Answer to Question #140133 in Trigonometry for chrisvel calva

Question #140133
Find the course and distance from a position 37⁰ 01’ N 9⁰ 00’W to a position 36⁰ 11’N
6⁰ 02’W.
1
Expert's answer
2020-10-26T19:12:25-0400

φA=3701φ_A=37^{\circ}01' N; -the geographical latitude of position A

λA=900λ_A=9^{\circ}00' W; -the geographical longitude of position A

φB=3611φ_B=36^{\circ}11' N; -the geographical latitude of position B

λB=602λ_B=6^{\circ}02' W; -the geographical longitude of position B

Δσ=arccos(sin(φA)sin(φB)+\Delta σ=arccos(sin(φ_A)sin(φ_B)+

+cos(φA)cos(φB)cos(λBλA))=+cos(φ_A)cos(φ_B)cos(λ_B-λ_A))=

=arccos(sin(3701)sin(3611)+=arccos(sin(37^{\circ}01')sin(36^{\circ}11')+

+cos(3701)cos(3611)cos(602900))=+cos(37^{\circ}01')cos(36^{\circ}11')cos(6^{\circ}02'-9^{\circ}00'))=

=arccos(0.60200.5904+=arccos(0.6020\cdot 0.5904+

+0.79850.8071cos(258))=+0.7985\cdot 0.8071\cdot cos(-2^{\circ}58'))=

=arccos(0.3554+0.64450.9986)==arccos(0.3554+0.6445\cdot 0.9986)=

=arccos(0.9990)=2.52=arccos(0.9990)=2.52^{\circ} - the central angle between A and B;

the great-circle distance between two points AB=602.52=151.2AB=60\cdot2.52^{\circ}=151.2‬ nautical miles,

or AB=R2.52π180=63710.0440=280.3AB=R\cdot2.52^{\circ}\frac{π}{180^{\circ}}=6371\cdot0.0440=280.3 km, where R=6371 km - the mean earth radius.


α=arccos(sin(φB)sin(φA)cos(Δσ)cos(φA)sin(Δσ))=α=arccos(\frac{sin(φ_B)-sin(φ_A)\cdot cos(\Delta σ)}{cos(φ_A)\cdot sin(\Delta σ)})=

=arccos(sin(3611)sin(3701)cos(2.52)cos(3701)sin(2.52))==arccos(\frac{sin(36^{\circ}11')-sin(37^{\circ}01')\cdot cos(2.52^{\circ})}{cos(37^{\circ}01')\cdot sin(2.52^{\circ})})=

=arccos(0.59040.60200.99900.79850.0439)==arccos(\frac{ 0.5904-0.6020\cdot 0.9990}{0.7985\cdot 0.0439})=

arccos(0.3137)=108.3arccos(-0.3137)=108.3^{\circ} - bearing from point A to point B


Answer: the distance between positions is about 151.2 nautical miles (280.3 km),

course (bearing from point A to point B) is about 108.3⁰.



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