Given angles/sides, is there a necessary and sufficient condition on whether polygon exists?

We are given the angles and side lengths of a polygon $a,b,...$ and $A,B,...$ like so: (the polygon has non-negative side lengths and is convex)

How can we find out whether a polygon with these angles/sides actually exists?

We could start at one vertex and go around the corners of the polygon using trig functions making sure that the final coordinates match the initial coordinates.

e.g. for the quadrilateral above

$c+d \cos(\pi-C) + a \cos(2\pi-C-D) + b \cos(3\pi-C-D-A)=0$ $d \sin(\pi-C)+a \sin(2\pi-C-D)+b\sin(3\pi-C-D-A)=0$

We could square and add these to get a single equation.

Are there any other necessary and sufficient conditions like that? I am especially interested in conditions that don't use trig functions, if there are any. Thanks!

• Angles should add up to $(n-2)\pi$, where n is the number of edges. As for the sides, any side should be shorter than the rest of them combined. These are the necessary conditions, albeit pretty trivial. – Ivan Neretin Sep 27 '15 at 18:02
• @IvanNeretin Yes, but they are not sufficient. $A=B=C=D=90^\circ$ and $a=b=c=1,d=2$. – Alex Sep 27 '15 at 18:07
• Sure, I never said they are. The sufficient conditions are presented in your post, I don't think they can be made any simpler. – Ivan Neretin Sep 27 '15 at 18:13
• By "polygon" are we to assume that degenerate polygons (such as a line segment traversed in both directions or even a single point joined to itself) and self-intersecting polygons are allowed? Are negative side lengths and angles outside the range $[0,2\pi)$ allowed? If so, the only requirement is that the initial point and terminal point are equal, which is true if their two Cartesian coordinates are equal. That is what you gave in your equations, so they would be necessary and sufficient. But you may want additional requirements, such as non-negative side lengths. – Rory Daulton Sep 27 '15 at 18:27
• @RoryDaulton Let's say the polygons have non-negative side-lengths and are convex. – Alex Sep 27 '15 at 18:34