Q68 / 160 · TS EAMCET 2024 (09 May Shift 2) · Mathematics

In a triangle $A B C$, if $\tan \frac{A}{2}: \tan \frac{B}{2}: \tan \frac{C}{2}=15: 10: 6$, then $\frac{a}{b-c}=$

  • A
    $\frac{8}{3}$
  • B
    $\frac{7}{3}$
  • C
    5
  • Correct answer
    D
    4

Solution

We know that $\begin{aligned} & \tan \frac{A}{2}=\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}=15 \mathrm{~K} \quad ...\mathrm{(i)} \\ & \tan \frac{B}{2}=\sqrt{\frac{(s-a)(s-c)}{s(s-b)}}=10 \mathrm{~K} \quad ...\mathrm{(ii)} \\ & \tan \frac{C}{2}=\sqrt{\frac{(s-a)(s-b)}{s(s-c)}}=6 K \quad ...\mathrm{(iii)} \end{aligned}$
Divide (i) by (ii), we get $\begin{aligned} & \sqrt{\frac{(s-b)(s-c)}{s(s-a)} \times \frac{s(s-b)}{(s-a)(s-c)}}=\frac{15}{10} \\ & \Rightarrow \frac{s-b}{s-a}=\frac{3}{2} \Rightarrow s=3 a-2 b \quad ...\mathrm{(iv)} \end{aligned}$
Divide (ii) by (iii), we get $\begin{aligned} & \sqrt{\frac{(s-a)(s-c)}{s(s-b)}} \times \frac{s(s-c)}{(s-a)(s-b)} \end{aligned}=\frac{10}{6}$ $\begin{aligned} & \Rightarrow 5 b-3 c=25 \Rightarrow 5 b-3 c=a+b+c \\ & \Rightarrow 4 b-4 c=9 \Rightarrow \frac{a}{b-c}=4\end{aligned}$

Hence, the correct answer is option: 'D'.

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