User:DrBeWa/sandbox

Cite web

Template:Cite web, parms copy-and-paste, parms description, Wizzard for this template: Templator

date: 2024-04-27 -- |accessdate=27 April 2024 -- |date=20 May 2010 -- x . . . .

Cite web w/ archiveurl

Help:Using the Wayback Machine, Template:Cite web, Usage/copy-and-paste, Examples, Wizzard for this template: Templator

date: 2024-04-27 -- |accessdate=27 April 2024 -- |archive-date=19 March 2015 . .

Details

Template:Details

Further information: How To Derive The Equations For Triangulation

Fiddling

${\begin{aligned}-2x&&\;+\;&&y&&\;+\;&&2z&&\;=\;&&-3&\qquad (L_{3}){\text{xx}}\end{aligned}}$

Calculation

Triangulation may be used to calculate the coordinates and distance from the shore to the ship.

$\ell =\ell _{1}+\ell _{2},~~~\tan \alpha ={\frac {d}{\ell _{1}}}~\Rightarrow ~\ell _{1}={\frac {d}{\tan \alpha }}$

$\ell ={\frac {d}{\tan \alpha }}+{\frac {d}{\tan \beta }}$

Therefore

$d=\ell \,/\,({\tfrac {1}{\tan \alpha }}+{\tfrac {1}{\tan \beta }})$

Using the trigonometric identities tan α = sin α / cos α and sin(α + β) = sin α cos β + cos α sin β, this is equivalent to:

$d={\frac {\ell \sin \alpha \sin \beta }{\sin(\alpha +\beta )}}$

From this, it is easy to determine the distance of the unknown point from either observation point, its north/south and east/west offsets from the observation point, and finally its full coordinates.