**Exercise:**

Prove that the celestial
equator cuts the horizon at azimuth 90° and 270°,

at any
latitude (except at the North and South Poles).

Draw
"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
?

Use
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".