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The measure of the acute angle between the lines given by the equation
$3 x^{2}-4 \sqrt{3} x y+3 y^{2}=0$ is
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$3 x^{2}-4 \sqrt{3} x y+3 y^{2}=0$ is
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Verified Answer
The correct answer is:
$30^{\circ}$
Comparing $3 x^{2}-4 \sqrt{3} x y+3 y^{2}=0$, with $a x^{2}+2 h x y+b y^{2}=0$, we get
$\mathrm{a}=3, \mathrm{~h}=-2 \sqrt{3}, \mathrm{~b}=3$
We know that, $\tan \theta=\left|\frac{2 \sqrt{h^{2}-a b}}{a+b}\right|$
$\therefore \tan \theta=\left|\frac{2 \sqrt{12-9}}{3+3}\right|=\left|\frac{2 \sqrt{3}}{6}\right|=\left|\frac{1}{\sqrt{3}}\right|$
$\therefore \quad \theta=30^{\circ}$
$\mathrm{a}=3, \mathrm{~h}=-2 \sqrt{3}, \mathrm{~b}=3$
We know that, $\tan \theta=\left|\frac{2 \sqrt{h^{2}-a b}}{a+b}\right|$
$\therefore \tan \theta=\left|\frac{2 \sqrt{12-9}}{3+3}\right|=\left|\frac{2 \sqrt{3}}{6}\right|=\left|\frac{1}{\sqrt{3}}\right|$
$\therefore \quad \theta=30^{\circ}$
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