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(D) The given equation is $2x^2 - 5xy + 2y^2 - 3x + 3y + 1 = 0$.Comparing this with the general equation $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$,we g

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$	an^{-1} \frac{3}{4}$

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(D) The given equation is $2x^2 - 5xy + 2y^2 - 3x + 3y + 1 = 0$.Comparing this with the general equation $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$,we get $a = 2$,$2h = -5$ (so $h = -5/2$),and $b = 2$.The angle $	heta$ between the pair of straight lines is given by the formula $	an 	heta = \left| \frac{2\sqrt{h^2 - ab}}{a + b} ight|$.Substituting the values: $	an 	heta = \left| \frac{2\sqrt{(-5/2)^2 - (2)(2)}}{2 + 2} ight|$.$	an 	heta = \left| \frac{2\sqrt{25/4 - 4}}{4} ight| = \left| \frac{2\sqrt{9/4}}{4} ight| = \left| \frac{2(3/2)}{4} ight| = \frac{3}{4}$.Therefore,$	heta = 	an^{-1} \frac{3}{4}$.

The angle between the two straight lines represented by the equation $2x^2 - 5xy + 2y^2 - 3x + 3y + 1 = 0$ is

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