Answer
$m \angle 1 = 77^{\circ}$
$m \angle 2 = 103^{\circ}$
$m \angle 3 = 103^{\circ}$
Work Step by Step
According to theorem 6-19, the base angles of an isosceles trapezoid are congruent; therefore, if one of the base angles is $77^{\circ}$, the other base angle, $\angle 1$, is $77^{\circ}$.
If we take a look at the diagram, we see that we essentially have two transversals cutting a pair of parallel lines. The angles formed by each transversal are, in actuality, same-side interior angles; these types of angles are supplementary. But first, we need to find the sum of the interior angles of a trapezoid. We do this using the following formula:
sum of interior angles of a polygon = $(n - 2)180^{\circ}$, where $n$ is the number of sides in the polygon
Replace the $n$ with $4$:
sum of interior angles of a polygon = $(4 - 2)180^{\circ}$
Evaluate what is in parentheses first:
sum of interior angles of a polygon = $(2)180^{\circ}$
Multiply to simplify:
sum of interior angles of a polygon = $360^{\circ}$
Let's set up the equation to find the other two base angles, $\angle 2$ and $\angle 3$, which are also base angles and, thus, are congruent:
$m \angle 2 + m \angle 3 = 360^{\circ} - (77 + 77)^{\circ}$
Evaluate what is in parentheses first:
$m \angle 2 + m \angle 3 = 360^{\circ} - (154^{\circ})$
Subtract to solve:
$m \angle 2 + m \angle 3 = 206^{\circ}$
Since $m \angle 2 = m \angle 3$, we just divide to get the measures of both angles:
$m \angle 2 = m \angle 3 = 206^{\circ} \div 2$
Divide to solve:
$m \angle 2 = m \angle 3 = 103^{\circ}$
