In triangle $ABC$, with the usual notations, $a > b$, and the segments $f_{1}$ and $f_{2}$, which are the distances from the internal and external angle bisectors of angle $C$ to the line $AB$, respectively, satisfy

$$
\frac{f_{2}}{f_{1}}=\frac{a+b}{a-b} \sqrt{3}
$$

Determine the angle at $C$.
Let the intersection points of $f_{1}$ and $f_{2}$ with the line $A B$ be $D$ and $E$, respectively. (Note that $E$ does not exist when $a=b$.) Then, the area of triangle $A B C$ can be expressed as the sum of the areas of triangles $C D A$ and $C D B$, and also as the difference of the areas of triangles $C E B$ and $C E A$.

![](https://cdn.mathpix.com/cropped/2024_05_02_4605f5b35ebf6a602a7dg-1.jpg?height=274&width=759&top_left_y=302&top_left_x=672)

If the angle to be found is $\gamma$, these two equalities become:

$$
\begin{gathered}
a f_{1} \sin \frac{\gamma}{2} + b f_{1} \sin \frac{\gamma}{2} = a b \sin \gamma \\
a f_{2} \sin \left(90^{\circ} + \frac{\gamma}{2}\right) - b f_{2} \sin \left(90^{\circ} - \frac{\gamma}{2}\right) = a b \sin \gamma
\end{gathered}
$$

from which

$$
f_{1} = \frac{2 a b}{a + b} \cos \frac{\gamma}{2}, \quad f_{2} = \frac{2 a b}{a - b} \sin \frac{\gamma}{2} \quad \text{and}
$$

$$
\frac{f_{2}}{f_{1}} = \frac{a + b}{a - b} \tan \frac{\gamma}{2}
$$

These are valid in every triangle.

Comparing with the condition:

$$
\tan \frac{\gamma}{2} = \sqrt{3}, \quad \frac{\gamma}{2} = 60^{\circ}, \quad \gamma = 120^{\circ}
$$

Remarks. 1. The segments $f$ can be considered as derived from areas. They can also be calculated from the triangles $C A C_{i}$ ($i=1,2$) using the sine theorem, taking into account the division ratio theorem:

$$
\begin{gathered}
C_{i} A : C_{i} B = C A : C B, \quad A C_{i} = \frac{b}{c \pm b} \cdot c \\
f_{i} \left\{\begin{array}{l}
\sin \\
\cos
\end{array} \frac{\gamma}{2} = A C_{i} \sin \alpha = \frac{b}{a \pm b} c \sin \alpha = \frac{a b \sin \gamma}{a \pm b}\right.
\end{gathered}
$$

2. Let us be aware: we did not determine a specific measurement of a specific triangle - as if the other measurements did not interest us for some reason. The given condition - as the only known relationship for the triangle - is naturally not sufficient for the complete determination of the triangle, but it is sufficient for calculating the specific angle. The equation (1) characterizes a whole class of triangles, and interestingly, the angle at $C$ has a common value for this class.

At first glance, 2 measurements appear on the right side of (1) for the triangles, and 3 in (2). The following transformation better shows that it is about the relationship of ratios:

$$
\frac{f_{2}}{f_{1}} = \frac{1 + \frac{b}{a}}{1 - \frac{b}{a}} \tan \frac{\gamma}{2}
$$