This cube is drawn in perspective so don't expect all 
the right angles to look like right angles.



Suppose each edge of the cube is 1 unit long. Then since 
triangle ABC is a right triangle, its hypotenuse AC, the 
red line, is  by the Pythagorean theorem.

The triangle ACD is also a right triangle, and by the
Pythagorean theorem on it we get that AD²=AC²+CD² which
leads to AD = .

Now we draw in the green diagonal CE, and also the red line DE.



Now let's remove rectangle ACDE.

We want to find angle AOE.



tan(∠DAC) = 

∠DAC = 35.26438968°

∠EAD = 90°-35.26438968° = 54.73561032°

Since triangle AOE is isosceles, so ∠AEO = ∠EAD = 54.73561032°.

So Since the sum of the three angles of triangle AOE, 

∠AOE = 180° - ∠AEO - ∠EAD = 180° - 2(54.73561032°) = 70.52877937°

That's the answer, and we can draw in the other two diagonals of the cube.

All three diagonals intersect at the same point. And the acute angle 
between any two of the three adjacent diagonals is 70.52877937°.  



Edwin