Question

The greatest angle of a triangle whose sides are $x^2+x+1,2x+1$ and $x^2-1$, is:

• A. ${60}^o$
• B. ${90}^o$
• C. ${120}^o$
• D. none of these

Answer 1

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

Let $a=x^2-1;b=2x+1,c=x^2+x+1$

Here the greatest side is $c=x^2+x+1$

The greatest angle $C$ will be opposite the greatest side.

$c=x^2+x+1$

Use the cosine rule

$\Rightarrow c^2=a^2+b^2-2bc.\cos C$

$\Rightarrow (x^2+x+1{)}^2=(x^2-1{)}^2+(2x+1{)}^2-2(x^2-1\left)\right(2x+1)\cos C$

$\Rightarrow 2(x^2-1)(2x+1)cosC=\left[\right(x^2-1{)}^2-(x^2+x+1{)}^2]+(2x+1{)}^2$

$\Rightarrow 2(x^2-1)(2x+1)\cos C=(x^2-1+x^2+x+1)(x^2-1-x^2-x-1)+(2x+1{)}^2$

$\Rightarrow 2(x^2-1)(2x+1)\cos C=x(2x+1)(-2-x)+(2x+1{)}^2$

$\Rightarrow 2(x^2-1)cosC=x(-2-x)+(2x+1)=-x^2+1$

$\Rightarrow 2\cos C=-1$

$\Rightarrow cosC=-1/2$

$\Rightarrow C={120}^0$