The acute angle between the lines given by y | vedclass

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(D) The given equations of the lines are $y = \sqrt{3}x - 1$ and $y = \frac{1}{\sqrt{3}}x - \frac{7}{\sqrt{3}}$.Comparing these with $y = mx + c$,the

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$30$

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(D) The given equations of the lines are $y = \sqrt{3}x - 1$ and $y = \frac{1}{\sqrt{3}}x - \frac{7}{\sqrt{3}}$.Comparing these with $y = mx + c$,the slopes are $m_{1} = \sqrt{3}$ and $m_{2} = \frac{1}{\sqrt{3}}$.The angle $	heta$ between the lines is given by $	an 	heta = \left| \frac{m_{1} - m_{2}}{1 + m_{1}m_{2}} ight|$.Substituting the values: $	an 	heta = \left| \frac{\sqrt{3} - \frac{1}{\sqrt{3}}}{1 + (\sqrt{3})(\frac{1}{\sqrt{3}})} ight| = \left| \frac{\frac{3-1}{\sqrt{3}}}{1 + 1} ight| = \left| \frac{2/\sqrt{3}}{2} ight| = \frac{1}{\sqrt{3}}$.Since $	an 	heta = \frac{1}{\sqrt{3}}$,we have $	heta = 30^{\circ}$.

The acute angle between the lines given by $y - \sqrt{3}x + 1 = 0$ and $\sqrt{3}y - x + 7 = 0$ is (in $^{\circ}$)

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