Let ABC be a triangle such that \angleACB = \frac{\pi}{6} and let a, b and c denote the lengths of the sides opposite to A, B and C respectively. The value(s) of x for which a = x2 + x + 1, b = x2

Engineering

Mathematics

Basic Rules of Properties of Triangle

Question

Let ABC be a triangle such that $∠$ ACB = $\frac{π}{6}$ and let a, b and c denote the lengths of the sides opposite to A, B and C respectively. The value(s) of x for which a = x² + x + 1, b = x² – 1 and c = 2x + 1 is/are

$- (2 + \sqrt{3})$

$2 + \sqrt{3}$

$1 + \sqrt{3}$

$4 \sqrt{3}$

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$\cos \frac{π}{6} = \frac{a^{2} + b^{2} - c^{2}}{2 ab} = \frac{\sqrt{3}}{2} \Rightarrow a^{2} + b^{2} - c^{2} = \sqrt{3} ab or {(x^{2} + x + 1)}^{2} + {(x^{2} - 1)}^{2} - {(2 x + 1)}^{2}$

$= \sqrt{3} (x^{2} + x + 1) (x^{2} - 1) or {{(x^{2} + x + 1)}^{2} - {(2 x + 1)}^{2}} + {(x^{2} - 1)}^{2} = \sqrt{3} (x^{2} + x + 1) (x^{2} - 1)$

$or (x^{2} + 3 x + 2) (x^{2} - x) + {(x^{2} - 1)}^{2} = \sqrt{3} (x^{2} + x + 1) (x^{2} - 1)$

$or (x + 2) (x + 1) x (x - 1) + {(x^{2} - 1)}^{2} = \sqrt{3} (x^{2} + x + 1) (x^{2} - 1)$

$or x^{2} + 2 x + (x^{2} - 1) = \sqrt{3} (x^{2} + x + 1)$

$or 2 x^{2} + 2 x - 1 = \sqrt{3} (x^{2} + x + 1)$

$or (2 - \sqrt{3}) x^{2} + (2 - \sqrt{3}) x - (\sqrt{3} + 1) = 0$

only positive value of $x = 1 + \sqrt{3}$

Note: Other value of x = – $(2 + \sqrt{3})$ is rejected (think!)