Question

The sides of a triangle are $x^2+x+1,2x+1$, and $x^2-1$ Prove that the greatest angle is ${120}^{\circ }$

Answer 1

Let a $a=x^2+x+1,b=2x+1,c=x^2-1$
First, we have to decide which side is the greatest. We know that in a triangle, the length of each side is greater than zero.
Therefore, we have $b=2x+1>0$ and $c=x^2-1>0$

Thus, $x>-\frac{1}{2}$ and
$x^2>1$ $\Rightarrow x>-\frac{1}{2}$ and $x<-1$ or $x>1\Rightarrow x>1$ $a=x^2+x+1={(x+\frac{1}{2})}^2+\left(\frac{3}{4}\right)$ is always positive. Thus, all sides a,b and c are positive when $x>1$

Now, $x>1$ or $x^2>x$ or $x^2+x+1>2x+1\Rightarrow a>b$ Also, when $x>1$ $x^2+x+1>x^2-1\Rightarrow a>c$ Thus,$a=x^2+x+1$ is the greatest side and the angle A opposite to this side is the greatest angle.
$\therefore \cos A=\frac{b^2+c^2-a^2}{2bc}$ $=\frac{(2x+1{)}^2+(x^2-1{)}^2-(x^2+x+1{)}^2}{2(2x+1)(x^2-1)}$
$=\frac{-2x^3-x^2+2x+1}{2(2x^3+x^2-2x-1)}=-\frac{1}{2}=\cos {120}^{\circ }$ $\Rightarrow A={120}^{\circ }$