a while back I created a tile made of arrows: 
I did it using a vector graphics software, without really understanding the properties of this shape.
Now, let's say I want to write a program to generate the shape, and not do it manually, how would I go about it?
So far, I managed to figure out some of the shape properties through measuring:
It is clear that $a=a$ & $b=b$, also, I noticed that a perfect hexagon blocking a perfect hexagon that in turn blocks a third perfect hexagon, define the $``width"$ of the "arrow head" (the image shows it more clearly).
So let's say I'm given the length of the arrow: $x$, how do I represent all the sides & angles with relation to $x$?
Those two numbers $a,b$ are the lengths of two sides of a triangle whose third side has length $\sqrt{3} x / 3$. This is enough to tell you that, within certain bounds, the numbers $a$ and $b$ can be chosen independently, and that all angles are determined by the choices of $a,b$.
First, to justify this statement about the triangle, you need to picture a fundamental domain $\triangle ABC$ for the symmetries of your tiling. In your picture with the colored $a,b,x$ labels, consider light brown colored $x$ segment, oriented towards the right, let its initial vertex be $A$ and let its terminal vertex be $B$. The fundamental domain is an isosceles triangle with base $\overline{AB}$, angles $\pi/6$ at $A$ and $B$, and opposite vertex $C$ with angle $2/\pi/3$. That point $C$ is located in your picture with the colored $a,b,c$ labels as follows: start at $A$, move along the adjacent edge labelled $b$ to its opposite endpoint which I'll denote $D$, then take a sharp left turn and move along the adjacent unlabelled edge of length $a$ to its opposite endpoint which is $C$. You can see, by following these directions in the big picture of the tiling, that $C$ is a point of order 3 rotational symmetry of the picture. The points $A,B$ are each points of order 6 dihedral symmetry.
With this picture of the fundamental domain in mind, you can compute using trigonometry that: $$\overline{AC} =\sqrt{3} x / 3 $$ You can also see from this description that there is a triangle with one side $AC$, one side $AD$ of length $b$, and one side $CD$ of length $a$.
Finally, using trigonometry you can compute what all the angles have to be. For instance, the exterior angle of the arrow at $D$ is the same as the interior angle $\theta$ of the $ACD$ triangle at $D$ which you can compute with the law of cosines $$\overline{AC}^2 = \overline{AD}^2 + \overline{CD}^2 - 2 \overline{AD} \, \overline{CD} \cos(\theta) $$ $$3x^2 = b^2 + a^2 - 2 ab \cos(\theta) $$ which you can solve for $\theta$.