What is the sum of the angles of a triangle on Earth orbit? Gauss went out and measured triangles made up of mountain peaks to show that the angles sum up to 180 degree.
However, general relativity leads to non-Euclidian space and I would like to get a better feeling for it. 
What would be the sum of angles in a triangle that has all corners on earth's orbit (and with sides of equal length)? The sides are defined by light beams traveling between the corners.
 A: Calculating the sum of the interior angles precisely woud be a big task as we'd need to compute the trajectory of the light ray and there isn't a convenient analytic expression for this. However we can easily calculate an upper limit for the interior angles.
The key fact we need to know is that the deflection angle $\theta$ of a light ray in the Schwarzschild metric is given by:

$$ \theta = \frac{4GM}{c^2r} \tag{1} $$
Strictly speaking this is an approximation that fails when the curvature is high, but at the Earth-Sun distance the curvature is so small the the approximation is essentially exact.
The reason this applies to your question can be seen by drawing out the experiment you describe:

We take three points around the Earth's orbit that form the corners of an equilateral triangle and shine the light ray between the corners. Because the light ray is deflected by the Sun we have to shine the light ray outwards a bit (I've grossly exaggerated in the diagram!). Equation (1) gives us the total deflection of a light ray that comes in from infinity and departs to infinity. Our light ray will have a deflection smaller than this because it neither comes from infinity nor ends up at infinity, but we can use equation (1) as an upper limit for the deflection angle we observe. So let's calculate that upper limit.
The angle between the light ray and the side of the triangle is just $\theta/2$, and there are six such angles, so the sum of the interior angles exceeds $\pi$ by $3\theta$. The upper limit for the sum $\Theta$ is therefore:
$$ \Theta \le \pi + \frac{24GM}{c^2r} $$
where $r$ is the radius of the Earth's orbit and I've made the approximation that the distance of closest approach is $r/2$. It remains only to feed in the values for the various parameters, which gives us:
$$ \Theta \le \pi + 0.00000024 $$
A: I was writing my answer for this question, but before I could save it the question got closed because it was a duplicate of this one here and also somehow scrawled. But since the other question is about black holes, and this here for an orbit around the earth, I'll still post my answer here since the the effects around a black hole are much more visible.
The setup is this: observer $\rm A$ hovers at the photon sphere at $\rm r=3 GM/c²$ and sends a tangential light ray to observer $\rm B$ and a radial one to observer $\rm D$. When observer $\rm B$ receives the signal, he reflects it by $\rm 90°$. Observer $\rm C$ receives this signal and again reflects it by $\rm 90°$ to send it to observer $\rm D$.
In an euclidean space, the angle between the rays $\rm D$ receives from $\rm A$ and $\rm C$ would also be $\rm 90°$; in the curved spacetime around a black hole it is obviously less. The exact angle difference depends on the distance from the black hole and the area covered between the rays (the closer to the central mass and the larger the covered area, the larger the difference).
Raytracing in Schwarzschild-Droste coordinates:

On the other hand, if you encompass the black hole instead of setting up the experiment next to it, the sum of the angles would be greater than in euclidean space (see John Rennie's answer), and when you get as close as the photon sphere, the angle in every corner will blow up to $180°$ (wich means the corners will disappear), for the reason that the straight line of a light path is curved to a circle there.
