The acute angle between the lines 6x^2 + 11xy | vedclass

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(B) The acute angle $	heta$ between the pair of straight lines represented by $ax^2 + 2hxy + by^2 = 0$ is given by $	an 	heta = \left| \frac{2\sqrt

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$	an^{-1}\left(\frac{\sqrt{361}}{4}ight)$

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(B) The acute angle $	heta$ between the pair of straight lines represented by $ax^2 + 2hxy + by^2 = 0$ is given by $	an 	heta = \left| \frac{2\sqrt{h^2 - ab}}{a+b} ight|$.Comparing the given equation $6x^2 + 11xy - 10y^2 = 0$ with $ax^2 + 2hxy + by^2 = 0$,we get $a = 6$,$b = -10$,and $2h = 11$,so $h = \frac{11}{2}$.Substituting these values into the formula:$	an 	heta = \left| \frac{2\sqrt{(\frac{11}{2})^2 - (6)(-10)}}{6 + (-10)} ight| = \left| \frac{2\sqrt{\frac{121}{4} + 60}}{-4} ight| = \left| \frac{2\sqrt{\frac{121 + 240}{4}}}{-4} ight| = \left| \frac{2\sqrt{\frac{361}{4}}}{-4} ight| = \left| \frac{2 \cdot \frac{\sqrt{361}}{2}}{-4} ight| = \left| \frac{\sqrt{361}}{-4} ight| = \frac{\sqrt{361}}{4}$.Therefore,$	heta = 	an^{-1}\left(\frac{\sqrt{361}}{4}ight)$.

The acute angle between the lines $6x^2 + 11xy - 10y^2 = 0$ is

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