Finding lengths in a diagram. Find $x$ and $y$.

Given that:
$$a_1+a_2+a_3-a_4+a_5=180^\circ$$
$$\text{cos}(180^\circ-a_5)=0.4$$
I managed to solved this problem using Mathematica based on an equation with whole bunch of ArcTan. However, I am looking for an easier way to solve it based on its geometrical properties.
Thank you!.
Remark: 


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*Thanks @Blue for editing the question!

*code for solving the problem in Mathematica:


ratio = Tan[ArcCos[0.4]];
Solve[ArcTan[ratio*x/(10-x)]+ArcTan[ratio*x/(3-x)]+Pi-ArcTan[ratio*x/(x-1)]+Pi-ArcCos[0.4]-(Pi-ArcTan[ratio*x/(x-0.1)])==Pi&&x>1&&x<3,x]

$$x=t\cos a_5,\\y=t\sin a_5 $$ where $a_5$ is given by the second equation.
Then there remains a sum of four angles which can indeed be expressed as arc tangents of terms $t_k:=y/(x-x_k)$.
As
$$\tan(a_1+a_2+a_3-a_4)=\frac{\tan(a_1+a_2)+\tan(a_3-a_4)}{1-\tan(a_1+a_2)\tan(a_3-a_4)}
=\frac{\dfrac{t_1+t_2}{1-t_1t_2}+\dfrac{t_3-t_4}{1+t_3t_4}}{1-\dfrac{t_1+t_2}{1-t_1t_2}\dfrac{t_3-t_4}{1+t_3t_4}}$$ you will get a ratio of cubic/quartic polynomials in $t$.
I don't see a possible simplification among the terms so this problem seems to be equivalent to the resolution of a quartic equation. There will be no easy shortcut.
