Q85 / 160 · TS EAMCET 2023 (12 May Shift 1) · Mathematics
In a triangle $\mathrm{ABC}$, if $\mathrm{r}_1=2 \mathrm{r}_2=3 \mathrm{r}_3$, then $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}=$
- A$\frac{75}{60}$
- B$\frac{155}{60}$
- C$\frac{176}{60}$
- Correct answerD$\frac{191}{60}$
Solution
$r_1=2 r_2=3 r_3$
$\begin{aligned}
\Rightarrow & \frac{\Delta}{s-a}=\frac{2 \Delta}{s-b}=\frac{3 \Delta}{s-c} \\
& 2(s-a)=s-b \\
\Rightarrow & s=2 a-b ... (i)\\
& 3(s-a)=s-c \\
\Rightarrow & 2 s=3 a-c ... (ii) \\
& 2(s-c)=3(s-b) \\
\Rightarrow & s=3 b-2 c ... (iii)
\end{aligned}$
From (i), $a+b+c=4 a-2 b$
$3 a-3 b=c ... (iv)$
From (ii), $a+b+c=3 a-c$
$\begin{aligned}
& 2 a-b=2 c ... (v) \\
\therefore \quad & 6 a-6 b=2 a-b \\
\Rightarrow & 4 a=5 b \\
\Rightarrow & \frac{a}{b}=\frac{5}{4}... (vi)
\end{aligned}$
$\begin{aligned} & \text { From (iii), } a+b+c=6 b-4 c \\ & \Rightarrow \quad a+5 c=5 b \\ & \therefore \quad a+5 c=4 a \\ & \Rightarrow 5 c=3 a \\ & \Rightarrow \quad \frac{c}{a}=\frac{3}{5} ... (vii) \\ & \because \quad a=\frac{5 b}{4}=\frac{5 c}{3} \\ & \Rightarrow \quad 3 b=4 c \\ & \therefore \quad \frac{b}{c}=\frac{4}{3} ... (viii) \\ & \therefore \quad \frac{a}{b}+\frac{b}{c}+\frac{c}{a}=\frac{5}{4}+\frac{4}{3}+\frac{3}{5}=\frac{75+80+36}{60}=\frac{191}{60} .\end{aligned}$
Hence, the correct answer is option: 'D'.