If theta_1 and theta_2 are the acute angles b | vedclass

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(A) The angle $	heta$ between the pair of lines $ax^2 + 2hxy + by^2 = 0$ is given by $	an 	heta = \left| \frac{2\sqrt{h^2 - ab}}{a+b} ight|$.For

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$	heta_1 = 	heta_2$

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(A) The angle $	heta$ between the pair of lines $ax^2 + 2hxy + by^2 = 0$ is given by $	an 	heta = \left| \frac{2\sqrt{h^2 - ab}}{a+b} ight|$.For the first equation $3x^2 - 7xy + 4y^2 = 0$,we have $a=3, 2h=-7 \implies h=-7/2, b=4$.$	an 	heta_1 = \left| \frac{2\sqrt{(-7/2)^2 - 3(4)}}{3+4} ight| = \left| \frac{2\sqrt{49/4 - 12}}{7} ight| = \left| \frac{2\sqrt{1/4}}{7} ight| = \frac{2(1/2)}{7} = \frac{1}{7}$.For the second equation $6x^2 - 5xy + y^2 = 0$,we have $a=6, 2h=-5 \implies h=-5/2, b=1$.$	an 	heta_2 = \left| \frac{2\sqrt{(-5/2)^2 - 6(1)}}{6+1} ight| = \left| \frac{2\sqrt{25/4 - 6}}{7} ight| = \left| \frac{2\sqrt{1/4}}{7} ight| = \frac{2(1/2)}{7} = \frac{1}{7}$.Since $	an 	heta_1 = 	an 	heta_2$,we have $	heta_1 = 	heta_2$.

If $\theta_1$ and $\theta_2$ are the acute angles between the pairs of lines $3x^2 - 7xy + 4y^2 = 0$ and $6x^2 - 5xy + y^2 = 0$ respectively,then:

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