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`

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

If the measures of the sides of triangle are (x^2 - 1), (x^2 + 1) and 2x cm, then the triangle would be

If the sides of a triangle are in the ratio x:y:sqrt(x^(2)+xy+y^(2)) ,then the greatest angle is

if length of the sides of a Delta ABC are x^(2)+x+1,2x+1,x^(2)-1 then half of the largest angle of a Delta ABC is

If the sides of a triangle are in A.P., and its greatest angle exceeds the least angle by alpha , show that the sides are in the ration 1+x :1:1-x , , where x = sqrt((1- cos alpha)/(7 - cos alpha))

The equations of the three sides of a triangle are x=2,y+1=0 and x+2y=4 .The coordinates of the circumcentre of the triangle are

If the sides of a triangle are x, y and sqrt((x^(2) + xy + y^(2))) , then then measure of its greatest angle is....

Statement-1: lf the measures of two angles of a triangle are 45 ° and 60 °, then the ratio of the smallest and the greatest sides are (sqrt3-1) : 1 Statement-2: The greatest side of a triangle is opposite to its greatest angle.