Effect of curvature of spacetime on intrinsic geometric properties (under general relativity) If a space has curvature, then the curvature can be seen intrinsically by finding sums of angles in triangles made of geodesics. Under general relativity, space-time is curved on local scales. On global scales, experiments have determined that space-time is almost totally flat.
If one were to, for each of three points in a gravitational gradient (like near a black hole), find the angle between the sight lines to the other two points and then add up all of the angles, would it add up to 180 (degrees) or perhaps it would show intrinsic curvature? 
Is the way of thinking about this good, because it is quite general and conceptual. I don't actually know the equations for general relativity, but this is what I thought of when I was thinking about intrinsic geometric properties.
 A: your question tackles some basic concepts of general relativity and hence pseudo-Riemannian manifolds. To answer the several parts your question, we will have to go a little into detail.
Limiting behaviour of general relativity
First of all, I have to confess that you somehow got the limiting behaviour of Einstein's theory of gravitation the other way round.
Locally, you will always be able to construct a Minkowski spacetime with metric
$$\eta = -c^2 dt^2 + d\mathbf{r}^2$$
such that your original spacetime metric
$$g=\eta + \text{higher order terms}_{ij}dx^i dx^j$$
is kind of Taylor expanded locally. This linearization works pretty well for "normal life" gravitation and you can explain gravitational waves or get Newtonian gravitation. How good a linear approximation can be will tell you the value of the scalar Riemann curvature $R = R^a_{\,a}$.
For long-range interactions, this picture fails as a Taylor expansion will be more and more inaccurate. Then, you have to take the full theory into account and can, under further assumptions, explain things corresponding to cosmology.
Measuring angles
First of all, we have to make a simplification. If you would assume that you want to measure the angles of your triangle on a dynamic spacetime, you would have some conceptual problems defining how to return to the point where you started, and, maybe worse: what is an angle then?
So, it is convenient to assume that you want to do that in a stationary spacetime. That means there exists a local isometry of the metric $\varphi$ and a corresponding Killing field $\xi = \dot{\varphi}(\tau)$ which is properly parametrized, say $g\left( \xi,\xi \right) = -c^2\,\forall\tau$, $$\varphi^\star g = g$$
Now we can define something like an ordinary angle which I guess you had in mind. Consider the metric $$\tilde{g}\left(X,Y\right) = g\left(X,Y\right) - \frac{1}{g(\xi,\xi)}g\left(\xi,Y\right)g\left(\xi,X\right)$$
which only takes the spatial part of the scalar product of the vectors $X$ and $Y$, then you can assign an angle $\theta$ to these quantities in the usual sense as
$$\cos \theta = \frac{\tilde{g}\left(X,Y\right)}{\sqrt{\tilde{g}\left(X,X\right)\cdot\tilde{g}\left(Y,Y\right)}}$$
Now you have to think of what $X$ and $Y$ should be. Since you want to compare these quantities at three different points, you will have to parallel transport them between these points using a geodesic motion, again in the 4D setup.
Light geodesics will be totally different from e.g. human ones. So, generally, the angles will depend on how you transport.
So, DarthPickley, if you have a light ray travelling on the event horizon of a Schwarzschild black hole, what would be the sum of the angles of a corresponding "triangle"?
Sincerely
Robert
