Let's construct a triangle, given one side with the angle lying on it, as well as the sum of the other two sides.
In triangle $ABC$, let the given side be $a$, and the angle opposite to it be $\gamma$.

![](https://cdn.mathpix.com/cropped/2024_05_02_e998f505c06ee6835b79g-1.jpg?height=436&width=355&top_left_y=229&top_left_x=880)

If side $c$ is rotated around vertex $A$ onto the ray $CA$, we obtain triangle $CDB$. This is constructible since two sides are known: $a$ and $b+c$, as well as the angle $\gamma$ between them.

Triangle $ADB$ is isosceles, and thus $\angle ADB = \angle DBA$. This angle must be measured on the ray $BD$ from vertex $B$, specifically on the side of the ray $BD$ that contains $C$. If the condition $b+c > a$ is satisfied, the angle opposite to $b+c$ is greater than the angle at $D$, and thus $A$ indeed lies within the segment $CD$.

The solvability condition: $b+c > a$ and $0^{\circ} < \gamma < 180^{\circ}$.