Caption: A diagram giving---for the cognoscenti---a proof by inspection that the altitude above due north of the north celestial pole (NCP) A_N_NCP equals the observer's latitude on Earth L: i.e.,
Note that a right angle can be rotated into a right angle.
Features:
L = δ ± [A(±) - 90°]: For finding latitude with south latitude counted as a negative number. This formula is used in celestial navigation.
λ = (360°/P)(t_UT - t_UT_G) where: P is the period: sidereal day for fixed stars, apparent solar day for the Sun. t_UT is the marine chronometer time of the transit of meridian. t_UT_G is the prime meridian (Greenwich meridian) time of the transit of meridian which you read from some source (e.g., a nautical almanac) but is about your local time as appropriate for the astronomical objects???.