Motivation: Let $a = \cos(\alpha), b = \cos(\beta),c = \cos(\gamma)$ where $\alpha,\beta,\gamma$ are the angles in a triangle. Then by $$\alpha+\beta+\gamma = \pi$$ and using $\cos(x+y) = \cos(x)\cos(y)-\sin(x)\sin(y)$, $\sin(\operatorname{acos}(x) = \sqrt{1-x^2}$ we find that the $a,b,c$ are points on the surface:
$$a^2+b^2+c^2+2abc-1=0$$
What is known about this surface, or what is it called?
a plot of the surface with Wolfram Alpha
Edit: With the help of @GEdgar I think I found the name of this surface by looking at the plots of both surfaces. (Cayley's nodal cubic surface).
Now the question remains how to show algebraically that they are the same?
Edit: Since every three point metric space can be isometrically emdedded in $\mathbb{R}^2$, we can build the possibly to a line degenerated triangle from these three points:
Using the law of cosines to define angles, given distances, we find that the quantities:
$$a:=\frac{d(y,z)^2+d(y,x)^2-d(x,z)^2}{2d(y,z)d(x,y)}, $$ $$b:=\frac{d(y,z)^2+d(z,x)^2-d(x,y)^2}{2d(y,z)d(z,x)},$$ $$c:=\frac{d(x,z)^2+d(y,x)^2-d(y,z)^2}{2d(x,z)d(x,y)}$$ satisfy by what was given above the following equation:
$$a^2+b^2+c^2+2abc-1=0$$
hence are points on the Cayley's nodal cubic surface. For instance for the metric on natural numbers $$d(x,y) = \sqrt{\sigma(x)+\sigma(y)-2\sigma(\gcd(x,y))}$$ and for three pairwise distinct primes $p,q,r$ we get the following nice formula:
$$\pi = \operatorname{acos}(\frac{r}{\sqrt{(p+r)(q+r)}})+\operatorname{acos}(\frac{q}{\sqrt{(p+q)(q+r)}})+\operatorname{acos}(\frac{p}{\sqrt{(p+r)(p+q)}})$$
Setting
$$a = \frac{r}{\sqrt{(p+r)(q+r)}}, b= \frac{q}{\sqrt{(p+q)(q+r)}}, c=\frac{p}{\sqrt{(p+r)(p+q)}}$$
we see that those are points on the Cayley nodal cubic surface.