Astronomy: Sun azimuth from altitude, hour angle, and declination

[Frank Einstein]

Homework Statement

Hi everybody, I am trying to do a numerical aplication of change of coordinate system to another; the objective is to calculate the azimut of the sun knowing the altitude, hour angle and declination .

Homework Equations

Cosh * Sina = Cosδ * Sin H
where a is the azimut
h is the altitude, in this case 27.5°
δ is the declination, in this case 7.2°
and H is the hour angle, in this case 10:00 on GTM+2; so 20h which is 300°

The Attempt at a Solution

Sina = [ Cosδ * Sin H ] / CosH

If I substitute the values, the azimut is -75.6°.
The problem is that if I check the value of the azimut on the program stellarium for august 12 2014, the date from where the data has been chosen, the azimut is of 104º, and I don't know why there is such a difference between my azimut and the program's, Can anyone help please?

 

[Simon Bridge]

What does a negative azimuthal angle indicate?
Note 180-75.4=104.6

 

[Frank Einstein]

A negative azimut would mean that the location of the sun is north west is stead of north east, which is where it sould be before 14h.
About 180ºThat's true, but that would mean that I am measuring my azimut from the south pole, not that I am doing it from the north.
I will study the deduction of the equation to see if I was doing measuring from the south all along.
Thank you very much.

 

[andrevdh]

I get a different value, 34^o, for the azimuth using wiki formula:

http://en.wikipedia.org/wiki/Celestial_coordinate_system#Equatorial_.E2.86.90.E2.86.92_horizontal

seems yours should then be

sin(a) = cos(δ) sin(H) / cos(h)

Also note that hour angle should be 10/24 x 360 degrees.