## Trigonometry 7th Edition

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