Finding side length from two scalene triangles with common angle and side and ratio between sides. Trigonometry I have two scalene triangles with a common angle and side. I would like to find the length of $x_2$.
I have all variables in green: the angles $\angle$ B,$\angle$ C,$\angle$ D,$\angle$ E and side "length". I also have the ratio $\frac{x_2}{x_1}$.
Stumped on this trigonometry question. I have tried using Sine rule and solving simultaneously but always end up canceling out all my terms. In my head there are enough equations to solve for the variables but can't seem to figure out how I would do it.
Any help greatly appreciated.

 A: There are three triangles. Two compose the largest one. I'll call the leftmost part $L,$ its counterpart on the right $R,$ and their composite $K.$
Looking at $R,$ we can write down an equation for our goal $x_2$ as follows: $$x_2^2=a_1^2+a_2^2-2a_1a_2\cos E.$$ Thus we only need find $a_1$ and $a_2,$ and we'd be done.
To find $a_1,$ look at $L,$ the we have that $$\frac{\sin B}{\text{Length}}=\frac{\sin A}{a_1}.$$ Similarly, by looking at $K,$ we find that $$\frac{\sin D}{\text{Length}}=\frac{\sin A}{a_2}.$$ This helps us to find $a_2.$
From the last two equations we eliminate $A$ to find an equation in $a_1,\,a_2.$
Now look at the triangle $R,$ from which we may write down $$\frac{\sin E}{x_2}=\frac{\sin D}{a_1}.$$
Together with the first equation, we find a $3×3$ system in $a_1,\,a_2\,x_2,$ which we can now solve for $x_2.$
OK, let's do this together. From the two middle equations we find that $$\sin A=\frac{a_1\sin B}{\text{Length}}=\frac{a_2\sin D}{\text{Length}},$$ so that $$a_1=\frac{\sin D}{\sin B}a_2.$$ From the last equation we also have $$a_1=\frac{\sin D}{\sin E}x_2,$$ so that we have that $$a_2=\frac{\sin D}{\sin E}\frac{\sin B}{\sin D}x_2=\frac{\sin B}{\sin D}x_2.$$
Substituting now into the first equation and solving for $x_2$ is straightforward, and gives $$x_2^2\left(\frac{\sin B+\sin D}{\sin E}+1\right)=\frac{2\sin B\sin D\cot E}{\sin E},$$ amd finally recall that $x_2>0.$
A: From the cosine and sine rules,
$$L^2 = x_1^2+ a_1^2 -2x_1a_1\cos B\tag{1}$$
$$a_1=\frac{\sin D}{\sin E}x_2\tag{2}$$
Plug the known ratio $r=\frac{x_1}{x_2}$ and (2) into (1),
$$L^2 = r^2x_2^2+ \frac{\sin^2 D}{\sin^2 E}x_2^2 -2r\cos B \frac{\sin D}{\sin E}x_2^2\tag{1}$$
Solve for $x_2$,
$$x_2= L \left(r^2+ \frac{\sin^2 D}{\sin^2 E} -2r\cos B \frac{\sin D}{\sin E}\right)^{-1/2}$$
