Prove that the celestial equator cuts the horizon at azimuth 90° and 270°,
at any latitude (except at the North and South Poles).

diagramDraw "the" triangle again.
We require the azimuth A of point X,
where X is on the horizon (i.e. a=0)
and also on the equator (i.e. δ=0)

Apply the cosine rule:
cos PX = cos PZ cos XZ + sin PZ sin XZ cos Z
to get 0 = 0 + sin (90-φ) cos A
Since 90°-φ is not 0 (we are not at the Poles),
cos A must be 0
so A = 90° or 270° .

At what angle does the celestial equator cut the horizon, at latitude φ ?

diagramUse the cosine formula:
cos SY = cos SW cos YW + sin SW sin YW cos W
This gives cos (90°-φ) = 0 + cos x
So the angle x is 90°-φ.
The celestial equator cuts the horizon at an angle of 90°-φ

Back to "coordinate conversion".