## Trigonometry 7th Edition

Published by Cengage Learning

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

#### Answer

$\angle$ 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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