Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 338 / 1 / a / ii

The great circle containing the pointing and celestial pole is the intersection of the unit sphere with the meridian plane $y=0$. The orthogonal projection of $\mathbf{s}$ onto that plane, normalized to unit length, isWithin the local meridian chart $\cos \Delta >0$, its declination is therefore $q$. Outside that chart, the same construction gives an oriented great circle parameter; the foot can lie on the opposite right-ascension half of the meridian, and $q$ need not be its ordinary declination.

To verify the right angle geometrically, the tangent to the great circle from $L$ toward $S$ is proportional to $\mathbf{s}-(\mathbf{s}\cdot \mathbf{\ell })\mathbf{\ell }$. Since $\mathbf{s}\cdot \mathbf{\ell }=\rho$, that tangent has only a $y$ component. The tangent from $L$ toward the pole, $\mathbf{P}-(\mathbf{P}\cdot \mathbf{\ell })\mathbf{\ell }$, lies in $y=0$. Their orthogonality proves the claimed spherical right angle. Thus within the local meridian chart, $q$ is the declination of the perpendicular foot on the pointing meridian.

There are antipodal perpendicular feet on the complete great circle. On the local meridian chart, the telescope field selects the foot near the pointing; the cotangent equation alone determines $q$ only modulo $\pi$. At a coincident foot or pole, the corresponding angle is understood by continuity rather than by a nonzero tangent vector.