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Calcular el acimut del Sol en el instante del ocaso verdadero desde la situación 61° 27,0' S, 122° 47' W, cuando la declinación del Sol es 21° 12,6' N.Calculate the Sun's azimuth at true sunset from position 61° 27.0' S, 122° 47' W, when the Sun's declination is 21° 12.6' N.
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Correct answer: d) 319°
Why it is correct
At true sunset, sin A = sin d / cos l = sin(21º 12.6′) / cos(61º 27.0′) = 0.75714, giving an amplitude of W 49.2º N since declination is north. As a circular azimuth that is 270º + 49.2º = 319.2º ≈ 319º.
Why the others are not
- a) There is no sunset/sunrise only when |l| + |d| ≥ 90º; here 61º 27′ + 21º 13′ = 82º 40′ < 90º.
- b) Subtracts 49.2º from 270º (220.8º) as if the declination were south.
- c) Forgets to divide by cos l and uses only the declination.
How to work it out in the exam
- Sunset with north declination: between West (270º) and North (360º) → 270º + A.
Where this question comes from
- Junta de Andalucía · 2023 · 1ª convocatoria · CY Módulo de navegación · question 15 Official paper · Official answer key
Pending instructor review. Machine translation, not reviewed.