Assume a simple graph, having a triangle in it. The points are $A, B$ and $C$. I know points $A(x,y)$, $B(x,y)$ and the distances between $A$, $B$ and $C$. I want to get the coordinates of $C(x,y)$.
Eg :- $A(x,y) = A(30,30)$, $B(x,y) = B(90,30)$. The distance between $A$ and $B$ is $60~\text{cm}$, $A$ and $C$ is $50~\text{cm}$ and $B$ and $C$ is $50~\text{cm}$. How to find $C(x,y)$?
Observe that point $C$ is $50~\text{cm}$ from both points $A$ and $B$.
The locus of points that is $50~\text{cm}$ from $A$ is the circle with radius $50~\text{cm}$ centered at $A$, which is $$(x - 30~\text{cm})^2 + (y - 30~\text{cm})^2 = (50~\text{cm})^2 \tag{1}$$ The locus of points that is $50~\text{cm}$ from $B$ is the circle with radius $50~\text{cm}$ centered at $B$, which is $$(x - 90~\text{cm})^2 + (y - 30~\text{cm})^2 = (50~\text{cm})^2 \tag{2}$$ Point $C$ must be located at one of the two points where these two circles intersect.
By subtracting equation 2 from equation 1, you can solve for the $x$-coordinate of $C$. You can then substitute your result into equation 1 or equation 2 to find the possible $y$-coordinates of $C$, of which there are two.
The two solutions to this problem are the points $(60, 70)$ and $(60, -10)$.
In the comments, you asked about solving the same problem for the points $A(20, 20)$, $B(60, 20)$ with $AB = 40$, $AC = 30$, and $BC = 60$.
To obtain the coordinates of point $C$, you need to solve the system of equations \begin{align*} (x - 20)^2 + (y - 20)^2 & = 30^2\\ (x - 60)^2 + (y - 20)^2 & = 60^2 \end{align*} You should obtain the points $(6.25, 46.66)$ and $(6.25, -6.66)$, where the $y$-coordinates are approximate.