Question

In a $\Delta ABC,$ If $\cot \frac{A}{2}:\cot \frac{B}{2}:\cot \frac{C}{2}=1:4:15,$ then the greatest angle is:

• A. $\frac{\pi }{3}$
• B. $\frac{\pi }{4}$
• C. $\frac{\pi }{6}$
• D. $\frac{2\pi }{3}$

Answer 1

The correct option is D $\frac{2\pi }{3}$

$\cot \frac{A}{2}:\cot \frac{B}{2}:\cot \frac{C}{2}=1:4:15.$
We know, $\cot \left(\frac{A}{2}\right)=\frac{s(s-a)}{\Delta }$
$\therefore 1:4:15=\frac{s(s-a)}{\Delta }:\frac{s(s-a)}{\Delta }:\frac{s(s-a)}{\Delta }$
$\Rightarrow 1:4:15=s-a:s-b:s-c$
$\Rightarrow 1:4:15=b+c-a:a+c-b:a+b-c$
$\Rightarrow b+c-a=k\cdots \left(i\right)$
$\Rightarrow a+c-b=4k\cdots \left(ii\right)$
$\Rightarrow a+b-c=15k\cdots \left(iii\right)$
Adding $\left(i\right),\left(ii\right)$
$c=\frac{5k}{2}$
Adding $\left(ii\right),\left(iii\right)$
$a=\frac{19k}{2}$
Adding $\left(i\right),\left(ii\right)$
$b=\frac{16k}{2}$
So, $\Rightarrow a:b:c=19:16:5$
$\therefore A$ is the greatest angle.
$\Rightarrow \cos A=\frac{b^2+c^2-a^2}{2bc}$
$\Rightarrow \cos A=\frac{{16}^2+5^2-{19}^2}{\mathrm{2.16.5}}=\frac{-1}{2}\Rightarrow A=\frac{2\pi }{3}$