# find the largest of the three angle of the triangles?

The sides of a triangle are given to be $$x^2 +x +1 ,$$ $$2x+1,$$ $$x^2-1$$Then find the largest of the three angle of the triangles .

I tried by applying hero's formula and then equating to the formula

$$0.5 bc \sin\theta$$. But using these methods , then i have to try 3 times for each angle . And still i cant get the value of x. I think i can do it throuh by randomly assigning a value to x and then get the value of the angle opposite to the longest side.

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Note that :

$\begin{cases} x^2+x+1 > 2x+1 > x^2-1, & \text{if } x\in(1,1+\sqrt 3) \\ x^2+x+1 > x^2-1 > 2x+1, & \text{if } x\in(1+\sqrt 3,+\infty) \end{cases}$

Hence , the largest angle of the triangle is angle opposite to the side $x^2+x+1$ . Let us denote that angle as $\alpha$ , then according to Cosine rule we can write following equality :

$(x^2+x+1)^2=(2x+1)^2+(x^2-1)^2-2(2x+1)(x^2-1)\cdot \cos \alpha$

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And $x^2-1>0$ implies $x>1$ or $x<-1$, while $2x+1>0$ implies $x>-\frac{1}{2}$, hence $x>1$. – Jonas Meyer Apr 1 '12 at 5:15
Or being a triangle means that $x^2+x+1>(x^2-1)+(2x+1)$ which simplifies to $1>x$. So there is no triangle. – Mark Bennet Apr 1 '12 at 5:38
@MarkBennet Actually being triangle means that $(x^2-1)+(2x+1) > x^2+x+1$ – pedja Apr 1 '12 at 5:49
@Pedja: Crass error-thanks - it's early in the morning. However the cosine formula seems to reduce to $cos (\alpha)=-\frac12$, though that may be subject to the same source of error. – Mark Bennet Apr 1 '12 at 6:46