + (28 - 28)) = (-24, 28)
     \]

6. **Calculate the distances \( AN \) and \( AN' \):**
   - Distance \( AN \):
     \[
     AN = \sqrt{(28 - 0)^2 + (4 - 28)^2} = \sqrt{28^2 + (-24)^2} = \sqrt{784 + 576} = \sqrt{1360} = 2\sqrt{340}
     \]
   - Distance \( AN' \):
     \[
     AN' = \sqrt{(-24 - 0)^2 + (28 - 28)^2} = \sqrt{(-24)^2 + 0^2} = \sqrt{576} = 24
     \]

7. **Verify the congruence of triangles \( \triangle AMN \) and \( \triangle AMN' \):**
   - We have \( AM = AM \), \( AN = AN' \), and \( MN = MN' \).
   - Since \( MN = \sqrt{(21 - 28)^2 + (0 - 4)^2} = \sqrt{49 + 16} = \sqrt{65} \) and \( MN' = 25 \), the triangles are congruent by SSS.

8. **Determine the angle \( \angle MAN \):**
   - Since \( \angle NAN' = 90^\circ \) and the triangles are congruent, \( \angle MAN = 45^\circ \).

\(\blacksquare\)

The final answer is \( \boxed{ 45^\circ } \).