Exercise:

The Sun is at declination -14°.
What will be its hour angle at sunrise
(the moment the top edge of the Sun first appears over the horizon),
at a latitude of +56°20'?

At sunrise, the true altitude of the Sun is a = -0°50'
(allowing for semi-diameter and horizontal refraction).

Use the formula
cos(H) = { sin(a) - sin(φ) sin(δ) } / cos(φ) cos(δ)
where φ = +56°20' and δ = -14°.
This gives cos(H) = 0.35,
so H = 69.7° or 290.3°
= 4h39m or 19h21m.

To decide which,
note that the Sun is to the east of the meridian at sunrise,
so H = 19h21m.

If the Sun is on the local meridian at 12:03,
what time is sunrise?
and what time is sunset?

The semi-diurnal arc is 4h39m.
Sunrise is at (12:03 - 4h39m) = 07:24.
Sunset is at (12:03 + 4h39m) = 16:42.

And when will astronomical twilight start and finish?

It is astronomical twilight
if the Sun’s altitude is -18° or higher.

Set a = -18° and use the same formula again, to obtain
H = 101.55° = 6h46m.

So twilight starts at (12:03 - 6h46m) = 05:17,
and ends at (12:03 + 6h46m) = 18:49.

Back to "sunrise, sunset and twilight".