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==In geodesy== {{Main|Inverse geodetic problem}} {{See also|Earth section paths#Inverse problem|Vincenty's formulae#Inverse problem|Geographical distance#Ellipsoidal-surface formulae}} [[File:Bearing and azimuth along the geodesic.png|thumb|The azimuth between [[Cape Town]] and [[Melbourne]] along the [[geodesic]] (the shortest route) changes from 141° to 42°. [[Orthographic projection in cartography|Azimuthal orthographic projection]] and [[Miller cylindrical projection]].]] We are standing at latitude <math>\varphi_1</math>, longitude zero; we want to find the azimuth from our viewpoint to Point 2 at latitude <math>\varphi_2</math>, longitude ''L'' (positive eastward). We can get a fair approximation by assuming the Earth is a sphere, in which case the azimuth ''α'' is given by :<math>\tan\alpha = \frac{\sin L}{\cos\varphi_1 \tan\varphi_2 - \sin\varphi_1 \cos L}</math> A better approximation assumes the Earth is a slightly-squashed sphere (an ''[[oblate spheroid]]''); ''azimuth'' then has at least two very slightly different meanings. '''[[Earth normal section|Normal-section]] azimuth''' is the angle measured at our viewpoint by a [[theodolite]] whose axis is perpendicular to the surface of the spheroid; '''geodetic azimuth''' (or '''geodesic azimuth''') is the angle between north and the [[ellipsoidal geodesic]] (the shortest path on the surface of the spheroid from our viewpoint to Point 2). The difference is usually negligible: less than 0.03 arc second for distances less than 100 km.<ref name="T&G">Torge & Müller (2012) Geodesy, De Gruyter, eq.6.70, p.248</ref> Normal-section azimuth can be calculated as follows:{{cn|date=January 2022}} : <math>\begin{align} e^2 &= f(2 - f) \\ 1 - e^2 &= (1 - f)^2 \\ \Lambda &= \left(1 - e^2\right) \frac{\tan\varphi_2}{\tan\varphi_1} + e^2 \sqrt{\frac{1 + \left(1 - e^2\right)\left(\tan\varphi_2\right)^2} {1 + \left(1 - e^2\right)\left(\tan\varphi_1\right)^2}} \\ \tan\alpha &= \frac{\sin L}{(\Lambda - \cos L)\sin\varphi_1} \end{align}</math> where ''f'' is the flattening and ''e'' the eccentricity for the chosen spheroid (e.g., {{frac|{{val|298.257223563}}}} for [[World Geodetic System|WGS84]]). If ''φ''<sub>1</sub> = 0 then : <math>\tan\alpha = \frac{\sin L}{\left(1 - e^2\right)\tan\varphi_2}</math> To calculate the azimuth of the Sun or a star given its [[declination]] and [[hour angle]] at a specific location, modify the formula for a spherical Earth. Replace ''φ''<sub>2</sub> with declination and longitude difference with hour angle, and change the sign (since the hour angle is positive westward instead of east).{{cn|date=January 2022}}
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