Let's construct a triangle, given two sides, $a$ and $b$, and knowing that (using the usual notation) $\alpha=2 \beta$.
![](https://cdn.mathpix.com/cropped/2024_05_02_1c7b69c064e4477e74f0g-1.jpg?height=247&width=382&top_left_y=131&top_left_x=861)

Figure 1

Solution. Let's write the sine rule for the sides of the triangle:

$$
\frac{a}{b}=\frac{\sin 2 \beta}{\sin \beta}=\frac{2 \sin \beta \cos \beta}{\sin \beta}
$$

From this,

$$
\cos \beta=\frac{a}{2 b}
$$

Since \(0 \leq \cos \beta \leq 1\), it follows that \(a \leq 2 b\). Moreover, the larger angle is opposite the larger side, so \(b < a\). Therefore, we can construct the desired triangle:

We construct a circular arc with the given segment \(a\) as the chord and an angle \(\alpha = 2 \beta\). We then intersect this arc with a segment of length \(b\). The solvability condition is \(b < a < 2 b\), and since \(3 \beta + \gamma = 180^\circ\), it follows that \(\beta < 60^\circ\). There is a unique solution because a triangle can always be uniquely constructed from two sides and the angle opposite the larger side.

![](https://cdn.mathpix.com/cropped/2024_05_02_1c7b69c064e4477e74f0g-1.jpg?height=395&width=377&top_left_y=1136&top_left_x=858)

Figure 2