Question

In $\Delta$ $ABC$ the sides opposite to angles $A,B,C$ are denoted by $a,b,c$ respectively.

If $r_1=2r_2=3r_3$, then $a+b+c$ is equal to?

• A. $3a$
• B. $3b$
• C. $3c$
• D. none of these

Answer 1

The correct option is B $3b$

Given $r_1=2r_2=3r_3.$

$\Rightarrow s\tan \frac{A}{2}=2s\tan \frac{B}{2}=3s\tan \frac{C}{2}$

$\Rightarrow \tan \frac{A}{2}=2\tan \frac{B}{2}=3\tan \frac{C}{2}=k$ (say)

$\Rightarrow \tan \frac{A}{2}=k;\tan \frac{B}{2}=\frac{k}{2};\tan \frac{C}{2}=\frac{k}{3}$

Now, $\sum \tan \frac{A}{2}.\tan \frac{B}{2}=1$ $($ In $△ABC)$

$\Rightarrow \frac{k^2}{2}+\frac{k^2}{6}+\frac{k^2}{3}=1\Rightarrow k^2=1$

$\Rightarrow k=1$ $\because \tan \frac{A}{2}$ etc are+ve$)$

$\Rightarrow \tan \frac{A}{2}=1,\tan \frac{B}{2}=\frac{1}{2},\tan \frac{C}{2}=\frac{1}{3}$

$\Rightarrow A=\frac{\pi }{2}\Rightarrow R=\frac{a}{2}$

$b=2R\sin B=a\left[\frac{2\tan \left(\frac{B}{2}\right)}{1+{\tan }^2\left(\frac{B}{2}\right)}\right]=a\left[\frac{1}{1+\frac{1}{4}}\right]=\frac{4a}{5}$ and

$c=2R\sin C=a\left[\frac{2\tan \left(\frac{C}{2}\right)}{1+{\tan }^2\left(\frac{C}{2}\right)}\right]=\frac{\frac{2}{3}a}{1+\frac{1}{9}}=\frac{3}{5}a$

$\therefore (a+b+c)=a+\frac{4a}{5}+\frac{3a}{5}+\frac{12a}{5}=\frac{12}{5}\left(\frac{5}{4}b\right)=\frac{12}{5}\left(\frac{5}{3}c\right)$

$\Rightarrow (a+b+c)=\frac{12}{5}a=3b=4c$