Trigonometry 7th Edition

Published by Cengage Learning
ISBN 10: 1111826854
ISBN 13: 978-1-11182-685-7

Chapter 1 - Section 1.1 - Angles, Degrees, and Special Triangles - 1.1 Problem Set - Page 13: 72


$\angle$ GDH = $45^{\circ}$

Work Step by Step

In triangle GHD- $\angle$ GHD = $90^{\circ}$ [ Because GHDC is a face of a cube and hence is a square and all angles of a square are right angles] GH = DH ( Both are edges of a cube) Therefore triangle GHD is an isosceles right triangle. Now We know that angles opposite to equal sides of an isosceles triangle are also equal. Hence- $\angle$ GDH = $\angle$ HGD = $x$ (Let's assume) We know that sum of acute angles of a right triangle is 90 degrees, Therefore- $ x + x$ = $90^{\circ}$ Or, $ 2x$ = $90^{\circ}$ Or, $ x$ = $45^{\circ}$ Thus $\angle$ GDH = $45^{\circ}$
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