[ \sin a = \sin \phi \sin \delta + \cos \phi \cos \delta \cos H ] Azimuth from cosine law: [ \cos A = \frac\sin \delta - \sin \phi \sin a\cos \phi \cos a ] or using sine law: [ \sin A = \frac\cos \delta \sin H\cos a ]
: On February 26, 2016, we observe a star rise at 2:13 AM and set at 4:27 PM. From Villanova ( ( \phi = 40^\circ02'14'' \textN ) ), calculate the time of culmination and the star's equatorial coordinates (right ascension and declination).
For azimuth (using the law of sines or cosines): [ \cos A = \frac\sin \delta - \sin \phi \sin h\cos \phi \cos h ] But careful: This gives ambiguous quadrant (azimuth can be north or south). Better to use the formula for (\sin A) and check signs: spherical astronomy problems and solutions
δ≥ϕ−90∘delta is greater than or equal to phi minus 90 raised to the composed with power Substitute the latitude value:
Projection of the Earth's equator onto space (Celestial Equator). Coordinates: Declination ( ): The angular distance north ( ) or south ( −negative ) of the celestial equator. Right Ascension ( [ \sin a = \sin \phi \sin \delta
sina=sin(45∘)sin(30∘)+cos(45∘)cos(30∘)cos(30∘)sine a equals sine open paren 45 raised to the composed with power close paren sine open paren 30 raised to the composed with power close paren plus cosine open paren 45 raised to the composed with power close paren cosine open paren 30 raised to the composed with power close paren cosine open paren 30 raised to the composed with power close paren
Total Daylight=2×8.209 hours=16.418 hoursTotal Daylight equals 2 cross 8.209 hours equals 16.418 hours Better to use the formula for (\sin A)
can yield values in multiple quadrants. Use physical context (e.g., whether an object is rising in the East or setting in the West) to choose the correct angle.