The sides of a triangle are `x^2+x+1,2x+1,a n dx^2-1` . Prove that the greatest angle is `120^0dot`

Explanation

Let `a = x^(2) + x + 1, b = 2x + 1, and 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 gt 0 and c = x^(2) - 1 0`. Thus,
`x gt -(1)/(2) and x^(2) gt 1`
`rArr x gt - (1)/(2) and x lt -1 " or " x gt 1`
`rArr x gt 1`
`a = x^(2) + x + 1 = (x + (1)/(2))^(2) + ((3)/(4))` is always positive.
Thus, all sides a, b, and c are positive when `x gt 1`. Now
`x gt 1 " or" x^(2) gt x`
or `x^(2) + x + 1 gt 2x + 1 rArr a gt b`
Also, when `x gt 1`
`x^(2) + x + 1 gt x^(2) - 1 rArr a gt c`
Thus, `a = x^(2) + x + 1` is the greatest side and the angle A opposite to this side is the greatest angle
`:. cos A = (b^(2) + c^(2) -a^(2))/(2bc)`
`= ((2x + 1)^(2) + (x^(2) - 1)^(2) - (x^(2) + x + 1)^(2))/(2(2x + 1) (x^(2) -1))`
`= (-2x^(3) - x^(2) + 2x + 1)/(2(2x^(3) + x^(2) - 2x -1)) = -(1)/(2) = cos 120^(@)`
`rArr A = 120^(@)`