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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
I have change the question sir
$$\mathrm{I}\:\mathrm{have}\:\mathrm{change}\:\mathrm{the}\:\mathrm{question}\:\mathrm{sir} \\ $$
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
nice
$$\mathrm{nice} \\ $$
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
thank you
$$\mathrm{thank}\:\mathrm{you} \\ $$
Commented by I want to learn more last updated on 08/Aug/20
Wow, I really appreciate sir. God bless you sir.
$$\mathrm{Wow},\:\mathrm{I}\:\mathrm{really}\:\mathrm{appreciate}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$
Commented by peter frank last updated on 08/Aug/20
Mr w help please qn 107073  ezz
$$\mathrm{Mr}\:\mathrm{w}\:\mathrm{help}\:\mathrm{please}\:\mathrm{qn}\:\mathrm{107073} \\ $$$$\mathrm{ezz} \\ $$
Commented by mr W last updated on 08/Aug/20
i would if i could...
$${i}\:{would}\:{if}\:{i}\:{could}… \\ $$

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