Find the angles between the lines sqrt{3}x + | vedclass

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(A) The given lines are $\sqrt{3}x + y = 1$ and $x + \sqrt{3}y = 1$. Converting to slope-intercept form $y = mx + c$: Line $(1): y = -\sqrt{3}x + 1$,s

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$30^{\circ}, 150^{\circ}$

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(A) The given lines are $\sqrt{3}x + y = 1$ and $x + \sqrt{3}y = 1$. Converting to slope-intercept form $y = mx + c$: Line $(1): y = -\sqrt{3}x + 1$,so $m_1 = -\sqrt{3}$. Line $(2): \sqrt{3}y = -x + 1 \implies y = -\frac{1}{\sqrt{3}}x + \frac{1}{\sqrt{3}}$,so $m_2 = -\frac{1}{\sqrt{3}}$. The acute 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} - (-1/\sqrt{3})}{1 + (-\sqrt{3})(-1/\sqrt{3})} ight| = \left| \frac{(-3 + 1)/\sqrt{3}}{1 + 1} ight| = \left| \frac{-2}{2\sqrt{3}} ight| = \frac{1}{\sqrt{3}}$. Thus,$	heta = 30^{\circ}$. The obtuse angle is $180^{\circ} - 30^{\circ} = 150^{\circ}$. Therefore,the angles between the lines are $30^{\circ}$ and $150^{\circ}$.

Find the angles between the lines $\sqrt{3}x + y = 1$ and $x + \sqrt{3}y = 1$.

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