Two-places-A-and-B-both-on-a-parallel-of-latitude-N-differs-in-longitudes-by-Show-that-the-shortest-distance-between-them-is-2-sin-1-cos-sin-2-360-2piR-Topic-

Question Number 106990 by I want to learn more last updated on 08/Aug/20

Two places A and B both on a parallel of latitude α°N differs in longitudes by θ°. Show that the shortest distance between them is: (([2 sin^(− 1) (cos α sin (θ/2))])/(360)) × 2πR Topic: Longitude and Latitude

$$\mathrm{Two}\:\mathrm{places}\:\:\mathrm{A}\:\:\mathrm{and}\:\:\mathrm{B}\:\:\mathrm{both}\:\mathrm{on}\:\mathrm{a}\:\mathrm{parallel}\:\mathrm{of}\:\mathrm{latitude}\:\:\alpha°\mathrm{N} \\ $$$$\mathrm{differs}\:\mathrm{in}\:\mathrm{longitudes}\:\mathrm{by}\:\:\theta°.\:\mathrm{Show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{shortest}\:\mathrm{distance} \\ $$$$\mathrm{between}\:\mathrm{them}\:\mathrm{is}:\:\:\:\:\:\frac{\left[\mathrm{2}\:\mathrm{sin}^{−\:\mathrm{1}} \left(\mathrm{cos}\:\alpha\:\mathrm{sin}\:\frac{\theta}{\mathrm{2}}\right)\right]}{\mathrm{360}}\:\:×\:\:\mathrm{2}\pi\mathrm{R} \\ $$$$ \\ $$$$\mathrm{Topic}:\:\:\mathrm{Longitude}\:\mathrm{and}\:\mathrm{Latitude} \\ $$

Commented by I want to learn more last updated on 08/Aug/20

$$\mathrm{I}\:\mathrm{have}\:\mathrm{change}\:\mathrm{the}\:\mathrm{question}\:\mathrm{sir} \\ $$

Commented by I want to learn more last updated on 08/Aug/20

Ok sir, i understand. Prove that shortest distance between two points = (([2 sin^(− 1) (cos α sin (θ/2))])/(360)) × 2πR Topic: Longitude and latitude.

$$\mathrm{Ok}\:\mathrm{sir},\:\mathrm{i}\:\mathrm{understand}. \\ $$$$ \\ $$$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{shortest}\:\mathrm{distance}\:\mathrm{between}\:\mathrm{two}\:\mathrm{points} \\ $$$$=\:\:\frac{\left[\mathrm{2}\:\mathrm{sin}^{−\:\mathrm{1}} \left(\mathrm{cos}\:\alpha\:\mathrm{sin}\:\frac{\theta}{\mathrm{2}}\right)\right]}{\mathrm{360}}\:\:×\:\mathrm{2}\pi\mathrm{R} \\ $$$$\mathrm{Topic}:\:\:\mathrm{Longitude}\:\mathrm{and}\:\mathrm{latitude}. \\ $$

Commented by I want to learn more last updated on 08/Aug/20

Thank you sir. Please help me see to it when you are chanced sir

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{Please}\:\mathrm{help}\:\mathrm{me}\:\mathrm{see}\:\mathrm{to}\:\mathrm{it}\:\mathrm{when}\:\mathrm{you}\:\mathrm{are}\:\mathrm{chanced}\:\mathrm{sir} \\ $$

Commented by Tawa11 last updated on 15/Sep/21

$$\mathrm{nice} \\ $$

Answered by mr W last updated on 08/Aug/20

Commented by mr W last updated on 08/Aug/20

OA=OB=R CA=CB=R cos α AB=2×CA×sin (θ/2)=2R cos α sin (θ/2) AB=2×OA×sin (ϕ/2)=2R sin (ϕ/2) ⇒sin (ϕ/2)=cos α sin (θ/2) ⇒ϕ=2 sin^(−1) (cos α sin (θ/2)) the shortest distance from A to B on the sphere surface is the great circle arc AB^(⌢) : AB^(⌢) =ϕR=2R sin^(−1) (cos α sin (θ/2)) if ϕ=sin^(−1) (cos α sin (θ/2)) is not in rad, but in degree, then AB^(⌢) =2R sin^(−1) (cos α sin (θ/2))×((2π)/(360°))

$${OA}={OB}={R} \\ $$$${CA}={CB}={R}\:\mathrm{cos}\:\alpha \\ $$$${AB}=\mathrm{2}×{CA}×\mathrm{sin}\:\frac{\theta}{\mathrm{2}}=\mathrm{2}{R}\:\mathrm{cos}\:\alpha\:\mathrm{sin}\:\frac{\theta}{\mathrm{2}} \\ $$$${AB}=\mathrm{2}×{OA}×\mathrm{sin}\:\frac{\varphi}{\mathrm{2}}=\mathrm{2}{R}\:\mathrm{sin}\:\frac{\varphi}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{sin}\:\frac{\varphi}{\mathrm{2}}=\mathrm{cos}\:\alpha\:\mathrm{sin}\:\frac{\theta}{\mathrm{2}} \\ $$$$\Rightarrow\varphi=\mathrm{2}\:\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{cos}\:\alpha\:\mathrm{sin}\:\frac{\theta}{\mathrm{2}}\right) \\ $$$${the}\:{shortest}\:{distance}\:{from}\:{A}\:{to}\:{B} \\ $$$${on}\:{the}\:{sphere}\:{surface}\:{is}\:{the}\:{great} \\ $$$${circle}\:{arc}\:\overset{\frown} {{AB}}: \\ $$$$\overset{\frown} {{AB}}=\varphi{R}=\mathrm{2}{R}\:\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{cos}\:\alpha\:\mathrm{sin}\:\frac{\theta}{\mathrm{2}}\right) \\ $$$${if}\:\varphi=\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{cos}\:\alpha\:\mathrm{sin}\:\frac{\theta}{\mathrm{2}}\right)\:{is}\:{not}\:{in}\:{rad}, \\ $$$${but}\:{in}\:{degree},\:{then} \\ $$$$\overset{\frown} {{AB}}=\mathrm{2}{R}\:\mathrm{sin}^{−\mathrm{1}} \left(\mathrm{cos}\:\alpha\:\mathrm{sin}\:\frac{\theta}{\mathrm{2}}\right)×\frac{\mathrm{2}\pi}{\mathrm{360}°} \\ $$

Commented by peter frank last updated on 08/Aug/20