Imagine a setup like this:
a b c are sides of a triangle,O is an arbitary point in 3D space,x y z are lines bewteen O and the vertex of the triangle, $\alpha$ $\beta$ $\gamma$ are angles between x y z.
From Law of cosines we have the following equations
$\frac{x^2 + y^2 -a^2}{2xy}=cos(\alpha)$
$\frac{y^2 + z^2 -b^2}{2yz}=cos(\beta)$
$\frac{z^2 + x^2 -c^2}{2zx}=cos(\gamma)$
Can we solve $x, y, z$ in terms of $a, b, c$ and $\alpha$, $\beta$, $\gamma$ ?
Edit: Here is a bit of more support information look at the 3 torus intersecting each other. The red one will be the x y which satisfy equation one, the green for equation two, blue for equation three. Where three color meets altogether is the answer we looking for. There are two on this side of the triangle and there is two mirror image on the other side. So out of 8 possible solutions, we should have two left. Yet still, how do we express them in mathematical terms.
