<s>[INST] 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 $a$.
 [/INST] The side of the triangle in G.P. with common ratio $\frac{2}{3}$ can be taken as $$a=\frac{3}{2}x, b=x, c=\frac{2}{3}x$$
Where, $x$ is a positive real number.
Hence, using cosine formula we get $$\cos C=\frac{a^2+b^2-c^2}{2ab}$$ $$=\frac{\frac{9}{4}x^2+x^2-\frac{4}{9}x^2}{2\cdot\frac{3}{2}x \cdot x }=\frac{101}{108}$$ $$\implies \sin C=\sqrt{1-\cos ^2 A}=\sqrt{1-\frac{101^2}{108^2}}=\frac{\sqrt{1463}}{108}$$ Hence, the circumscribed radius of the triangle is  given by the following formula $$\frac{c}{2\sin C}$$ $$=\frac{\frac{2}{3}x}{2\frac{\sqrt{1463}}{108}}=\frac{36}{\sqrt{1463}}x$$ $$\implies \frac{36}{\sqrt{1463}}x=6\sqrt{\frac{7}{209}}$$ $$x=\frac{7}{6}$$ Now, setting the value of $x$, all the sides of the triangle are calculated as follows   $$a=\frac{3}{2}x=\frac{3}{2}\cdot \frac{7}{6}=\frac{7}{4}$$ $$b=x=\frac{7}{6}$$ $$c=\frac{2}{3}x=\frac{2}{3}\cdot \frac{7}{6}=\frac{7}{9}$$
longest side is $a=\frac{7}{4}$
$$\bbox[5px, border:2px solid #C0A000]{\color{red}{\text{Longest side of the triangle},a=\color{blue}{\frac{7}{4}}}}$$
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