The sides of a triangle are in the ratio $3: 4: 5$. The sum of the areas of the squares that can be drawn over the sides of the triangle is $800 \mathrm{~m}^{2}$. What are the lengths of the sides and the angles of the triangle?
The triangle is right-angled because $3^{2}+4^{2}=5^{2}$.

Let the sides of the triangle be $3 x, 4 x$ and $5 x$, and let the angle opposite the smaller leg be $\alpha$, then

$$
\sin \alpha=\frac{3}{5}
$$

and thus

$$
\alpha=36^{\circ} 52^{\prime} 12^{\prime \prime}
$$

Furthermore,

$$
9 x^{2}+16 x^{2}+25 x^{2}=800
$$

so

$$
x=4
$$

therefore, the sides of the triangle are: $12 \mathrm{~m}, 16 \mathrm{~m}$ and $20 \mathrm{~m}$.

(Bertalan Lázár, Eger.)

Number of solutions: 29.