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 $E_1=(0;0)$ and the second $E_2=(3;0)$. 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.
Best Answer
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 $\pm$.