For a right-angled triangle with acute angles $\alpha$ and $\beta$:

$$
\operatorname{tg} \alpha+\operatorname{tg} \beta+\operatorname{tg}^{2} \alpha+\operatorname{tg}^{2} \beta+\operatorname{tg}^{3} \alpha+\operatorname{tg}^{3} \beta=70
$$

Determine the angles.
We know that the tangents of the acute angles of a right-angled triangle are reciprocals of each other, so with the notation $\operatorname{tg} \alpha=x$, our equation can be written as

$$
x+\frac{1}{x}+x^{2}+\frac{1}{x^{2}}+x^{3}+\frac{1}{x^{3}}=70
$$

This is a so-called reciprocal equation, and it is useful to introduce a new unknown $A$ for $x+\frac{1}{x}$. Then $x^{2}+\frac{1}{x^{2}}=A^{2}-2$ and $x^{3}+\frac{1}{x^{3}}=A^{3}-3 A$, so the equation becomes

$$
A+A^{2}-2+A^{3}-3 A=70
$$

Rearranging and factoring, we get

$$
(A-4)\left(A^{2}+5 A+18\right)=0
$$

Since $A^{2}+5 A+18=(A+2.5)^{2}+11.75>0$, the equation has only one real root $A=4$. From this, we get $x+\frac{1}{x}=4$, that is, $x^{2}-4 x+1=0$. Therefore, $x_{1,2}=2 \pm \sqrt{3}$.

Since $2+\sqrt{3}=\frac{1}{2-\sqrt{3}}$, the tangents of the two acute angles of the triangle are $2+\sqrt{3}$ and $2-\sqrt{3}$, so the acute angles of the triangle are $75^{\circ}$ and $15^{\circ}$. (We can determine the trigonometric functions of these angles precisely using the addition theorems, for example,

$$
\left.\operatorname{tg} 75^{\circ}=\operatorname{tg}\left(45^{\circ}+30^{\circ}\right)=\frac{\operatorname{tg} 45^{\circ}+\operatorname{tg} 30^{\circ}}{1-\operatorname{tg} 45^{\circ} \cdot \operatorname{tg} 30^{\circ}}=\frac{1+\frac{\sqrt{3}}{3}}{1-1 \cdot \frac{\sqrt{3}}{3}}=\frac{3+\sqrt{3}}{3-\sqrt{3}}=2+\sqrt{3} .\right)
$$

Based on the work of György Venter (Fazekas M. Főv. Gyak. Gimn., 11th grade)