= Solution
Rearrange the <first integral> from part (iv) and use the <half-angle identities>:
$$
\lambda-1=\frac{2y\sin^2(\psi/2)}{y_0-y}=\frac{2y_1\sin^2(\alpha/2)}{y_0-y_1}.
$$
Eliminating $\lambda$ gives
$$
\boxed{\sin^2\frac\psi2=\frac{y_1}{y}\frac{y_0-y}{y_0-y_1}\sin^2\frac\alpha2,\qquad y_1<y<y_0.}
$$
Squaring removes the sign of the slope; the rising and falling sides of the same profile have opposite tangent angles but obey the same height relation. Heights and angles must remain on the regular branch with $\cos\psi>0$.
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