In a disc model, would I be right in thinking that it is possible to have a triangle of total degrees 0 so long as each of it's vertices touch the boundary?
Explain what? The fact that the boundary circle is not included into the model? This holds by definition of the Poincaré Hyperbolic Disk: . I believe there are reasons for making such a definition; for example, according to the formula for distance, the distance between the origin and a point on the unit circle is infinity.
Therefore, the three arcs that are touching at points of the boundary circle are parallel lines in the model because they don't have common points.
Or do you need some explanation about the angles , , in the picture?
Another fact is about the area of triangles described in Wikipedia.
Therefore, the defect is always less than .Unlike Euclidean triangles whose angles always add up to 180 degrees or radians the sum of the angles of a hyperbolic triangle is always strictly less than 180 degrees. The difference is sometimes referred to as the defect. The area of a hyperbolic triangle is given by its defect multiplied by where . [K is a negative constant. --emakarov] As a consequence all hyperbolic triangles have an area which is less than .
That is not a picture of a triangle. The picture shows three "bounding parallels". By the definition of the Pioncare' Hyperbolic Disk, the points in the bounding circle are NOT part of the hyperbolic plane model so the three lines do not intersect and do not make a triangle. We can think of it as a "limit triangle" in the sense that, as you take the sides of an equilateral triangle larger and larger, that is what you get in the limit as the side lengths go to infinity.