Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 338 / 1 / a / i

Use azimuth measured from north toward west, and positive hour angle westward. These conventions are forced by the signs in the coordinate formulae. In the diagram the observer is at $O$, the local vertical meets the celestial sphere at the zenith $Z$, and the north celestial pole is $P$. The astronomical horizon is perpendicular to $OZ$; the celestial equator is perpendicular to $OP$. Their intersections with the celestial meridian give the north horizon point and the upper equatorial meridian point.

The north celestial pole has altitude $\theta$; thus the angle between $OP$ and $OZ$ is $90^{\circ }-\theta$. Project $X$ along its vertical great circle onto the astronomical horizon at $H$: the arc $HX$ is the altitude $\gamma$, and the westward horizon arc from north to $H$ is the azimuth $\eta$. Project $X$ along its hour circle onto the celestial equator at $Q$: the arc $QX$ is the declination $\delta$, and the westward equatorial arc from the upper celestial meridian to $Q$ is the hour angle $h$. Equivalently the triangle on the celestial sphere $PZX$ has sides $PZ=90^{\circ }-\theta$, $PX=90^{\circ }-\delta$ and $ZX=90^{\circ }-\gamma$. All five angles refer to arcs on their specified reference circles, not arbitrary angles in the projected drawing.