Question

the angle between the line √3x-y-2=0and x-√3y+1=0is​

Answer 1

$\large{ \underline{ \underline{ \bigstar \: \: \: \: \: \: { \pmb{ \sf{Conception \: : }}}}}}$

Angel b/w two lines is defined mathematically as, $\odot \: \: \: \: \sf \tan( \phi) = \tan( \pi - \theta) = - \tan( \theta) = - \bigg(\frac{m_1 - m_2}{1 + m_1m_2} \bigg)$$[ \sf{where, \: \: m_1 \: \: and \: \: m_2 \: \: are \: \: the \: \: slopes \: \: of \: \: intercepted \: \: lines}]$ $\\ \\ \large{ \underline{ \underline{ \bigstar \: \: \: \: \: \: { \pmb{ \sf{Solution \: : }}}}}}$

changing given lines into slope-intercept form, $\\ \\ \begin{array}{c|c} \dashrightarrow \sf x \sqrt{3} - y - 2 = 0 & \dashrightarrow \sf x - y \sqrt{3} + 1 = 0 \\ \\ \dashrightarrow \sf \: y = x \sqrt{3} - 2& \dashrightarrow \sf y = \frac{1}{ \sqrt{3} }x + \frac{1}{ \sqrt{3} } \\ \\ \sf \therefore \: \: slope \: \: is \: \: \sqrt{3} & \sf \therefore \: slope \: \: is \: \: \frac{1}{ \sqrt{3} } \\ \end{array}$Now, substituting into mathematical formula$\sf \dashrightarrow \tan( \theta) = \bigg(\frac{m_1 - m_2}{1 + m_1m_2} \bigg)$$\sf \dashrightarrow \tan( \theta) = \bigg(\frac{ \sqrt{3} - \frac{1}{ \sqrt{3} } }{1 + \sqrt{3} \times \frac{1}{ \sqrt{3} } } \bigg)$$\sf \dashrightarrow \tan( \theta) = \bigg( \frac{ \cancel{2 }}{{ \cancel{2} \: \sqrt{3} } } \bigg)$$\sf \dashrightarrow \tan( \theta) = \bigg( \frac{1}{ \sqrt{3} } \bigg)$$\dashrightarrow {\blue{{ \frak{\theta = \bigg(\frac{\pi}{6} } \bigg)}\: \: \: \: \: \&\: \: \: \:\:\: \phi = \bigg(\pi-\frac{\pi}{6}\bigg) =\bigg(\frac{5\pi}{6}\bigg)}}$

ʜᴇɴᴄᴇ, ᴀɴɢʟᴇs ʙ/ᴡ ᴛʜᴏsᴇ ᴛᴡᴏ ʟɪɴᴇs ᴡɪʟʟ ʙᴇ π/6 ᴀɴᴅ (π-π/6)= 5π/6

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