Question

In a triangle, $r_1=2r_2=3r_3$ then $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}$ is equal to:

• A. $\frac{75}{60}$
• B. $\frac{155}{60}$
• C. $\frac{176}{60}$
• D. $\frac{191}{60}$

Answer 1

The correct option is D $\frac{191}{60}$

Given that, $r_1=2r_2=3r_3$
$\Rightarrow \frac{\Delta }{s-a}=\frac{2\Delta }{s-b}=\frac{3\Delta }{s-c}=\frac{1}{k}\left(\text{say}\right)$
$s-a=k,s-b=2k,s-c=3k\cdots \left(i\right)$
After adding, we get
$3s-(a+b+c)=6k$
$\Rightarrow s=6k\cdots \left(ii\right)$
Solving $\left(i\right),\left(ii\right)$
$a=5k,b=4k,c=3k$
We get,
$\therefore \frac{a}{5}=\frac{b}{4}=\frac{c}{3}=k$
Now, $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}=\frac{5}{4}+\frac{4}{3}+\frac{3}{5}=\frac{191}{60}$