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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
TwoplacesAandBbothonaparalleloflatitudeα°Ndiffersinlongitudesbyθ°.Showthattheshortestdistancebetweenthemis:[2sin1(cosαsinθ2)]360×2πRTopic:LongitudeandLatitude
Commented by I want to learn more last updated on 08/Aug/20
I have change the question sir
Ihavechangethequestionsir
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.
Oksir,iunderstand.Provethatshortestdistancebetweentwopoints=[2sin1(cosαsinθ2)]360×2πRTopic:Longitudeandlatitude.
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
Thankyousir.Pleasehelpmeseetoitwhenyouarechancedsir
Commented by Tawa11 last updated on 15/Sep/21
nice
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=RCA=CB=RcosαAB=2×CA×sinθ2=2Rcosαsinθ2AB=2×OA×sinφ2=2Rsinφ2sinφ2=cosαsinθ2φ=2sin1(cosαsinθ2)theshortestdistancefromAtoBonthespheresurfaceisthegreatcirclearcAB:AB=φR=2Rsin1(cosαsinθ2)ifφ=sin1(cosαsinθ2)isnotinrad,butindegree,thenAB=2Rsin1(cosαsinθ2)×2π360°
Commented by peter frank last updated on 08/Aug/20
thank you
thankyou
Commented by I want to learn more last updated on 08/Aug/20
Wow, I really appreciate sir. God bless you sir.
Wow,Ireallyappreciatesir.Godblessyousir.
Commented by peter frank last updated on 08/Aug/20
Mr w help please qn 107073  ezz
Mrwhelppleaseqn107073ezz
Commented by mr W last updated on 08/Aug/20
i would if i could...
iwouldificould

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