Answer to Question #140133 in Trigonometry for chrisvel calva

Question

Find the course and distance from a position 37⁰ 01’ N 9⁰ 00’W to a position 36⁰ 11’N
6⁰ 02’W.

Answer 1

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

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

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

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

$\Delta σ=arccos(sin(φ_A)sin(φ_B)+$

$+cos(φ_A)cos(φ_B)cos(λ_B-λ_A))=$

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

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

$=arccos(0.6020\cdot 0.5904+$

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

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

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

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

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

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

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

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

$arccos(-0.3137)=108.3^{\circ}$ - 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.