Cubing Makes Perfect

Sorry guys for not posting my own problem for 3 month. I got a hard time on school.

Never mind about that, let's get started

Assume that the side of the cube is $a$

Now let's find $CB$

$CB = \sqrt {a^2 + a^2 }$

$CB = \sqrt{2a^2}$

$CB = a\sqrt{2}$

Now find $CA$

$CA = \sqrt{CB^2 + a^2}$

$CA= \sqrt{ a^2 + a^2 + a^2}$

$CA = a \sqrt{3}$

Now use Trigonometry to find $\measuredangle ACB$

$\frac{a \sqrt{2}}{a \sqrt{3}} = \cos \measuredangle ACB$

$\frac{\sqrt{2}}{\sqrt{3}} = \cos \measuredangle ACB$

$\measuredangle ACB = 35.26438968275...^\circ$

$\measuredangle ACB \approx 35.26^\circ$

A, B, C form a right triangle. Ratio of AB to BC (sidelength to diagonal) is 1/sqrt(2). Thus angle alpha equals inverse tangent of that ratio =35,26°

nice solution