## Geometry: Common Core (15th Edition)

$m \angle 1 = 77^{\circ}$ $m \angle 2 = 103^{\circ}$ $m \angle 3 = 103^{\circ}$
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: $m \angle 2 = m \angle 3 = 103^{\circ}$