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I can calutate the space diagonal of a cubic box using Pythagoras theorem twice. However, how can I find the space diagonal with trigonometry?

I tried this: A cubic box (picture) with sides $a$. The triangle in the bottom have hypotenuse $x$ \begin{align*} \cos(45^\circ)=\frac{a}{x} \\ x=\frac{a}{\cos(45^\circ)}=\sqrt{2}a \end{align*}

The triangle inscribed in the cube have height $a$ and base $x=\sqrt{2}a$. Call the space diagonal $y$ so \begin{align*} \cos(45^\circ)=\frac{x}{y} \\ y=\frac{x}{\cos(45^\circ)} = \frac{\sqrt{2}a}{\frac{\sqrt{2}}{2}}=2a \end{align*} This is wrong, what have I missed?

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  • $\begingroup$ $x$ is longer than $a$ so the final angle is not 45 degrees? $\endgroup$ – turkeyhundt Sep 14 '16 at 15:32
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long diagonal angle should be $$\tan { y } =\frac { a }{ x } =\frac { a }{ \sqrt { 2 } a } =\frac { 1 }{ \sqrt { 2 } } $$ $$\tan ^{ -1 }{ \left( \frac { 1 }{ \sqrt { 2 } } \right) ={ 35.264389682755 }^{ \circ } } $$ not ${ 45 }^{ \circ }$

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Your mistake was in assuming that angle is $45^{\circ}$. The "space cross section" forms a non-square rectangle ($x \neq a$). In a non-square rectangle, a triangular dissection via a diagonal does not give you $45^{\circ}$ corner angles (although they are complementary).

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