Complicated Trigonometry Math Problem (for me at least) How would I calculate the angle in the picture (REF 94.61) with all of the defined parameters (highlighted in red). 

I have tried to use right angled triangle everywhere in this shape, but I just could not work it out.
Also note that the 10mm long legs are joined to the same line, which is the line angled 94.61 degrees from the large leg. Furthermore, the thickness of these legs (1mm in this example) are equal, but can change (i.e. if one is 3mm other needs to be 3mm).
Note: All of the parameters can change, see image with different inputs/ouput here: Shape with different inputs
 A: Here is how you can do it: draw a line thru the lowest point parallel to the ground. You will get right isosceles right triangle with legs equal to $\frac{18}{\sqrt{2}} \approx 12.73$. Now draw line thru $10$ and $20$ marks parallel to the ground. You will get a right triangle with legs $60-12.73-1-1$ (horizontal leg) and $10$. Now you can find $\tan \alpha=\frac{10}{45.27}$. The angle of interest is $180°-45°-\arctan{\frac{10}{45.27}} \approx 122.54 °$
A: Here is my solution.  Here is a picture.

$AE', AG, EF, DD'$ are horizontal, $CD', CG, DF, EE'$ are vertical, $AB$ is at angle $\pi/4 = 45^\circ$, $\angle ABC$ is a right angle.  I will solve for angle $\theta=\angle DEF = \angle CAG$.  Then the angle we want is $3\pi/4-\theta$.
In the original problem, we are given lengths $AB=18, DD'=1,AE'=1, DF=10$.  Using these, we get:
$\angle ACB = \theta+\pi/4$, so $AC = 18 \csc(\theta+\pi/4)$ and
$CG = 18 \sin\theta\csc(\theta+\pi/4)$.
In right triangle $\triangle D D' C$ we have $DD'=1$ so $CD' = \tan(\theta)$; similarly $EE'=\tan(\theta)$.
Subtracting, $DF = CG - CD' - EE' = 18 \sin\theta\csc(\theta+\pi/4) - 2\tan(\theta)$
The problem to solve is
$$
18 \sin\theta\csc(\theta+\pi/4) - 2\tan(\theta) = 10
$$
This has two solutions,

The one near $\pi/4$ is
$$
\theta = -\arctan \frac{\left( 3\,\sqrt {2}-9+\sqrt {89-54\,\sqrt {2}}
 \right) \sqrt {2}}{2}
$$
(We have to solve a quadratic equation for $\tan\theta$.)
We get $\theta \approx 0.7049$ so the angle we want is 
$3\pi/4 - \theta \approx 1.6513 \approx 94.6106^\circ$.

In the other example, things are the same except $DF=13$ by subtracting.
We must solve
$$
18 \sin\theta\csc(\theta+\pi/4) - 2\tan(\theta) = 13
$$
The solution near $\pi/4$ is
$$
-\arctan \frac{\left( 15\,\sqrt {2}-36+\sqrt {1538-1080\,\sqrt 
{2}} \right) \sqrt {2}}{8} 
$$
We get $\theta \approx 1.1145$, so $3\pi/4-\theta \approx 1.2417 \approx 71.146^\circ$
