The acute angle between the lines (x^2+y^2) s | vedclass

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(C) The given equation is $(x^2+y^2) \sin 	heta + 2xy = 0$,which can be written as $(\sin 	heta)x^2 + 2xy + (\sin 	heta)y^2 = 0$. Comparing this wi

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$\frac{\pi}{2}-	heta$

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(C) The given equation is $(x^2+y^2) \sin 	heta + 2xy = 0$,which can be written as $(\sin 	heta)x^2 + 2xy + (\sin 	heta)y^2 = 0$. Comparing this with the general form $ax^2 + 2hxy + by^2 = 0$,we get $a = \sin 	heta$,$h = 1$,and $b = \sin 	heta$. Let $\alpha$ be the acute angle between the lines. The formula for the angle is $	an \alpha = \left| \frac{2\sqrt{h^2 - ab}}{a+b} ight|$. Substituting the values,we get $	an \alpha = \left| \frac{2\sqrt{1^2 - (\sin 	heta)(\sin 	heta)}}{\sin 	heta + \sin 	heta} ight|$. $	an \alpha = \left| \frac{2\sqrt{1 - \sin^2 	heta}}{2 \sin 	heta} ight| = \left| \frac{2\cos 	heta}{2 \sin 	heta} ight| = |\cot 	heta|$. Since $\alpha$ is the acute angle,$	an \alpha = 	an(\frac{\pi}{2} - 	heta)$. Therefore,$\alpha = \frac{\pi}{2} - 	heta$.

The acute angle between the lines $(x^2+y^2) \sin \theta+2xy=0$ is

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