The angle by which the coordinate axes are to | vedclass

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(D) The general second-degree equation is $ax^2 + 2hxy + by^2 = 0$.Here,$a = \sqrt{3}$,$2h = \sqrt{3}-1$,and $b = -1$.To eliminate the $xy$ term,the a

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$7.5$

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(D) The general second-degree equation is $ax^2 + 2hxy + by^2 = 0$.Here,$a = \sqrt{3}$,$2h = \sqrt{3}-1$,and $b = -1$.To eliminate the $xy$ term,the axes must be rotated by an angle $	heta$ given by the formula:$	an 2	heta = \frac{2h}{a-b}$Substituting the values:$	an 2	heta = \frac{\sqrt{3}-1}{\sqrt{3}-(-1)} = \frac{\sqrt{3}-1}{\sqrt{3}+1}$Multiplying the numerator and denominator by $(\sqrt{3}-1)$:$	an 2	heta = \frac{(\sqrt{3}-1)^2}{3-1} = \frac{3+1-2\sqrt{3}}{2} = \frac{4-2\sqrt{3}}{2} = 2-\sqrt{3}$We know that $	an 15^{\circ} = 	an(45^{\circ}-30^{\circ}) = \frac{	an 45^{\circ}-	an 30^{\circ}}{1+	an 45^{\circ}	an 30^{\circ}} = \frac{1-1/\sqrt{3}}{1+1/\sqrt{3}} = \frac{\sqrt{3}-1}{\sqrt{3}+1} = 2-\sqrt{3}$.Therefore,$	an 2	heta = 	an 15^{\circ}$.$2	heta = 15^{\circ} \implies 	heta = 7.5^{\circ}$.

The angle by which the coordinate axes are to be rotated about the origin so that the transformed equation of $\sqrt{3} x^2+(\sqrt{3}-1) x y-y^2=0$ is free from the $xy$ term is: (in $^{\circ}$)

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