This problem involves finding the angles in a rhombus with given diagonal properties.

Key information:

1. The rhombus is split by diagonals, which meet at a right angle (90°).
2. One of the angles is given as 53°.

Steps to Solve:

1. Angle at the intersection of diagonals: Diagonals of a rhombus bisect each other perpendicularly, meaning the angles they form at the intersection are always 90°.

   • Thus, $\angle 3 = 90^\circ$.

2. Using the property of the rhombus: Diagonals bisect opposite angles equally. So, each angle of the rhombus will be split into two equal parts by the diagonals.

3. Given angle $\angle 2 = 53^\circ$: This is half of the total angle at the vertex between sides 2 and 4.

   • Therefore, the total angle at that vertex is $2 \times 53^\circ = 106^\circ$.
   • This also means $m\angle 1 = 106^\circ$, because diagonals bisect angles of a rhombus equally.

4. Remaining angles: The other two angles (where sides 1 and 3 meet) must be supplementary to $106^\circ$, meaning they sum to $180^\circ$ with it.

   • The total opposite angle is $180^\circ - 106^\circ = 74^\circ$.
   • Since diagonals bisect the angles, $\angle 4 = \frac{74^\circ}{2} = 37^\circ$.

Final Angle Measures:

• $m\angle 1 = 106^\circ$
• $m\angle 2 = 53^\circ$
• $m\angle 3 = 90^\circ$
• $m\angle 4 = 37^\circ$

Let me know if you'd like further explanation! Here are some related questions:

1. What are the properties of the diagonals of a rhombus?
2. How do the diagonals of a rhombus bisect angles?
3. Why are diagonals in a rhombus perpendicular?
4. How can supplementary angles help in finding unknown angles in polygons?
5. How do you calculate angles in other quadrilaterals with diagonals?

Tip: In any rhombus, the diagonals bisect each other at 90°, a useful property for calculating angles.