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 379: 13


$m \angle 1 = 32^{\circ}$ $m \angle 2 = 90^{\circ}$ $m \angle 3 = 58^{\circ}$ $m \angle 4 = 32^{\circ}$

Work Step by Step

Diagonals of rhombuses bisect pairs of opposite angles, so $m \angle 3$ is also $58^{\circ}$. 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 $m \angle 2$, the angle that is bisected by the diagonal that also gave us $\angle 3$; therefore, we only have to find $m \angle 1$. 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: $m \angle 1 = 180 - (90 + 58)$ Evaluate parentheses first, according to order of operations: $m \angle 1 = 180 - (148)$ Subtract to solve: $m \angle 1 = 32^{\circ}$ Now that we know $m \angle 1$, we also know $m \angle 4$ because the diagonal bisected the angle containing both $\angle 1$ amd $\angle 4$, so $m \angle 4$ is the same as $m \angle 1$. $m \angle 4 = 32^{\circ}$
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