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 = $45^{\circ}$
