= Solution
With $f=3/4$,
$$
\cos\theta^*=\frac34\cos\theta-\frac14.
$$
Hydrophobicity requires $\theta^*>90^\circ$, hence
$$
\boxed{\theta>\cos^{-1}(1/3)\simeq70.53^\circ}.
$$
Equality is the $\theta^*=90^\circ$ boundary. A smooth-solid angle between about $70.53^\circ$ and $90^\circ$ can therefore become apparently hydrophobic if the suspended state persists.
For general $0<f\le1$, the criterion is $\cos\theta<1/f-1$. For $1/2<f\le1$, the threshold is $\cos^{-1}(1/f-1)$, reaching $90^\circ$ at $f=1$. For $f=1/2$, all $\theta>0$ satisfy it, with $\theta=0$ marginal. For $f<1/2$, the ideal equation predicts hydrophobicity for every original angle; formally $\theta^*\to180^\circ$ as $f\to0$.
\b[These conclusions require retention of gas beneath the liquid.] On intrinsically wettable textures the <Cassie–Baxter state> can be metastable. Pressure, condensation or defects may trigger invasion; penetration tests and contact-angle hysteresis distinguish that failure from the ideal suspended state. Roughness alone does not guarantee stable water repellency.
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