= Solution
Write $R=k_R\sqrt A$, $\bar h=k_H\sqrt A$, where the given geometry gives
$$
k_R=\tfrac12\sqrt{\sin\theta\cos\theta},\qquad k_H=\tfrac12\sqrt{\tan\theta}.
$$
Take a <Chézy drag law> $\tau=\rho c_du|u|$ and restrict to downstream flow $u>0$. Choose an <area> scale $A_0$, length scale $L$, and speed scale $U_0$ satisfying
$$
U_0^2=\frac{gSk_R\sqrt{A_0}}{c_d},\qquad t_0=L/U_0.
$$
This makes gravitational driving and drag equal in the reference uniform <steady flow>. Dividing the printed <momentum> equation by $gS$ gives
$$
\boxed{\epsilon F(u_t+uu_x)=1-\frac{u^2}{\sqrt A}-\epsilon(\sqrt A)_x,\quad \epsilon=\frac{k_H\sqrt{A_0}}{SL},\quad F=\frac{U_0^2}{gk_H\sqrt{A_0}}.}
$$
<Mass conservation> retains $A_t+(Au)_x=0$. In this notation $F$ is the square of the velocity-to-gravity-wave-speed ratio based on the chosen depth scale; authors sometimes instead denote its square root by <Froude number>. If the actual triangular <pressure> head from the preceding solution is used, replace $k_H$ by $2k_H$ in both parameter definitions. The displayed dimensionless form remains the same, with consistently redefined parameters.
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