Given: Observer at latitude 38° N. Sun’s declination = –10° (winter). Ignoring refraction, find the hour angle at sunrise (when Sun’s center is on the horizon). Solution: On the celestial sphere, at sunrise, the zenith distance = 90°. Use spherical cosine law: [ \cos(90°) = \sin(\textlat) \cdot \sin(\textdec) + \cos(\textlat) \cdot \cos(\textdec) \cdot \cos(H) ] [ 0 = \sin(38°)\sin(-10°) + \cos(38°)\cos(-10°)\cos(H) ] [ 0 = -0.1056 + 0.7660 \cdot 0.9848 \cdot \cos(H) ] [ 0.1056 = 0.7541 \cdot \cos(H) ] [ \cos(H) = 0.1400 \Rightarrow H = \pm 81.95° ] Sunrise is before noon, so (H = -81.95°) (or 5.46 hours before local solar noon). She looked up: “Sunrise in 5 hr 28 min.”

(Right Ascension and Declination), which is fixed against the stars. The Problem:

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Spherical Astronomy Problems And Solutions Extra Quality 【2025】

Given: Observer at latitude 38° N. Sun’s declination = –10° (winter). Ignoring refraction, find the hour angle at sunrise (when Sun’s center is on the horizon). Solution: On the celestial sphere, at sunrise, the zenith distance = 90°. Use spherical cosine law: [ \cos(90°) = \sin(\textlat) \cdot \sin(\textdec) + \cos(\textlat) \cdot \cos(\textdec) \cdot \cos(H) ] [ 0 = \sin(38°)\sin(-10°) + \cos(38°)\cos(-10°)\cos(H) ] [ 0 = -0.1056 + 0.7660 \cdot 0.9848 \cdot \cos(H) ] [ 0.1056 = 0.7541 \cdot \cos(H) ] [ \cos(H) = 0.1400 \Rightarrow H = \pm 81.95° ] Sunrise is before noon, so (H = -81.95°) (or 5.46 hours before local solar noon). She looked up: “Sunrise in 5 hr 28 min.”

(Right Ascension and Declination), which is fixed against the stars. The Problem: spherical astronomy problems and solutions

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