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: 17


$m \angle 1 = 90^{\circ}$ $m \angle 2 = 45^{\circ}$ $m \angle 3 = 45^{\circ}$

Work Step by Step

According to theorem 6-22, the diagonals of a kite form perpendicular angles. Therefore, $m \angle 1 = 90^{\circ}$. For the triangle containing $\angle 1$, we have the measure of two of the angles already. We have one angle measuring $90^{\circ}$ and another angle measuring $45^{\circ}$. Using the triangle sum theorem, we can subtract the sum of the two known angles from $180^{\circ}$ and get the measure of the third angle, $\angle 2$: $m \angle 2 = 180 - (90 + 45)$ Evaluate parentheses first: $m \angle 2 = 180 - (135)$ Subtract to solve: $m \angle 2 = 45^{\circ}$ We can see that we have two triangles that are congruent by SSS. Corresponding parts of congruent angles are congruent. Therefore: $m \angle 3 = 45^{\circ}$
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