Problem: An observer at latitude (\phi = 40^\circ) N sees a star with declination (\delta = 20^\circ) N at hour angle (H = 30^\circ) (west). Find its altitude and azimuth.
Solution:
Step 1 – Altitude (a): [ \sin a = \sin 40^\circ \sin 20^\circ + \cos 40^\circ \cos 20^\circ \cos 30^\circ ] Values: (\sin40\approx0.6428,\ \sin20\approx0.3420,\ \cos40\approx0.7660,\ \cos20\approx0.9397,\ \cos30\approx0.8660).
First term: (0.6428 \times 0.3420 = 0.2198)
Second term: (0.7660 \times 0.9397 = 0.7198); times (0.8660) = (0.6233)
Sum: (0.2198 + 0.6233 = 0.8431)
[
a = \arcsin(0.8431) \approx 57.5^\circ
]
Step 2 – Azimuth (A): Use sine formula: [ \sin A = \frac\cos\delta \sin H\cos a ] (\cos\delta = 0.9397,\ \sin H = 0.5,\ \cos a = \cos57.5^\circ \approx 0.5373) spherical astronomy problems and solutions
Numerator: (0.9397 \times 0.5 = 0.46985)
Divide: (0.46985 / 0.5373 \approx 0.8746)
[
A \approx \arcsin(0.8746) \approx 61.0^\circ \ \textor \ 119.0^\circ
]
Check (\cos A): (\cos A = (\sin\delta - \sin\phi\sin a)/(\cos\phi\cos a))
Numerator: (0.3420 - (0.6428\times0.8431) = 0.3420 - 0.5419 = -0.1999)
Denominator: (0.7660 \times 0.5373 = 0.4116)
(\cos A = -0.1999 / 0.4116 \approx -0.4857) → (A > 90^\circ).
Thus (A \approx 119^\circ) (measured from north through east).
Answer: (a \approx 57.5^\circ), (A \approx 119^\circ).
Question: A star has a declination $\delta = -10^\circ$. At what Hour Angle ($H$) does it set for an observer at Latitude $\phi = +40^\circ$?
Concept: When a star rises or sets, its altitude $h = 0^\circ$. Therefore, $\sin h = 0$. Problem: An observer at latitude (\phi = 40^\circ)
Formula: From the cosine formula, setting $h=0$: $$ 0 = \sin \phi \sin \delta + \cos \phi \cos \delta \cos H $$ $$ \cos H = - \frac\sin \phi \sin \delta\cos \phi \cos \delta $$ Or simplified: $$ \cos H = - \tan \phi \tan \delta $$
Solution: $$ \cos H = - \tan(40^\circ) \tan(-10^\circ) $$
Calculate:
Substitute: $$ \cos H = - (0.839 \times -0.176) $$ $$ \cos H = - (-0.147) $$ $$ \cos H = +0.147 $$ Substitute: $$ \cos H = - (0
Solve for $H$: $$ H = \arccos(0.147) \approx 81.5^\circ $$
Interpretation: The star sets at Hour Angle $H = 81.5^\circ$. Since $15^\circ = 1$ hour, the star sets $81.5 / 15 \approx 5.43$ hours after it crosses the meridian (Upper Culmination).
Answer: The Hour Angle at setting is $81.5^\circ$ (approx 5 hours 26 minutes).
For a spherical triangle with sides (a, b, c) (arc lengths in angular measure) and opposite angles (A, B, C):