Solve the right triangle if $\alpha=35^{\circ} 15^{\prime} 25^{\prime \prime}$ and the angle bisector of $\alpha$ is $t=15.3 \mathrm{~cm}$.
If the leg opposite to angle $\alpha$ is $a$, the adjacent leg is $b$, and the hypotenuse is $c$, then

$$
\begin{gathered}
b=t \cos \frac{\alpha}{2}=14.581 \text{ cm} \\
a=b \operatorname{tg} \alpha=t \cos \frac{\alpha}{2} \operatorname{tg} \alpha=10.307 \text{ cm} \\
c=\frac{b}{\cos \alpha}=t \frac{\cos \frac{\alpha}{2}}{\cos \alpha}=17.857 \text{ cm}
\end{gathered}
$$

The area of the triangle is

$$
T=\frac{a b}{2}=\frac{t^{2}}{2} \cos ^{2} \frac{\alpha}{2} \operatorname{tg} \alpha=75.143 \text{ cm}^{2}
$$

and finally the other acute angle

$$
\beta=90^{\circ}-\alpha=54^{\circ} 44' 35''
$$

(Kálmán Bendl, Budapest.)

The problem was also solved by: Bayer N., Csada I., Ehrenfeld N., Füstös P., Kirchknopf E., Kürth R., Lorenc A., Löwy J., Lusztig M., Paunz A., Sárközy P., Sebestyén I., Spitzer L., Szőke D., Wáhl V.