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Q.
Find the angle between the lines $y-\sqrt{3}x-5=0$ and $\sqrt{3}y-x+6=0$.
Straight Lines
Solution:
Given lines are
$y-\sqrt{3}x-5=0$ or $y=\sqrt{3}x+5 \quad\ldots\left(i\right)$
and $\sqrt{3}y-x+6=0$ or $y=\frac{1}{\sqrt{3}}x-2\sqrt{3}\quad\ldots\left(ii\right)$
Slope of line $\left(i\right)$ is $m_{1}=\sqrt{3}$ and slope of line $\left(ii\right)$ is $m_{2}=\frac{1}{\sqrt{3}}$.
The acute angle (say $\theta$)between two lines is given by
$tan\,\theta=\left|\frac{m_{2}-m_{1}}{1+m_{1}m_{2}}\right|=\left|\frac{\frac{1}{\sqrt{3}}-\sqrt{3}}{1+\sqrt{3}\times\frac{1}{\sqrt{3}}}\right|$
$=\left|\frac{1-3}{2\sqrt{3}}\right| = \pm\frac{1}{\sqrt{3}}$
which gives $\theta=30^{\circ}$. Hence, the angle between two lines is either $30^{\circ}$ or $\left(180 - 30\right)^{\circ} = 150^{\circ}$