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Find the angle theta between edge $AB$ and the diagonal $AD$ of the cube.

This was a rather confusing problem. Number $26$. A student asked my teacher the problem and he was unable to find it out, the class had a discussion too, but we ended up with the same result. The thing we were able to figure out is in the work above. Does anyone know how to continue this problem?

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    $\begingroup$ $\widehat{BAD}=\arctan\sqrt{2}$, what's to continue? $\endgroup$ – Jack D'Aurizio Apr 17 '17 at 22:19
  • $\begingroup$ I deleted my post. As Jack suggested... first try out! $\endgroup$ – imranfat Apr 17 '17 at 22:20
  • $\begingroup$ As you can probably see, the answers to 25 and 26 are complementary angles. $\endgroup$ – Joffan Apr 17 '17 at 23:55
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$\theta = \angle BAD $, so let's get some lengths of sides in this triangle. $\lvert AB \rvert = s$ By Pythagoras, $\lvert BD \rvert = s\sqrt{2}$. But $\angle ABD$ is a right angle. So $$\tan{\theta} = \frac{\lvert BD \rvert}{\lvert AB \rvert} = \sqrt{2}. $$ $\arctan{\sqrt{2}}$ is not rational, so that's the simplest way to write it.

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