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The sides of a triangle are $(x^2+x+1), (2x+1)$ and $(x^2-1)$. Then what is the largest of the 3 angles of triangle?

Rudstar
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1 Answers1

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Suppose $x>0$. As $(x^{2}-1)$ is a side of the triangle then $x^{2}-1>0$ , therefore $x>1$. Note that $(x^{2}+x+1)-(x^{2}-1)=x+2>0$, then $(x^{2}+x+1)>(x^{2}-1)$. Another hand, as $x>1$, $(x^{2}+x+1)-(2x+1)=x^{2}-x>0$. Therefore $(x^{2}+x+1)$ is the largest side of the triangle. The largest angle is the opposite to the side $(x^{2}+x+1).$

As $(x^{2}+x+1)$ is the largest side of the triangle, by the law of cosines, $$(x^{2}+x+1)^{2}=(2x+1)^{2}+(x^{2}-1)-2(2x+1)(x^{2}-1)\cos\theta$$ Therefore $\cos\theta=-\frac{1}{2}$, then $\theta=120^{\circ}$.

VJunior
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  • Yeah, right! This is similar to the approach with vectors. – Rudstar Apr 30 '14 at 09:22
  • You're assuming $X > 0$, but this is not necessary. Since all sides can be supposed to larger than $0$, this means that from $x^2-1>0$ it follows that $x>1\bigvee x<-1$. Also, from $2x+1>0$ follows that $x > -\tfrac{1}{2}$. Taking these together, we have $x>1$, without assuming anything. – SQB Apr 30 '14 at 11:42