This would be a pain to clearly write out, so I've made a picture of the exact set up:

I need to find the angle of elevation from point C. It's supposed to be $75^\circ$. I've tried using $\sin$, $\cos$ and $\tan$ and the given angles with the $20$m on the one side of each to slowly work my way through it, but I just can seem to get to that answer.


Denote the distance from point $C$ to the right angle by $x$. We can express length of the vertical segment in terms of $x$ in two different ways. We get an equation for $x$: $$x+40m=\sqrt{3}(x+20m).$$ Thus $x=10(\sqrt 3-1)m$ which implies that $\tan \alpha = \frac{x+20m}{x} = 2+\sqrt 3$ so $\alpha = 75^\circ$.

Pure geometrical approach is the following: consider square $ABCD$. Draw equilateral triangle $ADE$ such that $E$ lies inside square $ABCD$. Let $CE$ intersects $AB$ at $F$. Let $G$ be a point on the ray $BA$ such that $\angle BGC=30^\circ$.

enter image description here

We have $\angle BAE=90^\circ - 60^\circ=30^\circ=\angle BGC$ so $AE \parallel GC$ and since $E$ is midpoint of $FC$, $A$ must be midpoint of $GF$. You can easily see now that this is exactly the same configuration as on your drawing. The angle you look for is $\angle BFE$. Observe that triangles $BAE$ and $FEB$ are isosceles. Thus $\angle BFE = \angle EBF = \frac 12 \cdot \left( 180^\circ - \angle BAE \right) = \frac 12 \cdot \left( 180^\circ - 30^\circ\right) = 75^\circ$.

  • $\begingroup$ Eh... I kind of see where your going there, just getting a bit lost about half way through the last part on how that all ends up working out. :/ $\endgroup$ – windy401 Feb 12 '15 at 5:44
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    $\begingroup$ Where did you get lost? I'll try my best to explain it more clearly. $\endgroup$ – timon92 Feb 12 '15 at 10:38
  • $\begingroup$ Umm... well.. I see how you're drawing the new lines and triangles to get new stuff to use, but.. Ok.. I see little bit more now after staring at it longer but still not fully. I see how BAE and FEB are isosceles triangles, but on the last part... Yea, EFB and EBF are going to be equal, but how do you get that BAE is equal, the right thing to be taken out from 180 when figuring out the FEB/BEF angle so you can divide the rest by two to find the two equal angles? $\endgroup$ – windy401 Feb 12 '15 at 22:29
  • $\begingroup$ That part yes. I know why EBF and BFE are equal, and each half of what you get from (180-BAE), but why BAE. How do you know that's and equal angle to subtract from 180. $\endgroup$ – windy401 Feb 12 '15 at 22:36
  • $\begingroup$ Well, triangle $BAE$ is isosceles, so $\angle AEB=\angle EBA$. The sum of angles of any triangle is equal to $180^\circ$, so $\angle BAE + \angle AEB + \angle EBA = 180^\circ$. Since $\angle AEB = \angle EBA$, it simplifies to $2\angle EBA = 180^\circ-\angle BAE$, which gives desired equality. $\endgroup$ – timon92 Feb 12 '15 at 22:39

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