What can we say about the triangle, where in the usual notation, the following holds

$$
\frac{a}{b}=\sqrt{\frac{\tan \alpha}{\tan \beta}}, \quad a \neq b
$$
The first part of the assumption, equating it with the sine theorem,

$$
\frac{a^{2}}{b^{2}}=\frac{\sin ^{2} \alpha}{\sin ^{2} \beta}=\frac{\operatorname{tg} \alpha}{\operatorname{tg} \beta}=\frac{\sin \alpha \cos \beta}{\cos \alpha \sin \beta}
$$

Here, $\alpha \neq 0$ and $\beta \neq 0$, since these are angles of a triangle, and neither $\alpha$ nor $\beta$ can be $90^{\circ}$, otherwise the assumption would make no sense. Thus, by simplification or cross-multiplication,

$$
\sin \alpha \cos \alpha=\sin \beta \cos \beta \quad \sin 2 \alpha=\sin 2 \beta
$$

This can only hold for the angles of a triangle if $2 \alpha=2 \beta$, i.e., $\alpha=\beta$, or if $2 \alpha+2 \beta=180^{\circ}, \alpha+\beta=90^{\circ}$. The first possibility has been ruled out, so it must be a right-angled, non-isosceles triangle.

Peter Selényi (Budapest, Kvassay J. Bridge and Hydraulic Structures Techn., III. year)