If the lines y = (2 + sqrt{3})x + 4 and y = k | vedclass

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(C) The slopes of the given lines are $m_1 = 2 + \sqrt{3}$ and $m_2 = k$.Given the angle between the lines is $	heta = 60^{\circ}$,we use the formula

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$-1$

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(C) The slopes of the given lines are $m_1 = 2 + \sqrt{3}$ and $m_2 = k$.Given the angle between the lines is $	heta = 60^{\circ}$,we use the formula $	an 	heta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} ight|$.Substituting the values: $	an 60^{\circ} = \left| \frac{2 + \sqrt{3} - k}{1 + k(2 + \sqrt{3})} ight| = \sqrt{3}$.Case $1$: $\frac{2 + \sqrt{3} - k}{1 + k(2 + \sqrt{3})} = \sqrt{3} \implies 2 + \sqrt{3} - k = \sqrt{3} + k(2\sqrt{3} + 3) \implies 2 - k = k(2\sqrt{3} + 3) \implies k(3 + 2\sqrt{3} + 1) = 2 \implies k(4 + 2\sqrt{3}) = 2 \implies k = \frac{1}{2 + \sqrt{3}} = 2 - \sqrt{3}$.Case $2$: $\frac{2 + \sqrt{3} - k}{1 + k(2 + \sqrt{3})} = -\sqrt{3} \implies 2 + \sqrt{3} - k = -\sqrt{3} - k(2\sqrt{3} + 3) \implies 2 + 2\sqrt{3} = k - k(2\sqrt{3} + 3) \implies 2(1 + \sqrt{3}) = k(1 - 2\sqrt{3} - 3) \implies 2(1 + \sqrt{3}) = k(-2 - 2\sqrt{3}) \implies 2(1 + \sqrt{3}) = -2k(1 + \sqrt{3}) \implies k = -1$.

If the lines $y = (2 + \sqrt{3})x + 4$ and $y = kx + 6$ are inclined at an angle of $60^{\circ}$ to each other,then the value of $k$ will be

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