The measure of the acute angle between the li | vedclass

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(D) Comparing the given equation $3x^{2}-4\sqrt{3}xy+3y^{2}=0$ with the general form $ax^{2}+2hxy+by^{2}=0$,we get: $a=3, h=-2\sqrt{3}, b=3$. We know

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$30$

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(D) Comparing the given equation $3x^{2}-4\sqrt{3}xy+3y^{2}=0$ with the general form $ax^{2}+2hxy+by^{2}=0$,we get: $a=3, h=-2\sqrt{3}, b=3$. We know that the acute angle $	heta$ between the lines is given by $	an 	heta = \left| \frac{2\sqrt{h^{2}-ab}}{a+b} ight|$. Substituting the values,we get $	an 	heta = \left| \frac{2\sqrt{(-2\sqrt{3})^{2}-(3)(3)}}{3+3} ight|$. $	an 	heta = \left| \frac{2\sqrt{12-9}}{6} ight| = \left| \frac{2\sqrt{3}}{6} ight| = \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}}$. Since $	an 	heta = \frac{1}{\sqrt{3}}$,we have $	heta = 30^{\circ}$.

The measure of the acute angle between the lines given by the equation $3x^{2}-4\sqrt{3}xy+3y^{2}=0$ is (in $^{\circ}$)

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