Suppose the Cartesian coordinates of the three points are $A,B,C,\|A\|=\|B\|=\|C\|=1$.
Given $A$,$B$ and $b=dist(B,C)$, $c=dist(A,C)$, how to find $C$?
Here, $dist(\cdot,\cdot)$ is the distance on the sphere, which is $dist(A,B)=\cos^{-1}(A\cdot B)$.
Can we find an accurate analytical form, other than just projecting the points to the tangent plane as an approximation and then projecting the third point back to the sphere?
Let $V=A\times_3B$, so that $A,B,V$ form a basis for $\mathbb R^3$. That means you should be able to write
$$C=\alpha A+\beta B+\gamma V$$
for some scalar coefficients $\alpha,\beta,\gamma$. Taking dot products, we get three scalar equations:
$$A\cdot C=\alpha+\beta A\cdot B+0$$ $$B\cdot C=\alpha A\cdot B+\beta+0$$ $$V\cdot C=\;0\;+\;0\;+\gamma\lVert V\rVert^2$$
which you can solve for $\alpha,\beta,\gamma$ (e.g. using Cramer's rule), provided that you know all the other terms. But what are these other terms?
You already know $A\cdot B,\,A\cdot C,\,B\cdot C$, and the last term $\lVert V\rVert^2$ can be gotten from the identity
$$(A\cdot B)^2+\lVert A\times_3B\rVert^2=\lVert A\rVert^2\lVert B\rVert^2$$ $$\lVert V\rVert^2=1-(A\cdot B)^2.$$
Finally, $V\cdot C$ can be gotten from that same identity (along with bac-cab):
$$(C\cdot V)^2+\lVert C\times_3V\rVert^2=\lVert C\rVert^2\lVert V\rVert^2$$ $$(C\cdot V)^2+\lVert A(C\cdot B)-B(C\cdot A)\rVert^2=1-(A\cdot B)^2$$ $$(C\cdot V)^2+(C\cdot A)^2+(C\cdot B)^2-2(C\cdot A)(C\cdot B)(A\cdot B)=1-(A\cdot B)^2$$ $$C\cdot V=\pm\sqrt{1+2(A\cdot B)(A\cdot C)(B\cdot C)-(A\cdot B)^2-(A\cdot C)^2-(B\cdot C)^2}.$$
Putting all these pieces together, we find
$$C=$$ $$\frac{(A\cdot C)-(A\cdot B)(B\cdot C)}{1-(A\cdot B)^2}A$$ $$+\frac{(B\cdot C)-(A\cdot B)(A\cdot C)}{1-(A\cdot B)^2}B$$ $$\pm\frac{\sqrt{1+2(A\cdot B)(A\cdot C)(B\cdot C)-(A\cdot B)^2-(A\cdot C)^2-(B\cdot C)^2}}{1-(A\cdot B)^2}(A\times_3B).$$