= Solution
Use $B^0=d\bar b$ and $\bar B^0=\bar d b$. A <box Feynman diagram> with two <W bosons> and internal up-type <quarks> changes one neutral <quark flavor>[flavour] into the other. The <top quark> contribution dominates the dispersive <neutral B meson mixing>[mixing amplitude] and carries the squared CKM combination involving $V_{td},V_{tb}$. With a consistent meson phase convention,
$$
\frac qp\simeq\frac{V_{td}V_{tb}^*}{V_{td}^*V_{tb}}.
$$
The illustration in <4/b> shows this box and the dominant decay topology. A suitable common final state is $f=J/\psi\,K_S$, reconstructed for example through $J/\psi\to\mu^+\mu^-$ and $K_S\to\pi^+\pi^-$. The leading $\bar b\to\bar c c\bar s$ tree transition gives $B^0\to J/\psi K^0$, while the conjugate gives $\bar B^0\to J/\psi\bar K^0$. <Neutral kaon mixing> projects both kaon flavours onto the observed $K_S$.
The final state is approximately a <CP eigenstate> with $\eta_f=-1$: $J/\psi$ and $K_S$ each have CP approximately $+1$, and their relative <orbital angular momentum> is $L=1$ to couple the vector charmonium to the spin-zero parent. Neglecting the small kaon CP impurity and subleading <decay amplitude>[decay amplitudes],
$$
\boxed{\lambda_f=\frac qp\frac{\bar A_f}{A_f}\simeq-e^{-2i\beta},\qquad
\operatorname{Im}\lambda_f=\sin2\beta.}
$$
There are two indistinguishable paths, direct decay and mixing followed by decay, so their amplitudes interfere. For $\Delta\Gamma_d\simeq0$ and negligible direct <CP violation>, the <time-dependent neutral-B CP asymmetry> is
$$
\boxed{\frac{\Gamma(\bar B^0(t)\to f)-\Gamma(B^0(t)\to f)}
{\Gamma(\bar B^0(t)\to f)+\Gamma(B^0(t)\to f)}
=\sin2\beta\,\sin(\Delta m_d t).}
$$
Measure the production <quark flavor>[flavour] and proper decay time, fit the oscillation coefficient, and extract $\sin2\beta$. This determines $\beta$ up to the usual trigonometric ambiguities unless further information is used. The cancellation of the dominant decay phase is why this channel isolates the angle involving $V_{td}$ without needing $V_{ub}$.
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