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If-sides-of-triangle-are-x-2-x-1-2x-1-and-x-2-1-prove-that-greatest-angle-is-120-Also-find-the-range-of-x-such-that-triangle-exist-

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Question Number 15760 by Tinkutara last updated on 13/Jun/17
If sides of triangle are x^2 + x + 1, 2x + 1 and x^2 − 1, prove that greatest angle is 120°. Also find the range of x such that triangle exist.
$$\mathrm{If}\:\mathrm{sides}\:\mathrm{of}\:\mathrm{triangle}\:\mathrm{are}\:{x}^{\mathrm{2}} \:+\:{x}\:+\:\mathrm{1}, \\ $$$$\mathrm{2}{x}\:+\:\mathrm{1}\:\mathrm{and}\:{x}^{\mathrm{2}} \:−\:\mathrm{1},\:\mathrm{prove}\:\mathrm{that}\:\mathrm{greatest} \\ $$$$\mathrm{angle}\:\mathrm{is}\:\mathrm{120}°.\:\mathrm{Also}\:\mathrm{find}\:\mathrm{the}\:\mathrm{range}\:\mathrm{of}\:{x} \\ $$$$\mathrm{such}\:\mathrm{that}\:\mathrm{triangle}\:\mathrm{exist}. \\ $$
Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 14/Jun/17
x^2 +x+1−(2x+1)=x^2 −x>0 if:x>0 x^2 +x+1−(x^2 −1)=x+2>0 if:x>0 so: the greatest side is: x^2 +x+1 cosA=(((x^2 −1)^2 +(2x+1)^2 −(x^2 +x+1)^2 )/(2(x^2 −1)(2x+1)))= (x^2 −1)^2 +(2x+1)^2 −(x^2 +x+1)^2 = (x^2 −1−x^2 −x−1)(x^2 −1+x^2 +x+1)+(2x+1)^2 = =−x(x+2)(2x+1)+(2x+1)^2 = =(2x+1)(2x+1−x^2 −2x)=(2x+1)(1−x^2 ) ⇒cosA=(((2x+1)(1−x^2 ))/(2(x^2 −1)(2x+1)))=((−1)/2) ⇒∡A=120^• . 2) x^2 +x+1<2x+1+x^2 −1⇒x>1 2a) x^2 +x+1>2x+1−x^2 +1⇒2x^2 −x−1>0 x=((1±(√(1+8)))/4)=1,((−1)/2)⇒x<((−1)/2) ∨ x>1 2b) x^2 +x+1>x^2 −1−(2x+1)⇒3x>−3 ⇒x>−1 so: for: x>1 we have this triangle.
$${x}^{\mathrm{2}} +{x}+\mathrm{1}−\left(\mathrm{2}{x}+\mathrm{1}\right)={x}^{\mathrm{2}} −{x}>\mathrm{0}\:{if}:{x}>\mathrm{0} \\ $$$${x}^{\mathrm{2}} +{x}+\mathrm{1}−\left({x}^{\mathrm{2}} −\mathrm{1}\right)={x}+\mathrm{2}>\mathrm{0}\:{if}:{x}>\mathrm{0} \\ $$$${so}:\:{the}\:{greatest}\:{side}\:{is}:\:{x}^{\mathrm{2}} +{x}+\mathrm{1} \\ $$$${cosA}=\frac{\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} +\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} −\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{2}\left({x}^{\mathrm{2}} −\mathrm{1}\right)\left(\mathrm{2}{x}+\mathrm{1}\right)}= \\ $$$$\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} +\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} −\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)^{\mathrm{2}} = \\ $$$$\left({x}^{\mathrm{2}} −\mathrm{1}−{x}^{\mathrm{2}} −{x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} −\mathrm{1}+{x}^{\mathrm{2}} +{x}+\mathrm{1}\right)+\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} = \\ $$$$=−{x}\left({x}+\mathrm{2}\right)\left(\mathrm{2}{x}+\mathrm{1}\right)+\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} = \\ $$$$=\left(\mathrm{2}{x}+\mathrm{1}\right)\left(\mathrm{2}{x}+\mathrm{1}−{x}^{\mathrm{2}} −\mathrm{2}{x}\right)=\left(\mathrm{2}{x}+\mathrm{1}\right)\left(\mathrm{1}−{x}^{\mathrm{2}} \right) \\ $$$$\Rightarrow{cosA}=\frac{\left(\mathrm{2}{x}+\mathrm{1}\right)\left(\mathrm{1}−{x}^{\mathrm{2}} \right)}{\mathrm{2}\left({x}^{\mathrm{2}} −\mathrm{1}\right)\left(\mathrm{2}{x}+\mathrm{1}\right)}=\frac{−\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\measuredangle{A}=\mathrm{120}^{\bullet} \:. \\ $$$$\left.\mathrm{2}\right)\:{x}^{\mathrm{2}} +{x}+\mathrm{1}<\mathrm{2}{x}+\mathrm{1}+{x}^{\mathrm{2}} −\mathrm{1}\Rightarrow{x}>\mathrm{1} \\ $$$$\left.\:\mathrm{2}{a}\right)\:{x}^{\mathrm{2}} +{x}+\mathrm{1}>\mathrm{2}{x}+\mathrm{1}−{x}^{\mathrm{2}} +\mathrm{1}\Rightarrow\mathrm{2}{x}^{\mathrm{2}} −{x}−\mathrm{1}>\mathrm{0} \\ $$$${x}=\frac{\mathrm{1}\pm\sqrt{\mathrm{1}+\mathrm{8}}}{\mathrm{4}}=\mathrm{1},\frac{−\mathrm{1}}{\mathrm{2}}\Rightarrow{x}<\frac{−\mathrm{1}}{\mathrm{2}}\:\vee\:{x}>\mathrm{1} \\ $$$$\left.\mathrm{2}{b}\right)\:{x}^{\mathrm{2}} +{x}+\mathrm{1}>{x}^{\mathrm{2}} −\mathrm{1}−\left(\mathrm{2}{x}+\mathrm{1}\right)\Rightarrow\mathrm{3}{x}>−\mathrm{3} \\ $$$$\Rightarrow{x}>−\mathrm{1} \\ $$$${so}:\:\:{for}:\:{x}>\mathrm{1}\:{we}\:{have}\:{this}\:{triangle}. \\ $$
Commented by Tinkutara last updated on 14/Jun/17
Thanks Sir!
$$\mathrm{Thanks}\:\mathrm{Sir}! \\ $$

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