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Question: Answered & Verified by Expert
In $\triangle A B C$, if $\cot \frac{A}{2}: \cot \frac{B}{2}: \cot \frac{C}{2}=3: 7: 9$, then $a: b: c=$
MathematicsProperties of TrianglesAP EAMCETAP EAMCET 2023 (15 May Shift 1)
Options:
  • A $8: 6: 5$
  • B $5: 6: 8$
  • C $10: 8: 5$
  • D $5: 8: 10$
Solution:
1130 Upvotes Verified Answer
The correct answer is: $8: 6: 5$
$$
\begin{aligned}
& \text { } \cot \frac{A}{2}: \cot \frac{B}{2}: \cot \frac{C}{2}=3: 7: 9 \\
& \sqrt{\frac{s(s-a)}{(s-b)(s-c)}}: \sqrt{\frac{s(s-b)}{(s-a)(s-c)}}: \sqrt{\frac{s(s-c)}{(s-a)(s-b)}} \\
& =3: 7: 9 \\
& \Delta \sqrt{\frac{s(s-a)}{(s-b)(s-c)}}: \Delta \sqrt{\frac{s(s-b)}{(s-a)(s-c)}}: \Delta \sqrt{\frac{s(s-c)}{(s-a)(s-b)}} \\
& =3: 7: 9 \\
& \Rightarrow s(s-a): s(s-b): s(s-c)=3: 7: 9 \\
& \Rightarrow(s-a):(s-b):(s-c)=3: 7: 9
\end{aligned}
$$
So, we have:
$$
\begin{aligned}
& s-a=3 k \Rightarrow \frac{a+b+c-a}{2}=3 k \Rightarrow b+c-a=6 k ...(1)\\
& \text { and } s-b=7 k \Rightarrow a+c-b=14 k ...(2)\\
& s-c=9 k \Rightarrow a+b-c=18 k
...(3)\end{aligned}
$$
Solving equation (1), (2) and (3), we get:
$$
\begin{aligned}
& a=16 k, \quad b=12 k, \quad c=10 k \\
& \therefore a: b: c=16: 12: 10=8: 6: 5
\end{aligned}
$$

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