Editing Angle of parallelism (section)

==Demonstration==

[[Image:Angle of parallelism half plane model.svg|thumb|400px|right|The angle of parallelism, ''&Phi;'', formulated as: (a) The angle between the x-axis and the line running from ''x'', the center of ''Q'', to ''y'', the y-intercept of Q, and (b) The angle from the tangent of ''Q'' at ''y'' to the y-axis.<br>This diagram, with yellow [[ideal triangle]], is similar to one found in a book by Smogorzhevsky.<ref>A.S. Smogorzhevsky (1982) ''Lobachevskian Geometry'', §12 Basic formulas of hyperbolic geometry, figure 37, page 60, [[Mir Publishers]], Moscow</ref>]] 

In the [[Poincaré half-plane model]] of the hyperbolic plane (see [[Hyperbolic motion]]s), one can establish the relation of ''&Phi;'' to ''a'' with [[Euclidean geometry]]. Let ''Q'' be the semicircle with diameter on the ''x''-axis that passes through the points (1,0) and (0,''y''), where ''y'' > 1. Since ''Q'' is tangent to the unit semicircle centered at the origin, the two semicircles represent ''parallel hyperbolic lines''. The ''y''-axis crosses both semicircles, making a right angle with the unit semicircle and a variable angle ''&Phi;'' with ''Q''. The angle at the center of ''Q'' subtended by the radius to (0,&nbsp;''y'') is also ''&Phi;'' because the two angles have sides that are perpendicular, left side to left side, and right side to right side.  The semicircle ''Q'' has its center at (''x'',&nbsp;0), ''x'' < 0, so its radius is 1&nbsp;&minus;&nbsp;''x''. Thus, the radius squared of ''Q'' is

: <math> x^2 + y^2 = (1 - x)^2, </math>

hence

: <math> x = \tfrac{1}{2}(1 - y^2). </math>

The [[Metric (mathematics)|metric]] of the [[Poincaré half-plane model]] of hyperbolic geometry parametrizes distance on the ray {(0,&nbsp;''y'') : ''y'' > 0 } with [[logarithmic measure]]. Let the hyperbolic distance from (0,&nbsp;''y'') to (0,&nbsp;1) be ''a'', so: log&nbsp;''y'' &minus; log 1 = ''a'', so ''y'' = ''e<sup>a</sup>'' where [[e (mathematical constant)|''e'']] is the base of the [[natural logarithm]]. Then 
the relation between ''&Phi;'' and ''a'' can be deduced from the triangle {(''x'',&nbsp;0),&nbsp;(0,&nbsp;0),&nbsp;(0,&nbsp;''y'')}, for example:

: <math> \tan\phi = \frac{y}{-x} = \frac{2y}{y^2 - 1} = \frac{2e^a}{e^{2a} - 1} = \frac{1}{\sinh a}. </math>