The angle between the lines y - x + 5 = 0 and | vedclass

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(A) The given lines are $L_1: y = x - 5$ and $L_2: y = \sqrt{3}x + 7$. Comparing with $y = mx + c$,the slopes are $m_1 = 1$ and $m_2 = \sqrt{3}$. The

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$15$

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(A) The given lines are $L_1: y = x - 5$ and $L_2: y = \sqrt{3}x + 7$. Comparing with $y = mx + c$,the slopes are $m_1 = 1$ and $m_2 = \sqrt{3}$. The angle $	heta$ between two lines is given by $	an 	heta = |\frac{m_2 - m_1}{1 + m_1 m_2}|$. Substituting the values,$	an 	heta = |\frac{\sqrt{3} - 1}{1 + (1)(\sqrt{3})}| = |\frac{\sqrt{3} - 1}{\sqrt{3} + 1}|$. Rationalizing the denominator,$	an 	heta = |\frac{(\sqrt{3} - 1)^2}{3 - 1}| = |\frac{3 + 1 - 2\sqrt{3}}{2}| = |\frac{4 - 2\sqrt{3}}{2}| = |2 - \sqrt{3}|$. Since $	an 15^o = 2 - \sqrt{3}$,we have $	heta = 15^o$.

The angle between the lines $y - x + 5 = 0$ and $\sqrt{3}x - y + 7 = 0$ is: (in $^o$)

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