# Finding an angle in a triangle?

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

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.

• Hey and Welcome to MSE! Could you please tell us where are you stuck? What have you tried so far? – Vinyl_cape_jawa Mar 4 '19 at 12:08
• 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. – Vinyl_cape_jawa Mar 4 '19 at 12:14
• @Vinyl_coat_jawa I understand what you're saying, but there is no additional information that I have not already provided. – Ruby Pa Mar 4 '19 at 12:15
• But the how do you know these angles? – Vinyl_cape_jawa Mar 4 '19 at 12:26
• Where exactly is point $D$ ? – Pixel Mar 4 '19 at 13:51

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.
Note that $$\angle DAB = \arctan(\dfrac{1}{2})\\$$ (two units over four ones). Thus,
$$\angle ADC = 180°- (\angle DAB+ \angle CBA) = 180° - \arctan(\dfrac{1}{2})-\arctan(2) = 90°.$$