How to find length of the sides of a triangle given the ratio of the sines of the angles?

Question

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.

Answer 1

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 $\displaystyle \dfrac{\vert AC\vert\cdot \vert AB\vert}{\vert BC\vert} = \vert AB\vert\frac 65$

Answer 2

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:

$\frac{5}{9t}= \frac{5}{9|BC|}$