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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 equation of a pair of straight lines $ax^2 + 2hxy + by^2 = 0$,we have $a = \sin 	heta$,$h = 1$,and $b = \sin 	heta$.The angle $\alpha$ between the lines is given by $	an \alpha = \left| \frac{2\sqrt{h^2 - ab}}{a + b} ight|$.Substituting the values,$	an \alpha = \left| \frac{2\sqrt{1^2 - \sin^2 	heta}}{\sin 	heta + \sin 	heta} ight| = \left| \frac{2\cos 	heta}{2\sin 	heta} ight| = |\cot 	heta|$.Since $|\cot 	heta| = 	an(\frac{\pi}{2} - 	heta)$,the angle $\alpha = \frac{\pi}{2} - 	heta$.

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

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