Given the length of four sides of a convex quadrilateral and knowing that two adjacent angles are equal, the quadrilateral is determined. I want to know whether there's a formula representing the measure of angles with length of sides.
I've tried setting up simultaneous equations using cosine formula. Cannot solve them on paper because of the cosines.
I've also tried Mathematica. Use Reduce on those simultaneous equations. Mathematica didn't give a result after hours of running. After providing values of four length, Mathematica gave a solution depending on the root of a eight degree equation. But for the several group of values I experimented, the equation contains only even degree terms, which means it's actually a four degree equation.
In the above diagram, we can assign coordinates thusly:
$$O = (0,0) \qquad A = (a,0) \qquad B = ( a - b \cos\theta, b \sin\theta) \qquad C = ( d \cos\theta, d\sin\theta )$$
Then the key relation is $$\begin{align} c^2 = |\overline{BC}|^2 &= \left(a - b\cos\theta - d\cos\theta\right)^2 + \left(b \sin\theta - d\sin\theta\right)^2 \\ &= a^2 + ( b + d )^2\cos^2\theta - 2 a (b+d) \cos\theta + (b-d)^2\sin^2\theta \\ &= a^2 + b^2 + d^2 + 2 b d \cos 2\theta - 2 a(b+d) \cos\theta \end{align}$$
Therefore, $$4 b d\cos^2\theta - 2 a (b+d)\cos\theta + a^2 - c^2 + ( b - d )^2 = 0 $$ so that $$\cos\theta = \frac{a(b+d)\pm \sqrt{(a^2-4bd)(b-d)^2+4bc^2d}}{4bd}$$ where, presumably, one root is always extraneous.
Once you have $\theta$, you can find the lengths of the diagonals, and then also the measures of the remaining angles.