How to find lengths and corner coordinates of an irregular pentagon How do I find the side lengths and therefore corner coordinates of a pentagon with the following internal angles:
A = 140°, B = 60°, C = 160°, D = 80°, E = 100°

?
The pentagon is described here, but only by it's angles. I've tried to classify it into one of the 15 monohedral pentagons to get some relation between the sides, but with no luck.
Some background: I'm interested in writing some code to draw tessellating pentagons, but without a definition of the dimensions of the base shape I'm unable to get anywhere.
I'm particular interested in this pentagon from this paper:

(Thanks, and please be gentle, I haven't done much maths since university.)
 A: Angles alone are not enough to specify a pentagon; we also need three side lengths.
Now, from the picture here, these pentagons are equilateral; the rotational symmetry at the center tells us that the two sides at the $60^\circ$ angle are equal, the next ring tells us that those two sides are the same as the two sides adjacent to them, and the fifth side is determined by the other four. That's enough to specify them up to scaling.
Each pentagon can be realized as an equilateral triangle glued onto one side of a rhombus with $80^\circ$ and $100^\circ$ angles:

That was drawn with a programming language, Asymptote. The code is reproduced in the blockquote below:

size(6 cm);
pair A = origin;
  pair B = dir(30);
  pair C = dir(90);
  pair E = dir(170);
  pair D = C+E;
draw(A--B--C--D--E--cycle);
  draw(A--C,dotted);

Those have side length $1$ in the coordinate system used, although nothing was included to mark that scale.
A: Each pentagon consists of an equilateral triangle and a rhombus, see figure (and the Note at the end)  To obtain a desired tiling, you can consider the length of each side equal $1.$
The center of the configuration is in $0.$ Starting with two points (here $0$ and $1$), any further point is obtained by a rotation of a point we already have. With the use of complex numbers has a rotation a simple formula
$$z'-z_0=e^{i\varphi}(z-z_0),$$ where $z_0$ is ordinate of the center of the rotation, $z$ is that of the point you want to rotate, and $z'$ is the obtained image of $z.$
Complex ordinates of some points are enclosed.
Note added: there is a typo in the picture, the most right point is obtained by a clockwise rotation (negative angle), the right term is $e^{-i10\pi/18}.$
