Let's determine the area of the right-angled triangle, given that $\alpha=38^{\circ} 40^{\prime}$ and the angle bisector of $\alpha$ is $f_{a}=7.8 \mathrm{~cm}$.
Denoting the leg opposite to $\alpha$ by $a$, and the leg adjacent to $\alpha$ by $b$:

$$
b=f_{a} \cos \frac{\alpha}{2}, \quad a=b \operatorname{tg} \alpha=f_{a} \cos \frac{\alpha}{2} \cdot \operatorname{tg} \alpha
$$

and thus the sought area

$$
\begin{aligned}
& t=\frac{a b}{2}=\frac{f_{a}^{2} \cos ^{2} \frac{\alpha}{2} \operatorname{tg} \alpha}{2}=\frac{7.8^{2} \cos ^{2} 19^{\circ} 20^{\prime} \cdot \operatorname{tg} 38^{\circ} 40^{\prime}}{2}=\frac{60.84 \cdot 0.9436^{2} \cdot 0.8002}{2}= \\
& =60.84 \cdot 0.8904 \cdot 0.4001=21.67 \mathrm{~cm}^{2}
\end{aligned}
$$

Péter Tahy (Bp. II. Rákóczi g. II. o. t.)