How to find length of the sides of a triangle given the ratio of the sines of the angles? Consider $\triangle ABC$. 
Let $\dfrac{\sin A}{\sin B} = \dfrac56$ and 
$\dfrac{\sin B}{\sin C} = \dfrac45$. 
Find $\dfrac{\vert AC\vert\cdot \vert AB\vert}{\vert BC\vert}$.
If there is no definite answer, express in terms of one of the sides.
 A: Using the Law of sines: $\dfrac {\sin A}{|BC|} = \dfrac{\sin B}{|AC|} = \dfrac{\sin C}{|AB|}=D$ ($D$ is the diameter of the circumcircle)
we see that \begin{align}\dfrac{\sin A}{\sin B} =\dfrac{\vert BC\vert}{\vert AC\vert}= \dfrac56\\\dfrac{\sin B}{\sin C} =\dfrac{\vert AC\vert}{\vert AB\vert}= \dfrac45\end{align}
So $\displaystyle \dfrac{\vert AC\vert\cdot \vert AB\vert}{\vert BC\vert} = \vert AB\vert\frac 65$
A: From the Law of Sinus, we know that:
$\frac{\text{sin}(A)}{\text{sin}(B)}=\frac{|BC|}{|AC|}=\frac{5}{6}$
$\frac{\text{sin}(B)}{\text{sin}(C)}=\frac{|AC|}{|AB|}=\frac{4}{5}$
$\frac{\text{sin}(A)}{\text{sin}(C)}=\frac{|BC|}{|AB|}=\frac{2}{3}$
From this we get:
$6|BC|-5|AC|=0$
$5|AC|-4|AB|=0$
$3|BC|-2|AB|=0$
We can solve these equations with linear algebra:
$$
\left[
\matrix
{
6& -5&  0 & 0 \\
0&  5& -4 & 0 \\
3&  0& -2 & 0 
}
\right]\rightarrow\left[
\matrix
{
1& -\frac{5}{6}&  0 & 0 \\
0&  1& -\frac{4}{5} & 0 \\
0& 0&  0 & 0 
}
\right]
$$
Therefore, if t=|BC|, the solutions is given by:
$\left[
\matrix
{
|BC| \\
|AC| \\
|AB|
}
\right]=t\left[
\matrix
{
1 \\
\frac{5}{6} \\
\frac{2}{3}
}
\right]$
And the desired product is:
$\frac{5}{9t}= \frac{5}{9|BC|}$
