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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.

Thanks in advance.

1 Answers 1

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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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    @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.2012-04-01