Solve the right triangle if the hypotenuse $c=17.04 \mathrm{~m}$ and the bisector of angle $\beta$ is $t=12.96 \mathrm{~m}$.
If $a$ is the leg adjacent to angle $\beta$, then

$$
\frac{a}{c}=\cos \beta=2 \cos ^{2} \frac{\beta}{2}-1
$$

But

$$
\cos \frac{\beta}{2}=\frac{a}{t}
$$

so

$$
\frac{a}{c}=2 \frac{a^{2}}{t^{2}}-1
$$

or

$$
2 a^{2} c-a t^{2}-t^{2} c=0
$$

from which

$$
a=\frac{t}{4 c}\left(t+\sqrt{t^{2}+8 c^{2}}\right)
$$

The root must be taken with the positive sign, because $\sqrt{t^{2}+8 c^{2}}>t$. Substituting the given numerical values, we obtain

$$
a=11.95 \mathrm{~m}
$$

The angle $\beta$ can be found from the formula

$$
\cos \beta=\frac{a}{c}
$$

(Paunz Arthur, Pécs.)

The problem was also solved by: Csada I., Kirchknopf E., Kiss J., Kürth R., Sárközy P., Wáhl V.