Let's determine the acute angles that satisfy the following equation:

$$
1+\tan \alpha=\frac{35}{12} \sin \alpha
$$
$$
1+\tan \alpha=\frac{35}{12} \sin \alpha
$$

$$
\cos \alpha+\sin \alpha=\frac{35}{12} \sin \alpha \cos \alpha
$$

$$
\sin \alpha+\cos \alpha=\frac{35}{24} \sin 2 \alpha
$$

By squaring both sides of the equation:

$$
\sin ^{2} \alpha+\cos ^{2} \alpha+2 \sin \alpha \cos \alpha=\frac{1225}{576} \sin ^{2} 2 \alpha
$$

But

$$
\sin ^{2} \alpha+\cos ^{2} \alpha=1 \text { and } 2 \sin \alpha \cos \alpha=\sin 2 \alpha
$$

so

$$
\begin{gathered}
1+\sin 2 \alpha=\frac{1225}{576} \sin ^{2} 2 \alpha \\
\sin ^{2} 2 \alpha-\frac{576}{1225} \sin 2 \alpha=\frac{576}{1225}
\end{gathered}
$$

from which

$$
\sin 2 \alpha=\frac{288 \pm 888}{1225}
$$

Since we are only looking for acute angles, we must consider only the positive sign; thus

$$
\sin 2 \alpha=\frac{1176}{1225}=\frac{24}{25}
$$

Solving this equation, we get:

$$
2 \alpha_{1}=73^{\circ} 44^{\prime} 20^{\prime \prime} \quad \text { and } \quad 2 \alpha_{2}=106^{\circ} 15^{\prime} 40^{\prime \prime}
$$

from which

$$
\alpha_{1}=36^{\circ} 52^{\prime} 10^{\prime \prime} \quad \text { and } \quad \alpha_{2}=53^{\circ} 7^{\prime} 50^{\prime \prime}
$$

(Nándor Kántor, 7th grade, Budapest, Evangelical Gymnasium)

The problem was also solved by: Ervin Hofbauer and Pál István Langheim, Budapest; Aladár Visnya, Pécs; Bernát Friedmann, S.A.-Ujhely.