Geometry: Common Core (15th Edition)

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

Chapter 6 - Polygons and Quadrilaterals - 6-4 Properties of Rhombuses, Rectangles, and Squares - Practice and Problem-Solving Exercises - Page 380: 16


$m \angle 1 = 60^{\circ}$ $m \angle 2 = 90^{\circ}$ $m \angle 3 = 30^{\circ}$

Work Step by Step

Diagonals of rhombuses cross each other at right angles, so $m \angle 2 = 90^{\circ}$. Let's look at one of the smaller triangles. We have one angle measuring $90^{\circ}$ and an angle that measures $60^{\circ}$; therefore, we only have to find the angle adjacent to $\angle 3$. The interior angles of a triangle add up to $180^{\circ}$, so let's set up an equation where we can find the measure of one of the interior angles given the measures of the other two angles: $m$ third angle = $180 - (90 + 60)$ Evaluate parentheses first, according to order of operations: $m$ third angle = $180 - (150)$ Subtract to solve: $m$ third angle = $30$ Now that we know the angle adjacent to $\angle 3$, we also know $m \angle 3$ because the two angles are part of the larger angle that is bisected by a diagonal, so these angles are congruent: $m \angle 3 = 30^{\circ}$ For $m \angle 1$, it is also $60^{\circ}$ because it is adjacent to the angle marked $60^{\circ}$ and is the resulting angle when that bigger angle is bisected by the diagonal. $m \angle 1 = 60^{\circ}$
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