Question

If cot $\frac{A}{2}$:cot$\frac{B}{2}$:cot$\frac{C}{2}$=1:4:15, then largest angle is

• A. ${120}^0$
• B. ${135}^0$
• C. ${160}^0$
• D. ${10}^0$

Answer 1

The correct option is A ${120}^0$

We have,

$\begin{array}{c}wehaveratioisgiven,\\ \cot \frac{A}{2}:\cot \frac{B}{2}:\cot \frac{C}{2}=s-a:s-b:s-c[where,\tan \frac{A}{2}=\frac{\Delta }{s(s-a)}\\ Nowsimplify,\\ s-a:s-b:s-c=1:4:15weknow.....s=\frac{a+b+c}{2}\\ \frac{b+c-a}{2}:\frac{a+c-b}{2}:\frac{a+b-c}{2}=1:4:15\\ fromequ:...\\ \frac{b+c-a}{a+c-b}=\frac{1}{4},\\ Now,\\ 4b+4c-4a=a+c-b\\ 5b+3c=5a--------\left(i\right)\\ and\\ \frac{a+c-b}{a+b-c}=\frac{4}{15}\\ \Rightarrow 15a+15c-15b=4a+4b-4c\\ \Rightarrow 11a+19c-19b=0\\ \Rightarrow 19b-19c=11a\\ \therefore b-c=\frac{11a}{19}-------\left(ii\right)\\ ifwehaveaddequ\left(i\right)with3\left(ii\right).............\\ 5b+3b=5a+\frac{33a}{19}\\ b=\frac{16a}{19}andc=\frac{5a}{19}\\ andratioofabc,\\ a:b:c=19:16:5\end{array}$

$\begin{array}{c}And,ratioofabc.............\\ a:b:c=19:16:5\\ \begin{array}{c}\\ \Rightarrow \cos A=-\frac{1}{2}\\ \therefore A=\frac{2\pi }{3}={120}^0\\ so,correctoptionisA.\end{array}\end{array}$