Answer to Question #136179 in Trigonometry for jay

Question

A vessel steams a course 050°T, distance 500 miles, from position 40°25 ́S 175°50 ́E. Find a final position.

Answer 1

Sketch of the scenario above;

We find final latitude and final longitude

Final latitude

Apply the formula

$Distance = \frac{D. Lat}{cos C}$ And hence

$D. Lat = Distance × cosC$

$= 500 miles × cos(50°)$

$= 321.39'$

$= 5°21.39' (N)$

Therefore, $D. Lat = 5°21.39' (N)$

Now,

Final latitude = Initial latitude - D. Lat

$= 40° 25.00' (S) - 5° 21.39' (N) = 35° 03.61' (S)$

Final latitude $= 35° 03.61' (S)$

Final longitude

Here, we apply the formula;

$Departure= D. Lat. ×tanC$

$= 321.39' × tan(50°)$

$=383.02'$

Now finding;

Mean Latitude $(Lat_m) = 40°25' - \frac{5° 21.39'}{2} = 37° 44.31'$

Now finding

$D. Long. = \frac{Dep.}{cos(Lat_m)}$

$= \frac{383.02}{cos(37° 44.31')}$

$=484.34'$

$=8° 04.34' (E)$

Now final longitude = 360° 00.00' - (Initial Longitude + D. Long.)

$=360° 00.00' - (175° 50.00' (E) + 8° 04.34' (E))$

$=360° 00.00' - 183° 54.34'$

$= 176° 05.66' (W)$

Now,

The Final Position is $35° 03.61' (S) 176° 05.66' (W)$