If the angle 2 theta is acute,then the acute | vedclass

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(D) The given equation of the pair of lines is $Ax^2 + 2Hxy + By^2 = 0$,where $A = \cos 	heta - \sin 	heta$,$2H = 2 \cos 	heta$ (so $H = \cos 	het

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$	heta$

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(D) The given equation of the pair of lines is $Ax^2 + 2Hxy + By^2 = 0$,where $A = \cos 	heta - \sin 	heta$,$2H = 2 \cos 	heta$ (so $H = \cos 	heta$),and $B = \cos 	heta + \sin 	heta$.The formula for the acute angle $\alpha$ between the pair of lines is given by $	an \alpha = \left| \frac{2 \sqrt{H^2 - AB}}{A + B} ight|$.Substituting the values:$H^2 - AB = \cos^2 	heta - (\cos 	heta - \sin 	heta)(\cos 	heta + \sin 	heta) = \cos^2 	heta - (\cos^2 	heta - \sin^2 	heta) = \sin^2 	heta$.$A + B = (\cos 	heta - \sin 	heta) + (\cos 	heta + \sin 	heta) = 2 \cos 	heta$.Thus,$	an \alpha = \left| \frac{2 \sqrt{\sin^2 	heta}}{2 \cos 	heta} ight| = \left| \frac{2 \sin 	heta}{2 \cos 	heta} ight| = |	an 	heta|$.Since the angle $2 	heta$ is acute,$	heta$ is also acute,so $\alpha = 	heta$.

Process	Condition
$(I)$ Adiabatic	$(A)$ $\Delta W = 0$
$(II)$ Isothermal	$(B)$ $\Delta Q = 0$
$(III)$ Isochoric	$(C)$ $\Delta U \neq 0, \Delta W \neq 0, \Delta Q \neq 0$
$(IV)$ Isobaric	$(D)$ $\Delta U = 0$

If the angle $2 \theta$ is acute,then the acute angle between the pair of straight lines $x^2(\cos \theta - \sin \theta) + 2xy \cos \theta + y^2(\cos \theta + \sin \theta) = 0$ is:

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