Solving 4 unknown angles in quadrilateral possible?
Given a quadrilateral with 4 fixed lengths, is there a way to solve for its angles?
For example, I have a quadrilateral with 4 sides:
698.8m
512.5m
511.9m
695.8m
How do we solve for its unknown angles?
When you connect four rods of given length cyclically with knee-joints the resulting structure will not be rigid. This implies that the four lengths do not determine the angles between successive rods. Instead you will get $2$ equations for $\alpha$, $\beta$, $\gamma$, $\delta$, and there is the extra condition $\alpha+\beta+\gamma+\delta=2\pi$. So there is one degree of freedom.
You obtain the two equations by computing the length squares $e^2$ and $f^2$ of the two diagonals by means of the cosine theorem in two ways each, and then forgetting about $e$ and $f$.
Edit. When just one solution is needed we can arbitrarily give one of the angles a reasonable value and then numerically compute the other three angles. E.g., let the angle $\alpha_{23}$ between $a_2$ and $a_3$ be $90^\circ$. Then one diagonal $d$ can be computed by means of Pythagoras' theorem, and it remains to determine the angles of two triangles whose sides are known. This can be done using the cosine law. E.g., the cosine of the angle $\alpha_{41}$ is given by $$\cos\alpha_{41}={a_4^2+a_1^2-d^2\over 2 a_4 a_1}\ .$$
There is no way to find the angles without an additional constraint. To see this, lie one edge horizontally, and attach two other edges at each side. Each of the two edges has one degree of freedom (namely the angle), but the forth edge gives only one constraint - namely the distance between the two end points.
Thus, you have two unknowns and only one equation, which means there is no unique solution for the angles.
To find "a" solution, just choose one of the angles arbitrarily, so you know the position of three points (say $A,B,C$) in the plane. To find the fourth point $D$, you have to solve the equations:
$$L_1^2 = (C_x-D_x)^2 + (C_y-D_y)^2$$ $$L_2^2 = (B_x-D_x)^2 + (B_y-D_y)^2$$ In the unknowns $D_x,D_y$. Once you find the coordinates, you can use the dot product to find the rest of the angles.
I don't think there's enough information unless you exclude reflex angles. E.g. the sides above could make something looking like a kite or an arrow-head.
Edit: There are an infinite number of solutions. Joining four pencils / pens you can make any number of kites, from long skinny ones to ones that are nearly rectangular. I.e. what nbubis and Christian Blatter said.
There is a systematic way to find all such quadrilaterals using algebra (as mentioned in the responses, the solution is not unique). It is easier to work symbolically; we assume that we are given $a$, $b$, $c$ and $d$ and are looking for a quadrilateral $ABCD$ so that the lengths of the sides are $a$, $b$, $c$ and $d$ starting with $|AB|=a$ and continuing cyclically. Then there are unique $p$, $q$, $p_1$ and $q_1$ so that the quadrilateral is congruent to the one with $A=(0,0)$, $B=(a,0)$, $C=(p,q)$ and $D=(p_1,q_1)$---we translate $A$ to the origin and then rotate to position $B$ on the positve real axis. We use Mathematica to solve $$b^2=(p-a)^2+q^2,\quad c^2=(p_1-p)^2+(q_1-q)^2,\quad d^2=p_1^2+q_1^2.$$ Mathematica can deal with this equation and produces a family of solutions which depend on one parameter. For concrete data as in the question one need only substitute these values in the equation and solve again. The output is too complicated to reproduce here but it can be computed using Mathematica or Wolfram Alpha in a few minutes.
As mentioned above, one requires one additional piece of information to determine the quadrilateral---for example one angle or one diagonal: these follow easily from the usual triangle constructions. More interestingly, if one knows the area, then one gets two solutions, which sometimes coalesce to single one. (There are constraints on the data, of course, which are required for any solution to exist, as in the case of many triangle constructions). This can be seen by adding the fourth equation $$2 F=p_1 q_1 + (a-p)q + (p-p_1)(q+q_1) $$ where $F$ is the area. The form of the solution show that the resulting quadrilaterals can be constructed using ruler and compass. We are using signed areas which allows non-convex quadrilaterals as solutions in some cases.