Simple proof of existence of hyperbolic triangles I've studied the hyperbolic plane by analytically building up the hyperboloid model, the Klein—Beltrami disc, the Poincaré disc, and the half-plane model from scratch. Now I'd like to prove that, given $\alpha,\beta,\gamma>0$ with $\alpha+\beta+\gamma<\pi$, there exists a hyperbolic triangle with angles $\alpha$, $\beta$ and $\gamma$.
Is there an easy proof of this fact? I mean a proof that only starts from the basic Riemannian structure of the models: the knowledge what the geodesics are, an expression for the distance, the fact that some models are conformal. (Of course, I have found proofs in literature, but they tend to refer a lot to other results, requiring e.g. an extensive theory of trigonometry.)
 A: In the disc model take two lines $L_1,L_2$ emanating from the center $0$, and forming an angle $\alpha$ berween them. Now, consider the point $x_t\in L_1$ ad distance $t$ from $0$ and the line $L_3(t)$ emanating from $x_t$ and forming an angle $\beta$ with $L_1$. Let $y_t=L_3(t)\cap L_2$. Now, $y_t$ can be a single point or the empty set. If $y_t$ is a point, you get a triangle with angles $\alpha,\beta,\gamma(t)$.
You can check that $\gamma(t)$ varies continuously on $t$ and that
1) as $t\to 0$, $y_t$ is a point and the sum $\alpha+\beta+\gamma(t)\to\pi$
2) There is the first time $T$ such that $L_3(t)\cap L_2$ is empty.
As $t\to T$ you have $\gamma(t)\to 0$.
Given $\gamma<\pi-\alpha-\beta$, by means of the intermediate value theorem you see that there is $t$ so that $\gamma(t)=\gamma$
A: Yes you can 
Assuming the curvature to be $ \sqrt{-1 } $ the length of the side c is
$$ c = \operatorname{arcosh} \left( \frac{\cos \alpha \cos \beta  + \cos \gamma }{\sin \alpha \sin \beta} \right) $$
(found the formula in "the foundations of geometry and the non euclidean plane" by Martin corollary 32.16 )
so you can even construct your triangle :)
edit : corection of formula
The lenght of the sides are:
$$ a = \operatorname{arcosh} \left( \frac{\cos \beta \cos \gamma  + \cos \alpha }{\sin \beta \sin \gamma} \right) $$
$$ b = \operatorname{arcosh} \left( \frac{\cos \alpha \cos \gamma  + \cos \beta }{\sin \alpha \sin \gamma} \right) $$$$ c = \operatorname{arcosh} \left( \frac{\cos \alpha \cos \beta  + \cos \gamma }{\sin \alpha \sin \beta} \right) $$
then you can make the triangle (all sides and angles are known, what more do you need?)
