In $ riangle A B C$, if $\cot \frac{A}{2}: \cot \frac{B}{2}: \cot \frac{C}{2}=3: 7: 9$, then $a: b: c=$

Question

In $	riangle A B C$, if $\cot \frac{A}{2}: \cot \frac{B}{2}: \cot \frac{C}{2}=3: 7: 9$, then $a: b: c=$

Aicharya · Accepted Answer

$ \begin{aligned} & 	ext { } \cot \frac{A}{2}: \cot \frac{B}{2}: \cot \frac{C}{2}=3: 7: 9 \ & \sqrt{\frac{s(s-a)}{(s-b)(s-c)}}: \sqrt{\frac{s(s-b)}{(s-a)(s-c)}}: \sqrt{\frac{s(s-c)}{(s-a)(s-b)}} \ & =3: 7: 9 \ & \Delta \sqrt{\frac{s(s-a)}{(s-b)(s-c)}}: \Delta \sqrt{\frac{s(s-b)}{(s-a)(s-c)}}: \Delta \sqrt{\frac{s(s-c)}{(s-a)(s-b)}} \ & =3: 7: 9 \ & \Rightarrow s(s-a): s(s-b): s(s-c)=3: 7: 9 \ & \Rightarrow(s-a):(s-b):(s-c)=3: 7: 9 \end{aligned} $ So, we have: $ \begin{aligned} & s-a=3 k \Rightarrow \frac{a+b+c-a}{2}=3 k \Rightarrow b+c-a=6 k ...(1)\ & 	ext { and } s-b=7 k \Rightarrow a+c-b=14 k ...(2)\ & s-c=9 k \Rightarrow a+b-c=18 k ...(3)\end{aligned} $ Solving equation (1), (2) and (3), we get: $ \begin{aligned} & a=16 k, \quad b=12 k, \quad c=10 k \ & 	herefore a: b: c=16: 12: 10=8: 6: 5 \end{aligned} $

In $\triangle A B C$, if $\cot \frac{A}{2}: \cot \frac{B}{2}: \cot \frac{C}{2}=3: 7: 9$, then $a: b: c=$

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