Gauss has proven in his famous Theorema Egregium, that it is possible, to calculate the gaussian curvature from measuring angles and distances on the surface, irrespective of how the surface is embedded into space.
Question:
is it also possible, to calculate the euclidean distance of two points on the surface, also from distance- and angle-measurements on the surface, alone?
The answer is no. The fact that you can bend without stretching a piece of paper should convince you of that. Consider for instance a piece of a plane and piece of a cylinder.
The question you ask can be rephrased in the following way: take a 2 dimensional Riemannian manifold $(M^2,g)$. Under which condition is there a unique isometric embedding $f:M^2\to\mathbb{R}^3$ which is unique up to rigid motion ?
The only case where I know this is true is the case of convex surfaces: assume $M$ is diffeomorphic to $\mathbb{S}^2$ and has nonnegative Gauss curvature. Then Cohn-Vossen has proved that there exist a unique convex body $K\subset \mathbb{R}^3$ (up to rigid motion) such that $(M^2,g)$ is isometric to the boundary of $K$ with the induced Riemannian metric. This shows that, on a theoretical level the intrinsic metric on $M^2$ determines the distance of between two point in the extrinsic distance given by the embedding.
However, from what I know of the proof building the embedding from the metric is not easy at all and won't give any practical way to compute the extrinsic distance between two points in terms of the intrinsic geometry of the surface.
EDIT: I'll come back later to add some references.
My answer is contained in the one of Thomas which was posted when I wrote my answer.
No, consider the plane $\{(x,y,0)\mid x,y\in\mathbb,0<y<2\pi\}$ and the subset $\{(x,\cos y, \sin y)\mid x,y\in\mathbb,0<y<2\pi\}$ of the round cylinder in euclidean 3-space. Both are globally isometric, so all distances and angles are the same. Clearly, their euclidean distances are not the same.
