Q59 / 160 · TS EAMCET 2015 · Mathematics
In $\triangle A B C$, if $r_1=2 r_2=3 r_3$, then $b: c$ equals
- Correct answerA$4: 3$
- B$5: 4$
- C$2: 1$
- D$3: 2$
Solution
We have,
$$
\begin{aligned}
& r_1=2 r_2=3 r_3 \\
\Rightarrow & \frac{\Delta}{s-a}=\frac{2 \Delta}{s-b}=\frac{3 \Delta}{s-c}=\lambda \\
\Rightarrow & s-a=\frac{\Delta}{\lambda}, s-b=\frac{2 \Delta}{\lambda}, s-c=\frac{3 \Delta}{\lambda} \\
\Rightarrow & s-a+s-b+s-c=\frac{\Delta}{\lambda}+\frac{2 \Delta}{\lambda}+\frac{3 \Delta}{\lambda}=\frac{6 \Delta}{\lambda} \\
\Rightarrow & \quad s=\frac{6 \Delta}{\lambda}
\end{aligned}
$$
[say]
$\begin{aligned} & \text { Now, } s-a=\frac{\Delta}{\lambda} \Rightarrow \frac{6 \Delta}{\lambda}-a=\frac{\Delta}{\lambda} \Rightarrow a=\frac{5 \Delta}{\lambda} \\ & s-b=\frac{2 \Delta}{\lambda} \Rightarrow \frac{6 \Delta}{\lambda}-b=\frac{2 \Delta}{\lambda} \Rightarrow b=\frac{4 \Delta}{\lambda} \\ & s-c=\frac{3 \Delta}{\lambda} \Rightarrow \frac{6 \Delta}{\lambda}-c=\frac{3 \Delta}{\lambda} \Rightarrow c=\frac{3 \Delta}{\lambda} \\ & \therefore \quad b: c=\frac{4 \Delta}{\lambda}: \frac{3 \Delta}{\lambda} \\ & \Rightarrow \quad b: c=4: 3 \\ & \end{aligned}$
Hence, the correct answer is option: 'A'.