Hover or tap underlined Spanish terms to see their translation and exam tip.
Un buque que navega a RA 165° en l: 36° N, toma en el momento del ocaso del ʘ (sol) marcación a ese mismo astro 95° por Er (estribor), sabiendo que la declinación del ʘ en ese momento es de 13° 45’ S. Se pide calcular la Ct (corrección total):A vessel sailing on compass course Ra = 165° at latitude l = 36° N takes a relative bearing on the Sun at true sunset of 95° to starboard, when the Sun's declination is 13° 45’ S. Calculate the total compass error (Ct):
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Correct answer: a) -7.1º
Why it is correct
At true sunset, cos Z = −sin(13º 45′) / cos(36º) = −0.29382, giving Z = 107.1º from north towards west, or Zv = 360º − 107.1º = 252.9º. The compass azimuth is Za = Ra + M = 165º + 95º = 260º, so Ct = Zv − Za = 252.9º − 260º = −7.1º.
Why the others are not
- b) Would mean Zv and Za both equalled 260º.
- c) Subtracts Za − Zv, inverting the sign of Ct = Zv − Za.
- d) The calculation gives −7.1º exactly.
How to work it out in the exam
- At sunset with south declination, Zv = 270º − arcsin(sin |d| / cos l).
Where this question comes from
- Región de Murcia · 2025 · 2ª convocatoria · CY Tipo 1 · question 17 Official paper · Official answer key
Pending instructor review. Machine translation, not reviewed.