How to compute the side length of an equilateral hyperbolic triangle from its angle? In the hyperbolic plane we can have equilateral triangles with any angle smaller than $\pi/3$. The angle $\alpha$ determines their full shape. The area is easily obtained from $\alpha$ as $\pi - 3\alpha$. 
How can the side length be computed, or, essentially equivalently, how can such a triangle be constructed, e.g. in the Poincaré disk model of in the upper half plane?
 A: One of the Hyperbolic Laws of Cosines for triangles relates side-length $a$ to angles $\alpha$ (opposite $a$), $\beta$, $\gamma$ thusly:
$$\cos\alpha =-\cos\beta\cos\gamma + \sin\beta\sin\gamma \cosh a \tag{1}$$
So, for an equilateral triangle with $\alpha=\beta=\gamma$, we can solve to get
$$\cosh a = \frac{\cos\alpha(1+\cos\alpha)}{\sin^2\alpha} = \frac{\cos\alpha(1+\cos\alpha)}{1-\cos^2\alpha} = \frac{\cos\alpha}{1-\cos\alpha} \tag{2}$$
So, to construct such a triangle, we need to construct length $a$ satisfying $(2)$. At the moment, I can't think of a such a construction in the hyperbolic plane, so I'll provide a Euclidean construction in the Poincaré disk model, where one of the vertices of the triangle is taken to be the origin.
The key to constructing our side-length $a$ is constructing its corresponding "angle of parallelism". This angle, which we'll denote $\theta$, is adjacent to a leg of length $a$ in an infinite right triangle with convergently-parallel hypotenuse and "other" leg. If we can construct $\theta$, then it's straightforward to construct such a right triangle, hence the length $a$.
Length $a$ and its angle of parallelism $\theta$ are related by (among other identities)
$$\sin\theta = \operatorname{sech} a\tag{3}$$
Our length $a$ satisfies $(2)$, so we seek $\theta$ satisfying
$$\sin\theta = \frac{1-\cos\alpha}{\cos\alpha} \tag{4}$$
In the Poincaré Disk model (of radius $1$) with $\alpha = \angle ROP$, we readily determine such a $\theta$ by constructing a right triangle with leg $1-\cos\alpha$ and hypotenuse $\cos\theta$:

With $\theta$ in hand, we can rotate $R$ about $O$ by that angle to obtain $R'$. Let the tangent to the disk at $R'$ meet the extension of $\overline{OR}$ at $T$, and let the circle about $T$ through $R'$ meet $\overline{OR}$ at $U$. 

(An arc of) The circle $\bigcirc T$ represents a hyperbolic line convergently-parallel to $\overline{OR'}$, and this line is perpendicular to $\overline{OR}$. So, we have constructed an infinite right triangle $\triangle OUR'$ with acute angle $\theta$. The finite leg $\overline{OU}$ must have length equal to our desired $a$. 
A circle about $O$ transfers $U$ to $V$ on $\overline{OP}$, so that $\triangle OUV$ is our equilateral triangle with angle $\alpha$.
 
