<s>[INST] Find possible coordinates of a triangle that I know only the sides length I want to find the possible coordinates of each vertex of a triangle of which I know only the sides lengths like (3,4,5).
To find the first edge, I let E1(0;0) and the second E2(3;0). but I have a problem to find the 3rd vertex.How can I find the 3rd ? 
I know that we can find the angles with trigonometry but it's only for a right triangle, so I'm a little bit lost.
 [/INST] Let in triangle $ABC$, we know the side lengths $AB=c$, $AC=b$ and $BC=a$. We look for coordinates of vertices $A$, $B$ and $C$.
So, without loss of generality, we can take $B=(0,0)$, and then $C=(0,a)$ because $BC=a$.
We located to vertices $B$ and $C$ and we need only locate vertex $A$. To do so, suppose the circles $$x^2+y^2=c^2\quad \text{and}\quad (x-a)^2+y^2=b^2$$
then the intersection of this two circles gives the location of $A$.
Note that we have $$x^2+y^2=c^2\quad \text{and}\quad x^2+y^2-2ax+a^2=b^2$$
therefore $$c^2-2ax+a^2=b^2$$
or $$x=\frac{c^2+a^2-b^2}{2a}$$
and thus since $x^2+y^2=c^2$, $$y=\pm\sqrt{c^2-\frac{(c^2+a^2-b^2)^2}{4a^2}}$$
and the coordinates of $A$ is
$$\left(\frac{c^2+a^2-b^2}{2a},\pm\sqrt{c^2-\frac{(c^2+a^2-b^2)^2}{4a^2}} \right)$$
well, you can choose one of the signs $\pm$.
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