12.013. The ratio of the area of a right-angled triangle to the area of a square constructed on its hypotenuse is $k$. Find the sum of the tangents of the acute angles of the triangle.
## Solution.

In $\triangle A C B \quad \angle A C B=90^{\circ}$. Let $\angle C A B=\alpha, \angle A B C=\beta$, then $\operatorname{tg} \alpha+\operatorname{tg} \beta=\frac{\sin (\alpha+\beta)}{\cos \alpha \cdot \cos \beta}=\frac{1}{\cos \alpha \cdot \cos \beta} \cdot\left(\alpha+\beta=90^{\circ}\right)$. Denote $A B=B F=$ $=E F=A E=x \quad\left(A B F E-\right.$ square) (Fig. 12.16). $\frac{S_{\triangle A C B}}{S_{A B F E}}=\frac{A C \cdot B C}{2 \cdot x^{2}} ;$ $A C=A B: \cos \alpha, C B=A B \cdot \cos \beta$, then $\frac{S_{\triangle A C B}}{S_{A B F E}}=\frac{x^{2} \cos \alpha \cdot \cos \beta}{x^{2} \cdot 2} ; \frac{S_{\triangle A C B}}{S_{A B F E}}=k$ (by condition). We have that $\frac{1}{2 k}=\frac{1}{\cos \alpha \cdot \cos \beta}$. Therefore,

$$
\operatorname{tg} \alpha+\operatorname{tg} \beta=\frac{1}{2 k}
$$

Answer: $\quad \operatorname{tg} \alpha+\operatorname{tg} \beta=\frac{1}{2 k}$