I'm trying to determine the angle between an edge of a unit cube and either space diagonal which shares an end point with said edge (to the end that I may represent a vertex of the unit cube in spherical coordinates.)(adsbygoogle = window.adsbygoogle || []).push({});

My work: A face diagonal of the unit cube is of length [tex]\sqrt{1^2+1^2}=\sqrt{2}[/tex] by the Pythagorean theorem, now the right triangle having an edge and a face diagonal for its legs and a space diagonal as its hypotenuse gives the space diagonal's length as being [tex]\sqrt{1^2+( \sqrt{2} )^2}=\sqrt{3} [/tex] again by the Pythagorean theorem. Hence the desired angle is given by [tex] \mbox{arcsin} \left( \sqrt{\frac{2}{3}}\right)[/tex] or, equivalently, by [tex]\mbox{arccos} \left( \frac{1}{\sqrt{3}}\right)[/tex], yet I cannot determine an exact value for this angle (which I expect to be able to find given the simple geometric nature of the problem,) a numeric approximation of this angle is 54.7357... degrees.

Did I make a mistake? Is there some trick to this?

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: It should be simple: it's only trig, right?

**Physics Forums | Science Articles, Homework Help, Discussion**