What is the angle in degrees between the line | vedclass

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(D) The given lines are $y = 1x + 5$ and $y = \sqrt{3}x - 4$. Comparing these with $y = mx + c$,we get slopes $m_1 = 1$ and $m_2 = \sqrt{3}$. The angl

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$15$

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(D) The given lines are $y = 1x + 5$ and $y = \sqrt{3}x - 4$. Comparing these with $y = mx + c$,we get slopes $m_1 = 1$ and $m_2 = \sqrt{3}$. The angle $	heta$ between two 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{1 - \sqrt{3}}{1 + (1)(\sqrt{3})} ight| = \left| \frac{1 - \sqrt{3}}{1 + \sqrt{3}} ight|$. Rationalizing the expression: $	an 	heta = \left| \frac{(1 - \sqrt{3})^2}{1 - 3} ight| = \left| \frac{1 + 3 - 2\sqrt{3}}{-2} ight| = \left| \frac{4 - 2\sqrt{3}}{-2} ight| = | -2 + \sqrt{3} | = 2 - \sqrt{3}$. Since $	an(15^\circ) = 2 - \sqrt{3}$,we have $	heta = 15^\circ$.

What is the angle in degrees between the lines $y = x + 5$ and $y = \sqrt{3}x - 4$ (in $^\circ$)?

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