= Solution
Equal <radial velocity> amplitudes imply equal stellar masses. Under the circular, edge-on <eclipsing binary> interpretation, each star moves at $K=100\,\mathrm{km\,s^{-1}}$, and the relative speed is $2K$. Thus
$$
a=\frac{(K_1+K_2)P}{2\pi}=1.499\times10^{10}\,\mathrm m,
\qquad
M_1+M_2=\frac{P(K_1+K_2)^3}{2\pi G}.
$$
Using $P=5.45$ days gives
$$
\boxed{a\simeq0.100\,\mathrm{AU},\qquad M_1=M_2\simeq2.26M_\odot.}
$$
More generally, <spectroscopic binary> amplitudes determine $a\sin i=P(K_1+K_2)\sqrt{1-e^2}/(2\pi)$ and $(M_1+M_2)\sin^3i=P(K_1+K_2)^3(1-e^2)^{3/2}/(2\pi G)$. Thus these numerical masses and separation use $e=0$, $\sin i=1$; eclipses alone imply high inclination rather than its exact value.
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