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Given the following picture, how do I find $\angle CDA$?

enter image description here

Attempt:

I have found the following angles, which I think may be useful:
$$ \angle BAC = \arctan(3),\quad \angle ABC = \arctan(2),\quad \angle BCA = \arctan(1). $$
But I don't know how to go on.

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  • $\begingroup$ Hey and Welcome to MSE! Could you please tell us where are you stuck? What have you tried so far? $\endgroup$ – Vinyl_cape_jawa Mar 4 '19 at 12:08
  • $\begingroup$ So this is not the original problem? In this case you shouid post the whole question. Maybe there are some other information there we could use. $\endgroup$ – Vinyl_cape_jawa Mar 4 '19 at 12:14
  • $\begingroup$ @Vinyl_coat_jawa I understand what you're saying, but there is no additional information that I have not already provided. $\endgroup$ – Ruby Pa Mar 4 '19 at 12:15
  • $\begingroup$ But the how do you know these angles? $\endgroup$ – Vinyl_cape_jawa Mar 4 '19 at 12:26
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    $\begingroup$ Where exactly is point $D$ ? $\endgroup$ – Pixel Mar 4 '19 at 13:51
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Assuming the points are at the exact grid points where it looks like they are, the line $AD$ is orthogonal to $BC$. The line $BC$ goes $2$ units down for each unit we go to the right, and if we turn that $90^\circ$, we get a line which goes $1$ unit up for every $2$ units to the right. That's what $AD$ does.

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  • $\begingroup$ @RubyPa I'm not assuming that in any way. I'm showing it. There may be details I've glossed over, but this approach is definitely viable. $\endgroup$ – Arthur Mar 4 '19 at 12:33
  • $\begingroup$ Ahhh yes I see what you mean. Thanks for your help! :) $\endgroup$ – Ruby Pa Mar 4 '19 at 12:36
  • $\begingroup$ Haha that's one sweet question you could do with the converse of Pythagorean theorem. $\endgroup$ – user231094 May 27 '19 at 10:31
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Another approach:

Note that $\angle DAB = \arctan(\dfrac{1}{2})\\ $ (two units over four ones). Thus,

$$\angle ADC = 180°- (\angle DAB+ \angle CBA) = 180° - \arctan(\dfrac{1}{2})-\arctan(2) = 90°.$$

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