Let $\gamma$ be the circle of radius $5$ centered at $(0,0)$. Find the angle of parallelism between the point $(0,2)$ and the positive $x$ axis.
From this page, I understand that $$\sin(\pi(a))=\frac{1}{\cosh a}$$ where $a$ is the distance between a point and the foot of the perpendicular dropped to the ray. In my case, this would be the distance $(0,2)$ and $(0,0)$.
To find the angle of parallelism, would I need to compute the Euclidean distance, so $a=2?$ Or should I use the hyperbolic distance given here, which gives that $$a=|\ln(\frac{3\cdot5}{7\cdot5})|=|\ln(3/7)|$$ Does this formula for distance still apply given that we have modified the disk? Any help would be appreciated.
You can recover any information you want about lengths by explicitly finding the angle. I show how to do that below. In particular, after finding the angle, we get
$$ \cosh a = \frac{1}{\sin \pi(a)} $$ where my method recovers $\pi(a)$ as $\arctan$ of a rational number. A convenient alternative is $$ \sinh a = \frac{1}{\tan \pi(a)} $$
The original problem, point at $(0,2),$ we get Pythagorean Triple (20,21,29) (from parameters 5,2). The angle is $\arctan \frac{21}{20}$ so $\sinh a = \frac{20}{21}.$ My suggestion, point at $(0,4)$ we get Pythagorean Triple (9,40,41) (from parameters 5,4). The angle is $\arctan \frac{9}{40}$ so $\sinh a = \frac{40}{9}.$
The inverse function of $\sinh x$ is $$ \operatorname{arsinh} t = \log \left( t + \sqrt{t^2+1} \right), $$ using natural logarithm, base $e \approx 2.71828. \; $ The original problem asks for $ \operatorname{arsinh} \frac{20}{21},$ and this comes out to $$ \log \frac{49}{21} = \log \frac{7}{3}, $$ the absolute value you displayed in the question.
Conformal model. First, you need the geodesic passing through $(0,2)$ and tangent to the $x$ axis at $(5,0).$ Thus the center of the circular arc you need is along the line $x=5.$ Some $(5,c).$ The enclosing circle is $x^2 + y^2 = 25.$ Your circle, center at some $(5,c),$ has the same distance from $(5,0)$ and $(0,2).$ Meaning that $c^2 = 5^2 + ( c-2)^2,$ so that $c$ turns out to be rational, the squares cancel.
Next you need the geodesic through $(2,0)$ but meeting the $x$ axis orthogonally. Clear if you draw some pictures. What remains to be done is entirely in the ordinary Euclidean geometry of the diagram.
Oh, the angle of parallelism is the same as the angle between the two geodesics, some ordinary trigonometry. Again, conformal model, angles agree with angles
Here is how it looks if we move the point at $(0,2)$ to $(0,4).$ Everything still comes out rational, there are Pythagorean triples that are caused to appear...