Someone standing on the bank of a river sees a building on the opposite bank at an angle of $60^{\circ}$; if they move $30 \mathrm{~m}$ further, they see the building at an angle of $30^{\circ}$. Calculate the width of the river and the height of the building.
If the width of the river is $x$, and the height of the building is $y$, then

$$
y=x \operatorname{tg} 60^{\circ}=x \sqrt{3}
$$

and

$$
y=(30+x) \operatorname{tg} 30^{\circ}=(30+x) \frac{\sqrt{3}}{3}
$$

thus

$$
30+x=3 x
$$

from which

$$
x=15 \mathrm{~m}
$$

and therefore

$$
y=15 \sqrt{3}=25.98 \mathrm{~m}
$$

(Borota Braniszláv, Szeged.)