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El sábado 10 de Marzo de 2018, a hora civil de Greenwich (HcG)= 103000, están en situación estimada= 34º 18'N – 006º 36'W y tras calcular la declinación del Sol (d0) para ese instante= -4º 02.5’ y el ángulo en el Polo (P) con el Sol= 31º 40,2' (E); proceden a calcular el valor del azimut verdadero del Sol en ese instante (utilizando las fórmulas que definen los lados y ángulos del triángulo de posición), obteniendo el valor. :On Saturday 10 March 2018, at Greenwich civil time (GCT) 103000, they are at DR position 34º 18′N – 006º 36′W and, having worked out the Sun’s declination for that moment = -4º 02.5′ and the meridian angle (P) to the Sun = 31º 40.2′ (E), they work out the Sun’s true azimuth at that moment (using the formulas for the sides and angles of the position triangle), obtaining:
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Correct answer: b) Zv = 135.7º
Why it is correct
tan Z = sin P / (cos l tan d − sin l cos P) = 0.525 / −0.538. The denominator is negative, so Z = 180º − 44.3º = 135.7º, and with P east that is the true azimuth.
Why the others are not
- a) The formula does not give this value.
- c) The formula does not give this value.
- d) The formula does not give this value.
How to work it out in the exam
- South declination goes into the formula as negative.
- Negative denominator: add 180º. P east: keep the azimuth as it is.
Where this question comes from
- Región de Murcia · 2018 · 1ª convocatoria · CY Tipo 1 · question 16 Official paper · Official answer key
Pending instructor review. Machine translation, not reviewed.