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-

Question Number 15760 by Tinkutara last updated on 13/Jun/17

If sides of triangle are x^{2} + x + 1,

2 x + 1 and x^{2} - 1, prove that greatest

angle is 120 ° . Also find the range of x

such that triangle exist .

Answered by b.e.h.i.8.3.4.1.7@gmail.com last updated on 14/Jun/17

x^{2} + x + 1 - (2 x + 1) = x^{2} - x > 0 i f : x > 0

x^{2} + x + 1 - (x^{2} - 1) = x + 2 > 0 i f : x > 0

s o : t h e g r e a t e s t s i d e i s : x^{2} + x + 1

c o s A = \frac{{(x^{2} - 1)}^{2} + {(2 x + 1)}^{2} - {(x^{2} + x + 1)}^{2}}{2 (x^{2} - 1) (2 x + 1)} =

{(x^{2} - 1)}^{2} + {(2 x + 1)}^{2} - {(x^{2} + x + 1)}^{2} =

(x^{2} - 1 - x^{2} - x - 1) (x^{2} - 1 + x^{2} + x + 1) + {(2 x + 1)}^{2} =

= - x (x + 2) (2 x + 1) + {(2 x + 1)}^{2} =

= (2 x + 1) (2 x + 1 - x^{2} - 2 x) = (2 x + 1) (1 - x^{2})

\Rightarrow c o s A = \frac{(2 x + 1) (1 - x^{2})}{2 (x^{2} - 1) (2 x + 1)} = \frac{- 1}{2}

\Rightarrow ∡ A = 120^{∙} .

2) x^{2} + x + 1 < 2 x + 1 + x^{2} - 1 \Rightarrow x > 1

2 a) x^{2} + x + 1 > 2 x + 1 - x^{2} + 1 \Rightarrow 2 x^{2} - x - 1 > 0

x = \frac{1 \pm \sqrt{1 + 8}}{4} = 1, \frac{- 1}{2} \Rightarrow x < \frac{- 1}{2} \lor x > 1

2 b) x^{2} + x + 1 > x^{2} - 1 - (2 x + 1) \Rightarrow 3 x > - 3

\Rightarrow x > - 1

s o : f o r : x > 1 w e h a v e t h i s t r i a n g l e .

Commented by Tinkutara last updated on 14/Jun/17

Thanks Sir!