Angle of parallelism

Given a point not on a line, drop a perpendicular to the line from the point. Let a be the length of this perpendicular segment, and $Π (a)$ be the least angle such that the line drawn through the point does not intersect the given line. Since two sides are asymptotically parallel,

lim_{a \to 0} Π (a) = \frac{1}{2} π and lim_{a \to \infty} Π (a) = 0.

There are five equivalent expressions that relate $Π (a)$ and a:

\sin Π (a) = sech a = \frac{1}{\cosh a} = \frac{2}{e^{a} + e^{- a}},

\cos Π (a) = \tanh a = \frac{e^{a} - e^{- a}}{e^{a} + e^{- a}},

\tan Π (a) = csch a = \frac{1}{\sinh a} = \frac{2}{e^{a} - e^{- a}},

\tan (\frac{1}{2} Π (a)) = e^{- a},

Π (a) = \frac{1}{2} π - gd (a),

where sinh, cosh, tanh, sech and csch are hyperbolic functions and gd is the Gudermannian function.

Construction

János Bolyai discovered a construction which gives the asymptotic parallel s to a line r passing through a point A not on r. Drop a perpendicular from A onto B on r. Choose any point C on r different from B. Erect a perpendicular t to r at C. Drop a perpendicular from A onto D on t. Then length DA is longer than CB, but shorter than CA. Draw a circle around C with radius equal to DA. It will intersect the segment AB at a point E. Then the angle BEC is independent of the length BC, depending only on AB; it is the angle of parallelism. Construct s through A at angle BEC from AB.

\sin B E C = \frac{\sinh B C}{\sinh C E} = \frac{\sinh B C}{\sinh D A} = \frac{\sinh B C}{\sin A C D \sinh C A} = \frac{\sinh B C}{\cos A C B \sinh C A} = \frac{\sinh B C \tanh C A}{\tanh C B \sinh C A} = \frac{\cosh B C}{\cosh C A} = \frac{\cosh B C}{\cosh C B \cosh A B} = \frac{1}{\cosh A B} .

See Trigonometry of right triangles for the formulas used here.

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Construction

History

Demonstration

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