What is largest possible sum of all interoir angles of a triangle on a torus? As discussed in the question Triangles on a Torus, the outside of a torus has positive Gaussian curvature. A triangle inscribed on the outer surface will have interior angles which add to be greater than 180 degrees.
By adjusting the geometry of the Torus, and the points which define the triangle, what is the maximum possible sum of interior angles of the triangle?
I suspect the fringe case occurs with a horn torus with vertices that lie on the axis between positive and negative curvature.
Edit: “Inside of the Torus” refers to the side of the torus facing inward “toward the donut hole.” Alternatively, it is simply the surface of the torus with negative Gaussian curvature.
Edit 2: I’m referring to one single triangle drawn anywhere on the surface of the torus, not breaking the surface into a number of smaller triangles
 A: Consider three points on the positive Gauss curvature outer $K>0$ region. Points (T,M,B) are at top, middle and bottom in poloidal meridian principal direction respectively on a circular torus on  "outside" viewing. Alternately we can also say $(T,B)$ to be points on the crown where $ K=0$ and $M$ is on outer equator, with $K>0$.
As a first guess in the curvilinear toroidal triangle:
Two geodesics helically going all around the axis of symmetry fully for $ \theta_{max}=2 \pi$ rotation clockwise MT,MB.
and,
TMB,  the poloidal meridian  for the third geodesic side...
may make up a maximum angle sum.
By Gauss Bonnet thm
$$ \int K dS + \int k_g ds + \Sigma_i= 2 \pi$$
$$ \Omega +0 + \Sigma_i= 2 \pi$$
$$ \Omega +3 \pi -(A+B+C)=2 \pi$$
$$ (A+B+C)=\Omega + \pi $$
so that the geodesic triangle should also have maximum integral curvature $\Omega.$Accordingly $\theta$ is chosen as its maximum $ 2 \pi$  in a single cover mentioned as basis at start somewhat like in the following crude sketch.
At $T,B$ geodesics run along circumferential lines so from Clairaut's Law we have
$$\sin \alpha=\frac{r_T}{r_M} $$
The internal angles sum up to $\pi+ 2\alpha$ where $ 2\alpha $ is angle at wedge M.
$$ R=r_T,\; \frac{R+r}{2}=r_M,\text { when}\; R\to\infty, \alpha= 30^{\circ},\text{angle sum} 
=180^{\circ}+ 2 \cdot 30^{\circ}= 240^{\circ}.$$

