Question

In a triangle ABC, if $r_1=2r_2=3r_3$ then $[100-(4a/5b){]}^2$ is equal to

Answer 1

We are given $r_1=2r_2=3r_3$
$\Rightarrow s\tan \left(A/2\right)=2s\tan \left(B/2\right)=3s\tan \left(C/2\right)$
$\Rightarrow \frac{\tan \left(A/2\right)}{6}=\frac{\tan \left(B/2\right)}{3}=\frac{\tan \left(C/2\right)}{2}=k\left(say\right)$
Since $A/2+B/2+C/2={90}^o$
$\tan \left(A/2\right)\tan \left(B/2\right)+\tan \left(B/2\right)\tan \left(C/2\right)+\tan \left(C/2\right)\tan \left(A/2\right)=1$
$\Rightarrow (6.3+3.2+2.5)k^2=1\Rightarrow k=1/6$
So that $\sin A=\frac{2\tan \left(A/2\right)}{1+{\tan }^2\left(A/2\right)}=1$
Similarly $\sin B=4/5$
and $\frac{a}{b}=\frac{sinA}{\sin B}=\frac{5}{4}\Rightarrow \frac{4a}{5b}=1$
$\therefore [100-(4a/5b){]}^2=(1000-1{)}^2={99}^2$
$=9801$