Spherical Astronomy Problems And Solutions __full__ May 2026

[ \sin A = \frac\cos \delta \sin H\cos h ]

(\phi = 35°), (\delta = 16.333°), (H=63.724°). (\sin h = \sin35 \sin16.333 + \cos35 \cos16.333 \cos63.724) = (0.5736)(0.2813) + (0.8192)(0.9596)(0.4423) = 0.1613 + (0.8192 0.9596 0.4423) = 0.1613 + (0.7859*0.4423) = 0.1613 + 0.3476 = 0.5089. (h = \arcsin(0.5089) = 30.58^\circ). spherical astronomy problems and solutions

Apply the spherical law of cosines to the PZS triangle: [ \sin A = \frac\cos \delta \sin H\cos

At (\phi = 40^\circ N), (\delta = 20^\circ), (H = 30^\circ). (\sin h = \sin40 \sin20 + \cos40 \cos20 \cos30) (\sin h = (0.6428)(0.3420) + (0.7660)(0.9397)(0.8660)) (\sin h = 0.2198 + 0.6230 = 0.8428) → (h \approx 57.4^\circ). Apply the spherical law of cosines to the

Compute both (\sin A) and (\cos A) from: [ \sin A = -\frac\cos \delta \sin H\cos h ] (sign depends on convention; careful: some texts use azimuth from south) and [ \cos A = \frac\sin \delta - \sin \phi \sin h\cos \phi \cos h ] Then (A = \textatan2(\sin A, \cos A)) in radians. Part 4: Comprehensive Worked Example Problem: On 2024-10-15 at 4h UT, an observer at (\phi = 35^\circ N), longitude (= 75^\circ W) observes a star with (\alpha = 6h 45m 12s), (\delta = +16^\circ 20'). Find the star’s altitude and azimuth at that moment.

From the cosine law for side (PS) (which is (90° - \delta)): [ \sin \delta = \sin \phi \sin h + \cos \phi \cos h \cos A ]

Then determine (A) uniquely: If (\sin A > 0), (A) in (0°–180°); if (\sin A < 0), (A) in (180°–360°). Or use atan2.