= Solution
Assume a circular <orbit> and an <orbital inclination> $i\simeq90^\circ$, as appropriate for a simple <eclipsing binary> estimate. At quadrature the two <radial velocities> are opposite and equal to the <orbital speeds>. The more massive component has the smaller amplitude. With the <centre of mass> as origin,
$$
a_1=\frac{K_1P}{2\pi},\quad a_2=\frac{K_2P}{2\pi},\quad a=a_1+a_2,\quad \frac{M_1}{M_2}=\frac{K_2}{K_1}=2.
$$
Thus $a_1=0.1103\,\mathrm{AU}$, $a_2=0.2206\,\mathrm{AU}$ and <Kepler's third law> gives
$$
\boxed{a=0.3309\,\mathrm{AU},\quad M_1=8.06M_\odot,\quad M_2=4.03M_\odot.}
$$
The <orbital plane> contains the observer's line of sight; at quadrature the two <velocities> point towards and away from that observer, as labelled in the original schematic below. For an <elliptic orbit>, the measured amplitudes instead give $a\sin i=P(K_1+K_2)\sqrt{1-e^2}/(2\pi)$; the numerical answer requires the circular assumption.
\Image[/past-exam-of-the-short-options-paper-of-the-physics-prelims-of-the-university-of-oxford/2019-astrophysical-sketches.png]
{title=Binary geometry and an old globular-cluster Hertzsprung-Russell diagram; both panels are schematic}
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