Let's solve the right-angled triangle whose area is $504 \mathrm{~cm}^{2}$, and the hypotenuse is $65 \mathrm{~cm}$.
Let's denote the area by $t$, the hypotenuse by $c$, and the legs by $a$ and $b$. Then,

$$
a^{2}+b^{2}=c^{2}
$$

and

$$
a b=2 t
$$

By adding twice the second equation to the first, or subtracting it from the first, we obtain the following two equations:

$$
(a+b)^{2}=c^{2}+4 t
$$

and

$$
(a-b)^{2}=c^{2}-4 t
$$

from which

$$
a=\frac{\sqrt{c^{2}+4 t}+\sqrt{c^{2}-4 t}}{2} \quad \text { and } \quad b=\frac{\sqrt{c^{2}-4 t}-\sqrt{c^{2}-4 t}}{2}
$$

In this case,

$$
a=63 \mathrm{~cm}, \quad b=16 \mathrm{~cm}, \quad \text { or } \quad a=16 \mathrm{~cm}, \quad b=63 \mathrm{~cm} .
$$

The angles are provided by the formulas

$$
\sin \alpha=\frac{a}{c} \quad \text { and } \quad \sin \beta=\frac{b}{c}
$$

Substituting the values, we get

$$
\alpha=75^{\circ} 45^{\prime} \quad \text { and } \quad \beta=14^{\circ} 15^{\prime}
$$

Number of solutions: 35.