# Finding the angle of elevation, three points on ground know angles of two of them.

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 $\angle BGC=30^\circ$.
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 $\angle BFE = \angle EBF = \frac 12 \cdot \left( 180^\circ - \angle BAE \right) = \frac 12 \cdot \left( 180^\circ - 30^\circ\right) = 75^\circ$.
• 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. – timon92 Feb 12 '15 at 22:39