Question

lf in $\Delta$ $ABC$ $\cot \frac{A}{2}$ : $\cot \frac{B}{2}$ : $\cot \frac{C}{2}=1:4:15$, then the greatest angle of triangle is

• A. ${60}^0$
• B. ${90}^0$
• C. ${120}^0$
• D. ${135}^0$

Answer 1

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

By applying this equation

$\frac{s(s-a)}{△}$:$\frac{s(s-b)}{△}$:$\frac{s(s-c)}{△}$ = 1:4:15

$(s-a):(s-b):(s-c)=1:4:15$

now, let $s-a=k$ , $s-b=4k$ , $s-c=15k$

$s-a+s-b=8k=c$,

$s-b+s-c=19k=a$,

$s-a+s-c=16k=b$

now ratio of sides are $a:b:c=19:16:8$

now finding $\cos A$ because of a is largest side

$A={120}^{\circ }$