The acute angle between the lines x sin theta | vedclass

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Question

(A) The equation of the first line is $x \sin 	heta - y \cos 	heta = 5$. Its slope $m_1 = \frac{\sin 	heta}{\cos 	heta} = 	an 	heta$. The equati

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$|	heta - \alpha|$

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(A) The equation of the first line is $x \sin 	heta - y \cos 	heta = 5$. Its slope $m_1 = \frac{\sin 	heta}{\cos 	heta} = 	an 	heta$. The equation of the second line is $x \sin \alpha - y \cos \alpha + 11 = 0$. Its slope $m_2 = \frac{\sin \alpha}{\cos \alpha} = 	an \alpha$. Let $\beta$ be the acute angle between the lines. Then,$	an \beta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} ight| = \left| \frac{	an 	heta - 	an \alpha}{1 + 	an 	heta 	an \alpha} ight|$. Using the trigonometric identity $	an(	heta - \alpha) = \frac{	an 	heta - 	an \alpha}{1 + 	an 	heta 	an \alpha}$,we get $	an \beta = |	an(	heta - \alpha)|$. Therefore,$\beta = |	heta - \alpha|$.

The acute angle between the lines $x \sin \theta - y \cos \theta = 5$ and $x \sin \alpha - y \cos \alpha + 11 = 0$ is

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