The angle between the lines x - sqrt{3}y + 5 | vedclass

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(B) The given line is $x - \sqrt{3}y + 5 = 0$.Rearranging it into slope-intercept form $y = mx + c$,we get $\sqrt{3}y = x + 5$,which implies $y = \fra

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(B) The given line is $x - \sqrt{3}y + 5 = 0$.Rearranging it into slope-intercept form $y = mx + c$,we get $\sqrt{3}y = x + 5$,which implies $y = \frac{1}{\sqrt{3}}x + \frac{5}{\sqrt{3}}$.The slope of this line is $m_1 = \frac{1}{\sqrt{3}}$.The $y$-axis is a vertical line,so its slope $m_2$ is undefined (or we can consider the angle it makes with the $x$-axis as $90^o$).The angle $	heta$ between a line with slope $m$ and the $y$-axis is given by $	an 	heta = |\frac{1}{m}|$.Here,$	an 	heta = |\frac{1}{1/\sqrt{3}}| = \sqrt{3}$.Therefore,$	heta = 	an^{-1}(\sqrt{3}) = 60^o$.

The angle between the lines $x - \sqrt{3}y + 5 = 0$ and the $y$-axis is ....$^o$

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