Answer
$m \angle 1 = 90^{\circ}$
$m \angle 2 = 60^{\circ}$
$m \angle 3 = 60^{\circ}$
$m \angle 4 = 30^{\circ}$
Work Step by Step
Diagonals of rhombuses bisect pairs of opposite angles, so $m \angle 4$ is also $30^{\circ}$.
Diagonals of rhombuses cross each other at right angles, so $m \angle 1 = 90^{\circ}$.
Let's look at one of the smaller triangles. We have $m \angle 1 = 90^{\circ}$ and the angle that measures $30^{\circ}$; therefore, we only have to find $m \angle 2$.
The interior angles of a triangle add up to $180^{\circ}$:
$m \angle 2 = 180 - (90 + 30)$
Evaluate parentheses first, according to order of operations:
$m \angle 2 = 180 - (120)$
Subtract to solve:
$m \angle 2 = 60^{\circ}$
Now that we know $m \angle 2$, we also know $m \angle 3$ because the diagonal bisected the angle containing $\angle 2$ and the angle containing $\angle 3$. These bisected angles are opposite angles, so $m \angle 3$ is the same as $m \angle 2$.
$m \angle 3 = 60^{\circ}$
