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Día 18 de Abril de 2021. Hora civil de Greenwich (HcG)= 14:50, se encuentran en situación estimada= 28º 03,6´N – 070º 57,1´W y a esa hora, el Capitán tras haber hallado con ayuda del Almanaque Náutico la declinación del Sol= +11º 07,6´y el ángulo en el Polo con el Sol= 28º 16,4´ (E) y con las fórmulas que definen los ángulos y sus complementos del Triángulo de posición, calcula el valor circular del azimut verdadero del Sol en ese instante:On 18 April 2021 at HcG = 14:50, at DR position 28º 03.6' N, 070º 57.1' W, given the Sun's declination = +11º 07.6' and meridian angle P = 28º 16.4' (E), calculate the Sun's circular true azimuth:
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Correct answer: c) Zv = 117.0º
Why it is correct
Applying tan Z = sin P / (cos l tan d − sin l cos P) = 0.47368 / (0.17355 − 0.41418) = −1.9685 gives Z = 180º − 63.1º = 116.9º ≈ 117.0º. Since P is East, the circular azimuth is 117.0º.
Why the others are not
- a) Does not match the trigonometric solution of the position triangle.
- b) Does not match the trigonometric solution of the position triangle.
- d) Does not match the trigonometric solution of the position triangle.
How to work it out in the exam
- If the denominator (cos l tan d − sin l cos P) is negative, Z is in the southern quadrant (180º − |Z|).
Where this question comes from
- Región de Murcia · 2021 · 1ª convocatoria · CY Tipo 1 · question 15 Official paper · Official answer key
Pending instructor review. Machine translation, not reviewed.