Find the longest side of the triangle. The sides $a,b,c$ of a $\triangle ABC$ are in $GP$ whose common ratio is $\frac{2}{3}$ and the circumradius of the triangle is $6\sqrt{\frac{7}{209}}$.Find the longest side of the triangle.
I used law of sines $$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$$
I took $a=a,b=\frac{2}{3}a,c=\left(\frac{2}{3}\right)^2a$
which gives $$\frac{\sin A}{\sin B}=\frac{\sin B}{\sin C}$$
I am stuck now,how to find the longest side $a$.

Let $a,b,c$ be the side lengths. Then, $b,c$ can be written as $$b=\frac 23a,\quad c=\left(\frac 23\right)^2a$$
where $a$ is the longest side. So, we have, by the law of cosines,
$$a^2=\left(\frac 23a\right)^2+\left(\left(\frac 23\right)^2a\right)^2-2\cdot\frac 23a\cdot\left(\frac 23\right)^2a\cos A$$
$$\Rightarrow \cos A=-\frac{29}{48}.$$
Hence, 
$$a=2R\sin A=2R\sqrt{1-\cos^2A}=2\cdot 6\sqrt{\frac{7}{209}}\sqrt{1-\left(-\frac{29}{48}\right)^2}=\frac 74.$$
