I'm currently trying to find the values of two angles (a and b) in this quadrilateral, where I know the diagonal (dashed) length and the lengths of two of the sides. These two side lengths are equal in length to one another and the other diagonal:

enter image description here

So far I've been able to express the values of the sides opposite to a and b in terms of the known side lengths and the angles, but I don't know how I can relate this to the (dashed) diagonal length, because this quadrilateral will not always be a parallelogram or a kite. Furthermore, the two diagonals wouldn't always intersect at their midpoints.

Is there any way that I can calculate the values of a and b using the values that I already have?

  • $\begingroup$ I wouldn't tag this algebraic-geometry :) $\endgroup$ – M.B. Feb 7 '14 at 21:12
  • $\begingroup$ By symmetry, I'd guess that $a = b$ (if not, there won't be a unique solution). And again, in that case the diagonals bisect. If not, all you'll find is a relation between $a$ and $b$. $\endgroup$ – vonbrand Feb 8 '14 at 0:06
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    $\begingroup$ @vonbrand Unfortunately symmetry is not guaranteed, and I have now come to the conclusion that I need to know at least one of the angles in order to be able to solve it. $\endgroup$ – Thomas Denney Feb 8 '14 at 9:04

As stated the problem can have many solutions.

For example:

Let the quadrilateral $ABCD$ such that $AD=DB=BC=x$ and $AC=\sqrt{3}x$. See figure 1.

enter image description here

Figure 1

The angle $b$ depends on $a$ as the expression:

$$b = \frac{a}{2}- \frac {\pi}{2}+ \arccos(- \frac{1}{2 \sin {\frac{a}{2}}}+ \sin {\frac{a}{2}})$$

So we have many solutions, for example: $a=\frac{\pi}{3}$ and $b=\frac{\pi}{3}$ or $a=\frac{\pi}{4}$ and $b=\frac{\pi}{2}$ as shown in figure 2.

enter image description here

Figure 2

  • $\begingroup$ How does your formula for b take into account the value of AC? It could be anything between x and 3x. $\endgroup$ – Thomas Denney Feb 9 '14 at 13:52
  • $\begingroup$ @ProgrammingThomas The formula is derived from the example I have given. $\endgroup$ – RicardoCruz Feb 9 '14 at 13:55
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    $\begingroup$ @ProgrammingThomas If you want a general formula, you can take $AC=kx$ (for $1 \le k \le 3$) and you will get: $b = \frac{a}{2}- \frac {\pi}{2}+ \arccos(- \frac{(k^2 - 1)}{4 \sin {\frac{a}{2}}}+ \sin {\frac{a}{2}})$. $\endgroup$ – RicardoCruz Feb 9 '14 at 14:13

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