# (problem) Spherical trigonometry and star declination

1. Jan 26, 2008

### giann_tee

I am looking for a general approach for a type of problems as follows...

Certain circumpolar star has a maximum azimuth A given for example as an angle from North to East and from North to West. Whats the declination of that star?

The problem looks like this: draw a celestial sphere and add Pole and Zenith. Put some small circle around the Pole representing the path of the star.

The Pole as the circle's center is on the local meridian (standard Zenith-North_pole-South_pole great circle).

From Zenith to the horizon we draw any 90degree meridian that is touching the edge of the given circle around the Pole. Two such lines from Zenith to horizon are defining just how wide the circle around Pole is.

How do I solve this?

I took the place where meridians touch with the star path - where the star is in extreme position for measuring Azimuth; call that place X.

The spherical triangle PZX has PZ=90-latitude and angle(pZx) is the Azimuth. However, beyond this the problem seems unsolvable.

Using software to simulate numerical case, I get an equation of type cos x + sin x = ... which can be solved in a computer but I think thats no real solution.

2. Jan 26, 2008

### giann_tee

Here's a problem from "Astronomy - Principles and Practice" Roy/Clarke

"10. In north latitude 45◦ the greatest azimuth (east or west) of a circumpolar star is 45◦. Prove that the star’s
declination is 60◦ N."

(page 87)

3. Jan 26, 2008

### giann_tee

Here's one that sounds similar but I can't guarantee...

"8. Calculate the azimuths of the star Procyon (declination = 5◦ N) when its zenith distance is 80◦ as seen by
an observer in latitude 56◦ N."

(page 87, same book)

4. Jan 26, 2008

### giann_tee

In that one I got 261,2841degrees and 98.7159degrees for Azimuths.

5. Feb 5, 2008

### giann_tee

Today I returned to this problem and came to realize it is very easy. The solution follows:

given that X is the position of the star on celestial sphere,

sin(45) / sin(PXZ) = sin (90-d) / sin A

The point is that anglePXZ is 90 degrees because the radius of the star's path is PX and ZX is tangential to that circle.

(In mind however, the angleZPX varies creating impression that circle starting from P to the point of intersection X is of a small type circles.)