Let the vertices of the cube be (0,0,0) (1,0,0) (0,1,0) (0,0,1) (1,1,0) (0,1,1) (1,0,1) and (1,1,1).

Then the diagonal from (0,0,0) to (1,1,1) is in the direction (1,1,1), and

the edge from (0,0,0) to (1,0,0) is in the direction (1,0,0).

Now the cos(theta) = (1,1,1).(1,0,0)/[|(1,1,1)| |(1,0,0)| = 1/sqrt(3)

where theta is the angle getween the diagonal and the edge.

So: theta = arccos(1/sqrt(3)) ~= 0.955 radian or 54.7 degrees

RonL