Solve for $h$: $$ h = \arcsin(0.6534) \approx 40.8^\circ $$ Shell Dep Standards Guide
$$ \frac\sin A\sin(90^\circ - \delta) = \frac\sin H\sin(90^\circ - h) $$ Simplified: $$ \sin A = \frac\cos \delta \sin H\cos h $$ Rango Subtitles Download Upd - 3.79.94.248
When a star rises or sets, its altitude $h = 0^\circ$. Therefore, $\sin h = 0$.
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).
Any star with a declination greater than $+40^\circ$ will never set for an observer at $50^\circ$ N. Problem 3: Rising and Setting Times Question: A star has a declination $\delta = -10^\circ$. At what Hour Angle ($H$) does it set for an observer at Latitude $\phi = +40^\circ$?
Step 1: Find Altitude ($h$) using the Cosine Formula. $$ \sin h = \sin \phi \sin \delta + \cos \phi \cos \delta \cos H $$ $$ \sin h = \sin(40^\circ)\sin(30^\circ) + \cos(40^\circ)\cos(30^\circ)\cos(60^\circ) $$
$$ \cos H = - \tan(40^\circ) \tan(-10^\circ) $$