In a rhombus, the diagonals bisect each other at right angles and also bisect the angles at each vertex.
Given that \( \angle MNL = 55^\circ \), this angle is split into two equal angles by diagonal \( LN \). Therefore, we can find \( \angle NML \):
\[ \angle NML = \angle MNL = 55^\circ \]
Now, since \( \angle MNL \) and \( \angle NML \) are angles associated with vertex \( N \), the remaining angle \( \angle N \) (the angles at vertex \( N \) split by the diagonal) will be:
\[ \angle N = \angle MNL + \angle NML = 55^\circ + 55^\circ = 110^\circ \]
Since the diagonals of a rhombus bisect the angles, both angles \( MNO \) and \( NMO \) must be equal to \( \angle N \) but on the opposite side, and so:
\[ \angle MNO = \angle MNL = 55^\circ \]
Thus,
\[ m\angle MNO = 55^\circ \]