Geometry: Common Core (15th Edition)

Published by Prentice Hall
ISBN 10: 0133281159
ISBN 13: 978-0-13328-115-6

Chapter 6 - Polygons and Quadrilaterals - 6-6 Trapezoids and Kites - Practice and Problem-Solving Exercises - Page 394: 12


$m \angle 3 = 60^{\circ}$ $m \angle 2 = 120^{\circ}$ $m \angle 1 = 120^{\circ}$

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 $60^{\circ}$, the other base angle, $\angle 3$, is $60^{\circ}$. 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, alternate interior angles; these types of angles are supplementary. Let's set up the equation to find the other two base angles, $\angle 1$ and $\angle 2$: $m \angle 2 = 180 - 60$ Subtract to solve: $m \angle 2 = 120^{\circ}$ If $m \angle 2 = 120^{\circ}$, then $m \angle 1 = 120^{\circ}$ because these two angles are base angles of the isosceles trapezoid and, thus, are congruent.
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