Imagine such a triangle:
We know the differences in distances: $\overline{OA} - \overline{BO}$ and $\overline{CO} - \overline{BO}$, as well as the distances between the points on the base: $\overline{AB}$ and $\overline{BC}$.
How do we solve for the triangle (or the point $O$ relative to any of the point $A$, $B$, or $C$)?
And beyond this, is there a solution that is computationally simple (e.g. fast to compute in a program)?
Given distances $OA-OB=p,\ OC-OB=q,\ AB=n,\ BC=m$, solve the triangle $\triangle OAC$.
To solve the triangle $\triangle OAC$ it is sufficient to find $OB$, let's call it $r$, since it is a radius of the circle centered at (yet unknown) point $O$ that passes through the points $P,\,B,\,Q$.
Stewart's Theorem is of a great help here, it helps to bind a radius $r$ with the known values $p,q,m$ and $n$ simply as \begin{align} m(r+p)^2+n(r+q)^2=(m+n)(r^2+m n), \end{align} which gives a neat expression for the radius: \begin{align} r&=\frac{1}{2} \frac{m n (m+n)-m p^2-n q^2}{m p+n q}. \end{align}