Find the acute angle between the lines y = 3 | vedclass

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(B) The equation of the first line is $y = 3$,which is a horizontal line parallel to the $x$-axis. Its slope $m_1 = 0$.The equation of the second line

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(B) The equation of the first line is $y = 3$,which is a horizontal line parallel to the $x$-axis. Its slope $m_1 = 0$.The equation of the second line is $y = \sqrt{3}x + 9$. Comparing this with $y = mx + c$,the slope $m_2 = \sqrt{3}$.Let $	heta$ be the angle between the lines. Then $	an 	heta = |\frac{m_2 - m_1}{1 + m_1m_2}|$.$	an 	heta = |\frac{\sqrt{3} - 0}{1 + 0 	imes \sqrt{3}}| = |\sqrt{3}| = \sqrt{3}$.Since $	an 	heta = \sqrt{3}$,we have $	heta = 60^{\circ}$.

Find the acute angle between the lines $y = 3$ and $y = \sqrt{3}x + 9$. (in $^{\circ}$)

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