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Spherical trigonometry
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=== From latitude and longitude === The spherical excess of a spherical quadrangle bounded by the equator, the two meridians of longitudes <math>\lambda_1</math> and <math>\lambda_2,</math> and the great-circle arc between two points with longitude and latitude <math>(\lambda_1, \varphi_1)</math> and <math>(\lambda_2, \varphi_2)</math> is <math display="block"> \tan\tfrac12 E_4 = \frac {\sin\tfrac12(\varphi_2 + \varphi_1)}{\cos\tfrac12(\varphi_2 - \varphi_1)} \tan\tfrac12(\lambda_2 - \lambda_1). </math> This result is obtained from one of Napier's analogies. In the limit where <math>\varphi_1, \varphi_2, \lambda_2 - \lambda_1</math> are all small, this reduces to the familiar trapezoidal area, <math display=inline>E_4 \approx \frac12 (\varphi_2 + \varphi_1) (\lambda_2 - \lambda_1)</math>. The area of a polygon can be calculated from individual quadrangles of the above type, from (analogously) individual triangle bounded by a segment of the polygon and two meridians,<ref>{{cite conference |last1=Chamberlain |first1=Robert G. |last2=Duquette |first2=William H. |title=Some algorithms for polygons on a sphere. |date=17 April 2007 |url=https://trs.jpl.nasa.gov/handle/2014/41271 |conference=Association of American Geographers Annual Meeting |publisher=NASA JPL |access-date=7 August 2020 |archive-date=22 July 2020 |archive-url=https://web.archive.org/web/20200722072320/https://trs.jpl.nasa.gov/handle/2014/41271 |url-status=live }}</ref> by a [[line integral]] with [[Green's theorem]],<ref>{{cite web |title=Surface area of polygon on sphere or ellipsoid β MATLAB areaint |url=https://www.mathworks.com/help/map/ref/areaint.html |website=www.mathworks.com |access-date=2021-05-01 |archive-date=2021-05-01 |archive-url=https://web.archive.org/web/20210501073522/https://www.mathworks.com/help/map/ref/areaint.html |url-status=live }}</ref> or via an [[equal-area projection]] as commonly done in GIS. The other algorithms can still be used with the side lengths calculated using a [[great-circle distance]] formula.
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