Question

In a $△ABC$ if $r_1=2r_2=3r_3$ then which are correct?

• A. $\cos B=\frac{9}{15}$
• B. $\cos C=\frac{4}{5}$
• C. $△ABC$ is a right angle triangle
• D. None of these

Answer 1

Answer: (A), (B), (C)

$\cos B=\frac{9}{15}$
$\cos C=\frac{4}{5}$
$△ABC$ is a right angle triangle

$r_1=2r_2=3r_3=k$(say)

$\Rightarrow r_1=k;r_2=\frac{k}{2};r_3=\frac{k}{3}$

$\Rightarrow \frac{\Delta }{s-a}=k;\frac{\Delta }{s-b}=\frac{k}{2};\frac{\Delta }{s-c}=\frac{k}{3}$

$\Rightarrow (s-a)=\frac{\Delta }{k};(s-b)=\frac{2\Delta }{k};(s-c)=\frac{3\Delta }{k}$

$\Rightarrow (s-a)=\lambda ;(s-b)=2\lambda ;(s-c)=3\lambda$

where $\lambda =\frac{\Delta }{k}$

Adding we get $3s-(a+b+c)=6\lambda$

$\Rightarrow 3s-2s=6\lambda \Rightarrow s=6\lambda$

$\Rightarrow a=5\lambda ;b=4\lambda ;c=3\lambda$

$\therefore \cos A=\frac{b^2+c^2-a^2}{2bc};\cos B=\frac{a^2+c^2-b^2}{2ac};\cos C=\frac{a^2+b^2-c^2}{2ab}$

$\Rightarrow \cos A=\frac{9{\lambda }^2+16{\lambda }^2-25{\lambda }^2}{24{\lambda }^2};\cos B=\frac{25{\lambda }^2+9{\lambda }^2-16{\lambda }^2}{30{\lambda }^2};\cos C=\frac{25{\lambda }^2+16{\lambda }^2-9{\lambda }^2}{40{\lambda }^2}$

$\Rightarrow \cos A=0;\cos B=\frac{9}{15};\cos C=\frac{4}{5}$

as $\cos A=0\Rightarrow A=\frac{\pi }{2}$ Hence $△ABC$ is a right angled triangle