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(C) The general equation of a pair of straight lines passing through the origin is $ax^2 + 2hxy + by^2 = 0$.Comparing $\sqrt{3}x^2 - 4xy + \sqrt{3}y^2

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inclined at $30^\circ$ to each other.

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(C) The general equation of a pair of straight lines passing through the origin is $ax^2 + 2hxy + by^2 = 0$.Comparing $\sqrt{3}x^2 - 4xy + \sqrt{3}y^2 = 0$ with the general form, we get $a = \sqrt{3}$, $2h = -4 \implies h = -2$, and $b = \sqrt{3}$.The angle $	heta$ between the pair of lines is given by $	an 	heta = \left| \frac{2\sqrt{h^2 - ab}}{a + b} ight|$.Substituting the values: $	an 	heta = \left| \frac{2\sqrt{(-2)^2 - (\sqrt{3})(\sqrt{3})}}{\sqrt{3} + \sqrt{3}} ight|$.$	an 	heta = \left| \frac{2\sqrt{4 - 3}}{2\sqrt{3}} ight| = \left| \frac{2(1)}{2\sqrt{3}} ight| = \frac{1}{\sqrt{3}}$.Since $	an 	heta = \frac{1}{\sqrt{3}}$, we have $	heta = 30^\circ$.

The pair of straight lines represented by the equation $\sqrt{3}x^2 - 4xy + \sqrt{3}y^2 = 0$ are

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