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(D) Given the equation of the pair of lines is $5x^2 + \frac{40}{3}xy + ky^2 = 0$. Comparing this with the general form $ax^2 + 2hxy + by^2 = 0$,we ha

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$\frac{\pi}{2}$

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(D) Given the equation of the pair of lines is $5x^2 + \frac{40}{3}xy + ky^2 = 0$. Comparing this with the general form $ax^2 + 2hxy + by^2 = 0$,we have $a = 5$,$2h = \frac{40}{3} \Rightarrow h = \frac{20}{3}$,and $b = k$. Let the slopes of the lines be $m_1$ and $m_2$. We know that $m_1 + m_2 = -\frac{2h}{b} = -\frac{40}{3k}$ and $m_1m_2 = \frac{a}{b} = \frac{5}{k}$. Given $m_1 = 3$,we have $3 + m_2 = -\frac{40}{3k}$ and $3m_2 = \frac{5}{k} \Rightarrow m_2 = \frac{5}{3k}$. Substituting $m_2$ into the sum equation: $3 + \frac{5}{3k} = -\frac{40}{3k}$. Multiplying by $3k$: $9k + 5 = -40$ $\Rightarrow 9k = -45$ $\Rightarrow k = -5$. The angle $	heta$ between the lines is given by $	an 	heta = \left| \frac{2\sqrt{h^2 - ab}}{a + b} ight|$. Here,$a + b = 5 + (-5) = 0$. Since the sum of the coefficients of $x^2$ and $y^2$ is zero,the lines are perpendicular. Therefore,$	heta = \frac{\pi}{2}$.

If the slope of one of the lines represented by $5x^2 + \frac{40}{3}xy + ky^2 = 0$ is $3$,then the angle between the pair of lines is

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