Given a triangle, two angles are provided: $\alpha=43^{\circ} 36^{\prime} 10^{\prime \prime}, \beta=11^{\circ} 25^{\prime} 8^{\prime \prime}$, and the difference between the projections of sides $a$ and $b$ onto the third side: $p-q=d=78$. What is the area of the triangle?
If the height corresponding to side $a$ is $C D$, then

$$
\cot \alpha=\frac{q}{C D} \text { and } \cot \beta=\frac{p}{C D}
$$

so

$$
p: q=\tan \alpha: \tan \beta
$$

from which

$$
(p+q):(p-q)=(\tan \alpha+\tan \beta):(\tan \alpha-\tan \beta)
$$

thus

$$
p+q=c=d \frac{\tan \alpha+\tan \beta}{\tan \alpha-\tan \beta}=d \frac{\sin (\alpha+\beta)}{\sin (\alpha-\beta)}
$$

The area of the triangle is

$$
\begin{gathered}
t=\frac{c^{2} \sin \alpha \sin \beta}{2 \sin \gamma}=\frac{d^{2} \sin ^{2}(\alpha+\beta) \sin \alpha \sin \beta}{2 \sin (\alpha+\beta) \sin ^{2}(\alpha-\beta)} \\
t=\frac{d^{2}}{2} \cdot \frac{\sin \alpha \sin \beta \sin (\alpha+\beta)}{\sin ^{2}(\alpha-\beta)}
\end{gathered}
$$

Substituting the given values: $t=1199.6 \mathrm{~m}^{2}$.

(Tódor Kármán.)

The problem was also solved by: Bojedain F., Borbély Z., Brandt D., Dénes A., Fleischmann S., Laczkó E., Lukhaub Gy., Petrogalli G., Probst E., Szabó I., Szabó K.