NIMCET 2011 Question Paper with solutions

Question : 14 of 120

Marks: +1, -0

The greatest angle of the triangle whose sides are

x^{2}+x+1,2 x+1, x^{2}-1

Solution:

It is clear that, the greatest side is

x^{2}+x+.1

By cosine law

∴ \cos θ= {(2 x+1)^{2}+(x^{2}-1)^{2}-(x^{2}+x+1)^{2}}/{2(2 x+1)(x^{2}-1)}

[(4 x^{2}+1+4 x)+(x^{4}+1-2 x^{2})

= { -(x^{1}+x^{2}+1+2 x^{3}+2 x+2 x^{2})]}/{2(2 x^{3}-2 x+x^{2}-1)}

=\;{-2 x^{3}-x^{2}+2 x+1}/{2(2 x^{3}-2 x+x^{2}-1)}=- {1}/{2}

⇒ θ=120^{∘}

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