Solve the right triangle if its area is $1350 \mathrm{~cm}^{2}$ and its sides form an arithmetic progression.
Let $a-d, a, a+d$ be the three sides of the sought triangle. Then

$$
a^{2}=(a+d)^{2}-(a-d)^{2}=4 a d
$$

which means

$$
d=\frac{a}{4}
$$

According to the problem,

$$
a(a-d)=\frac{3}{4} a^{2}=2 \cdot 1350
$$

from which the positive value of $a$ is $60 \mathrm{~cm}$, so $a-d=45 \mathrm{~cm}$ and $a+d=75 \mathrm{~cm}$.

If the angle opposite the side $a$ is $\alpha$, then

$$
\begin{aligned}
\sin \alpha & =\frac{a}{a+d}=0.6 \\
\alpha & =53^{\circ} 7^{\prime} 48^{\prime \prime}
\end{aligned}
$$

(Borota Braniszláv, Szeged.)