The cube in B&B has vertices labeled like this.

Then the faces are squares, so using the Pythagorean Theorem applied to triangle ABC, we have, if AB = s, then AC = s*sqrt 2. Also, ACGE is a rectangle with sides s and s*sqrt 2, so again by Pythagoras, AG = s * sqrt 3.

The diagonals AG and CE are the diagonals of this rectangle. Let O be the intersection of AG and CE. Let X be the foot of the perpendicular from O to EA.

In this figure we see lots of similar and congruent triangles. Since the arguments in these problems are informal (since we have no axioms about 3-space yet), we will just assert that the angle between two diagonals is EOA = 2* angle EOU = 2 * angle ECA (and leave the rest to the reader).

The angle between AC and BD, two diagonals on a square face, is a right angle. If angle EOA were a right angle, then angle ECA would be 45 degrees and EA/AC would = 1. But EA/AC actually equals sqrt (1/sqrt 2). So the angles are not equal.

Any pair of diagonals of the cube meet at the same angle. To start with, we see that any diagonal intersects AG in the same angle as CE. Pick any diagonal; one of its end points is at distance s from A and the other at distance s from G. Then again the end points of the diagonals form a rectangle of the same dimensions as ACGE, so the angles of intersection of the diagonals are the same. We can pick any other diagonal besides AG to start with , for the cube can be relabeled with any point as A and its opposite vertex G.

The angle between diagonals is 2* arctan sqrt (1/2) = 70.53 (to the nearest hundredth).