The <projected separation> is
$$
\boxed{R_{\rm sky}=\sqrt{x^2+y^2}
=a\sqrt{\cos^2f+\cos^2I\sin^2f}
=a\sqrt{1-\sin^2I\sin^2f}}.
$$
Rotating the projected coordinates through $-\Omega$ gives components $a\cos f$ along the nodal line and $a\cos I\sin f$ perpendicular to it. The <position angle> is therefore most safely written with the quadrant-preserving function
$$
\boxed{\phi=\Omega+operatorname{atan2}(\cos I\sin f,\cos f)},
$$
or, wherever the corresponding tangent is finite,
$$
\boxed{\tan(\phi-\Omega)=\cos I\tan f}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 316 / 1 / iii / Solution

{title2=$R_{\rm sky}$}

Projected separation is the distance between two objects after their relative displacement is projected onto the plane of the sky.


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