What is the greatest angle of the triangle whose sides are $x^2+x+1,2 x+1, x^2-1$ ?

Question

Aicharya · Accepted Answer

Sides of triangle are $\begin{aligned} & x^2+x+1,2 x+1 	ext { and } x^2-1 \ & x^2+x+1 	ext { is the greatest side. }\end{aligned}$ $	herefore \cos A=\frac{b^2+c^2-a^2}{2 b c}$ Here, $a=x^2+x+1, b=2 x+1, c=x^2-1$ $	herefore \cos A=\frac{(2 x+1)^2+\left(x^2-1ight)^2-\left(x^2+x+1ight)^2}{2(2 x+1)\left(x^2-1ight)}$ $\Rightarrow \cos A=\frac{\left.(2 x+1)^2+(x^2-1+x^2+x+1(x^2-1-x^2-x-1ight)}{2(2 x+1)\left(x^2-1ight)}$ $\begin{aligned} & \Rightarrow \cos A=\frac{(2 x+1)^2+\left(2 x^2+xight)(-x-2)}{2(2 x+1)\left(x^2-1ight)} \ & \Rightarrow \cos A=\frac{(2 x+1)\left(2 x+1-x^2-2 xight)}{2(2 x+1)\left(x^2-1ight)}\end{aligned}$ $\Rightarrow \cos A=\frac{-(2 x+1)\left(x^2-1ight)}{2(2 x+1)\left(x^2-1ight)}=\frac{-1}{2}$ $	herefore \quad A=120^{\circ}$

What is the greatest angle of the triangle whose sides are $x^2+x+1,2 x+1, x^2-1$ ?

Solution