You start by constructing a segment with length 'a' then draw an angle of measure 'α', another segment with length 'b' and another angle of measure 'β' and a segment with length c.What is the distance between the points a and c, which we will call it d.
Note that by b and d I mean the length of the entire segment and not a portion of it.I've tried solving it in many different ways I noted $b_1$,$b_2$ and $d_1$,$d_2$ the upper and lower part of b and same for d and I managed to express d only after $d_2$ using the law of cosines and sines but then I had the problem that in order to calculate $d_2$ we need to express $b_2$ and I got nowhere. After that I tried extending segments angles and nothing worked.I imagine you could work out a system of equations and work d, implying more algreba than geometry.I would appreciate a full solution and answer, thank you.
Applying the law of cosines to $a$ and $b$, you can obtain the opposite side (i call it $e$):
\begin{equation} e^2=a^2+b^2-2ab\cos\alpha\\ e=\sqrt{a^2+b^2-2ab\cos\alpha} \end{equation}
Now you can use the law of sines getting the opposite angle (i call it $\gamma$) to the side $a$:
\begin{equation} \frac{\sin\gamma}{a}=\frac{\sin\alpha}{e}\\ \sin\gamma=a\frac{\sin\alpha}{e}\\ \gamma=\arcsin({a\frac{\sin\alpha}{e}})\\ \gamma=\arcsin\left({a\frac{\sin\alpha}{\sqrt{a^2+b^2-2ab\cos\alpha}}}\right) \end{equation}
Finally you can get $d$ applying the law of cosines again:
\begin{equation} d=\sqrt{e^2+c^2-2ec\cos(\gamma+\beta)}\\ d=\sqrt{a^2+b^2-2ab\cos\alpha+c^2-2ec\cos(\gamma+\beta)}\\ d=\sqrt{a^2+b^2+c^2-2ab\cos\alpha-2ec\cos(\gamma+\beta)}\\ d=\sqrt{a^2+b^2+c^2-2ab\cos\alpha-2c\sqrt{a^2+b^2-2ab\cos\alpha}\cos\left[\arcsin\left({a\frac{\sin\alpha}{\sqrt{a^2+b^2-2ab\cos\alpha}}}\right)+\beta\right]} \end{equation}