We are given a rhombus $QRST$ and need to find the measures of angles 1, 3, and 4. We are given that $m\angle2 = 31^\circ$.

1. Problem Description

We are given a rhombus $QRST$ and need to find the measures of angles 1, 3, and

4. We are given that $m\angle2 = 31^\circ$.

2. Solution Steps

A rhombus has the following properties:

* All four sides are congruent.

* The diagonals are perpendicular bisectors of each other.

* The diagonals bisect the angles.

Since the diagonals are perpendicular, we know that $m\angle1 = 90^\circ$.

Since the diagonals bisect the angles, $m\angle RTS = m\angle QTS$. So, $m\angle QTS = m\angle 2 = 31^\circ$.

Then, $\triangle RTS$ is an isosceles triangle because $RT=ST$. Also $m\angle TRS = m\angle 2 = 31^\circ$.

In rhombus $QRST$, we know that adjacent angles are supplementary. Also, opposite angles are congruent.

In triangle $RTS$, we have $m\angle TRS = m\angle RTS = 31^\circ$.

We can find $m\angle RST$:

$m\angle RST + m\angle TRS + m\angle RTS = 180^\circ$

$m\angle RST + 31^\circ + 31^\circ = 180^\circ$

$m\angle RST = 180^\circ - 62^\circ = 118^\circ$.

Since the diagonals bisect the angles, $m\angle3 = \frac{1}{2} m\angle RST = \frac{1}{2}(118^\circ) = 59^\circ$.

Since diagonals bisect the angles, $m\angle4 = m\angle3$, so $m\angle4 = 59^\circ$.

3. Final Answer

$m\angle1 = 90^\circ$

$m\angle3 = 59^\circ$

$m\angle4 = 59^\circ$