Hover or tap underlined Spanish terms to see their translation and exam tip.
Calcular el rumbo inicial de la derrota ortodrómica entre la situación de salida l= 36º 41,8’N; L= 008º 56,9’ W y la situación de llegada l’= 03º59,8’S L’= 033º 43,7’W.Calculate the initial great-circle course from departure l = 36º 41.8’ N, L = 008º 56.9’ W to arrival l’ = 03º 59.8’ S, L’ = 033º 43.7’ W.
Show the answer
Correct answer: b) Ri = 215º.
Why it is correct
With ΔL = 24º 46.8′ W and south arrival latitude (negative), tan Ri = sin ΔL / (cos l tan l′ − sin l cos ΔL) = 0.41913 / −0.59852, giving S 35.0º W, or a circular initial course Ri = 180º + 35º = 215º.
Why the others are not
- a) Adds the two denominator terms as if both latitudes were in the same hemisphere.
- c) Subtracts 35º from 360º (N 35º W), forgetting that a negative denominator puts the course south.
- d) Places the course in the fourth quadrant with the wrong angle.
How to work it out in the exam
- An arrival latitude of opposite name enters tan l′ with a negative sign.
Where this question comes from
- Ministerio de Transportes (Madrid) · 2024 · Convocatoria de noviembre · CY Test 1 · question 14 Official paper · Official answer key
- Ministerio de Transportes (Madrid) · 2024 · Convocatoria de noviembre · CY Test 2 · question 19 Official paper · Official answer key
Pending instructor review. Machine translation, not reviewed.