= Solution
Put $c_0=\sqrt{g'h_0/2}$. The undisturbed reservoir has $u=0$ and <Riemann invariants> $u\pm4c=\pm4c_0$. The right-going <characteristic curves> supply the invariant $u+4c=4c_0$ throughout the release. The spreading disturbance is the other family, whose speed is $u-c$. In its <rarefaction wave> let $\xi=x/t=u-c$. Then
$$
c=\frac{4c_0-\xi}{5},\qquad u=\frac{4(c_0+\xi)}5.
$$
At the reservoir edge $u=0,c=c_0$, so the back of the <rarefaction wave> travels at $-c_0$. At the dry edge $c=0$ the <Saint-Venant dry-front condition> gives
$$
\boxed{x_f(t)=4c_0t,\qquad u_f=4c_0=2\sqrt{2g'h_0}.}
$$
Thus the fan occupies $-c_0t\leq x\leq4c_0t$, while $x<-c_0t$ remains undisturbed. Its full length grows at $5c_0$; there is no finite-depth uniform shelf between the fan and the dry front. The optional interior depth is $h=h_0[(4c_0-x/t)/(5c_0)]^2$.
The fan's minus-family <characteristic curves> are rays $x=\xi t$. For the plus family in the fan, $dx/dt=u+c=(8c_0+3x/t)/5$; <integration> gives $x=4c_0t+C t^{3/5}$, with constants selected at entry from the reservoir. These curved paths transport the constant plus <Riemann invariant>.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2002/iii/paper-52-triangular-dam-break.png]
{title=Triangular-channel dry-bed release with its rarefaction rays and right-going characteristics}
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