#### Answer

$m \angle 1 = 55^{\circ}$
$m \angle 2 = 35^{\circ}$
$m \angle 3 = 55^{\circ}$
$m \angle 4 = 90^{\circ}$

#### Work Step by Step

Diagonals of rhombuses cross each other at right angles, so $m \angle 4 = 90^{\circ}$.
Let's look at one of the smaller triangles. We have one angle measuring $90^{\circ}$ and an angle that measures $35^{\circ}$; therefore, we only have to find $m \angle 1$.
The interior angles of a triangle add up to $180^{\circ}$, so let's set up an equation where we can find the measure of one of the interior angles given the measures of the other two angles:
$m \angle 1 = 180 - (90 + 35)$
Evaluate parentheses first, according to order of operations:
$m \angle 1 = 180 - (125)$
Subtract to solve:
$m \angle 1 = 55^{\circ}$
Now that we know $m \angle 1$, we also know $m \angle 3$ because the two angles are alternate interior angles, and alternate interior angles are congruent:
$m \angle 3 = 55^{\circ}$
We can get $m \angle 2$ by using the triangle-sum theorem, which states that the sum of the interior angles of a triangle equal $180^{\circ}$. One angle is congruent to $\angle 1$ because they are the angles resulting from a diagonal bisecting one of the angles of a rhombus. This angle has a measure of $55^{\circ}$. The other angle is formed by the crossing of the two diagonals of a rhombus and is a right angle. Now that we have two angles, we can figure out the third angle:
$m \angle 2 = 180 - (90 + 55)$
Evaluate what's in parentheses first, according to order of operations:
$m \angle 2 = 180 - (145)$
Subtract to solve:
$m \angle 2 = 35^{\circ}$