How to calculate the angles from angle-side couples in a series of triangles (with an algebraic solution)? In this image, the red and black labels are known, and the aim is to calculate one of the blue angles or sides (the rest will follow). 

There is no standard trigonometric solution, as we know only one angle and one side for each triangle. However, I believe there should be an algebraic solution for the common variables we have for different angles.
For example,
$$\frac{\sin b}{2A} = \frac{\sin c}{C} = \frac{\sin d}{D} $$
$$\frac{\sin b1}{A} = \frac{\sin c}{B} = \frac{\sin a}{D} $$
$$\frac{\sin b2}{A} = \frac{\sin e}{C} = \frac{\sin d}{B} $$
When knowing only $2A$ and $b$ for the main triangle, there are unlimited solutions. However, the given $a$ restricts to one single solution. This is why I believe this problem is solvable.
 A: It is difficult to analyze this symbolically, but at least I can give the equations and you can check if they have solutions for certain values of the given quantities. The strategy is to obtain two equations relating $D$ and $C$. Let $M=\cos\left(b_{1}+b_{2}\right)$ and $S=\sin\left(b_{1}+b_{2}\right)$. By inspection of the figure, we easily obtain the following equation: 
$$ C=\frac{2A\sin\left(b_{1}+a\right)}{\sin\left(b_{1}+b_{2}\right)} \tag{1}$$ Now $$\cos\left(a-b_{2}\right)=\cos\left(\left(a+b_{1}\right)-\left(b_{1}+b_{2}\right)\right)=M\cos\left(a+b_{1}\right)+S\sin\left(a+b_{1}\right) \tag{2}$$
We also have: $$B^{2}=A^{2}+C^{2}-2AC\cos\left(a-b_{2}\right) \tag{3}$$
$$B^{2}=A^{2}+D^{2}+2AD\cos\left(a+b_{1}\right) \tag{4}$$ Equating $(5)$ and $(6)$ gives
$$C^{2}-D^{2}=2A\left(C\cos\left(a-b_{2}\right)+D\cos\left(a+b_{1}\right)\right) \tag{5}$$ We also have $$ 4A^{2}=D^{2}+C^{2}-2DC\cos\left(b_{1}+b_{2}\right) \tag{6}$$ Here comes the tedious part, you substitute from $(2)$ into $(5)$, use the identity $\cos\left(a+b_{1}\right)=\sqrt{1-\sin^{2}\left(a+b_{1}\right)}$, and substitute for $\sin\left(a+b_{1}\right)$ from $(1)$ and finally solve $(5)$ and $(6)$ simultaneously. You can do that because both equations will only have $C$ and $D$ as unknowns in terms of known quantities then. You will get a monstrous polynomial and the situation will probably be quite complicated even if you substitute numbers for the known quantities. It would be interesting to know if this gives the solution(s) though.
One more thing, note that there will be  more to do if you were able to find suitable solutions for $(5)$ and $(6)$, namely, you will have to deal with the following system of equations, with $b$ given, you have: $$b_{1}+b_{2}=b \tag{7}$$
$$\sin\left(b_{1}\right)=\frac{A\sin a}{D} \tag{8} $$
$$\sin\left(b_{2}\right)=\frac{A\sin\left(a\right)}{C} \tag{9}$$
So $a$ can't be given arbitrarily. This says that this arrangement you are considering is quite special, if it exists (After playing around with some values, I think this can't exist due to the presence of some heavy constraints on $A$, $C$ and $a$, but attempting to prove this in general would be tedious).
A: Apply the sine rule to the two divided triangles to get
$$D = \frac{\sin a}{\sin b_1} A, \>\>\>\>\> C = \frac{\sin a}{\sin b_2} A$$
Substitute above $D$ and $C$ into the cosine formula below
$$4A^2 = D^2+C^2 -2DC \cos b$$
to obtain
$$\frac4{\sin^2a} = \frac1{\sin^2 b_1} + \frac1{\sin^2 b_2}  -\frac{2\cos b}{\sin b_1\sin b_2} \tag1$$
Along with
$$b_1+b_2=b\tag2$$
the two equations (1) and (2) determine $b_1$ and $b_2$. All other blue angles and sides can then be calculated.
