Open main menu
Home
Random
Recent changes
Special pages
Community portal
Preferences
About Wikipedia
Disclaimers
Incubator escapee wiki
Search
User menu
Talk
Dark mode
Contributions
Create account
Log in
Editing
Angle of parallelism
(section)
Warning:
You are not logged in. Your IP address will be publicly visible if you make any edits. If you
log in
or
create an account
, your edits will be attributed to your username, along with other benefits.
Anti-spam check. Do
not
fill this in!
==Construction== [[János Bolyai]] discovered a construction which gives the asymptotic parallel ''s'' to a line ''r'' passing through a point ''A'' not on ''r''.<ref>"Non-Euclidean Geometry" by Roberto Bonola, page 104, Dover Publications.</ref> 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''. :<math> \sin BEC = \frac{ \sinh {BC} }{ \sinh {CE} } = \frac{ \sinh {BC} }{ \sinh {DA} } = \frac{ \sinh {BC} }{ \sin {ACD} \sinh {CA} } = \frac{ \sinh {BC} }{ \cos {ACB} \sinh {CA} } = \frac{ \sinh {BC} \tanh {CA} }{ \tanh {CB} \sinh {CA} } = \frac{ \cosh {BC} }{ \cosh {CA} } = \frac{ \cosh {BC} }{ \cosh {CB} \cosh {AB} } = \frac{ 1 }{ \cosh {AB} } \,.</math> See [[Hyperbolic triangle#Trigonometry of right triangles|Trigonometry of right triangles]] for the formulas used here.
Edit summary
(Briefly describe your changes)
By publishing changes, you agree to the
Terms of Use
, and you irrevocably agree to release your contribution under the
CC BY-SA 4.0 License
and the
GFDL
. You agree that a hyperlink or URL is sufficient attribution under the Creative Commons license.
Cancel
Editing help
(opens in new window)