Find angles between the lines $\sqrt{3} x+y=1$ and $x+\sqrt{3} y=1$
The given lines are $\sqrt{3} x+y=1$ and $x+\sqrt{3} y=1$.
$y=-\sqrt{3} x+1$ $\ldots(1)$
and
$y=-\frac{1}{\sqrt{3}} x+\frac{1}{\sqrt{3}}$$\ldots(2)$
The slope of line (1) is $m_{1}=-\sqrt{3}$,
while the slope of line
(2) is $m_{2}=-\frac{1}{\sqrt{3}}$.
The acute angle i.e., $\theta$ between the two lines is given by
$\tan \theta=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right|$
$\tan \theta=\left|\frac{-\sqrt{3}+\frac{1}{\sqrt{3}}}{1+(-\sqrt{3})\left(-\frac{1}{\sqrt{3}}\right)}\right|$
$\tan \theta=\left|\frac{\frac{-3+1}{\sqrt{3}}}{1+1}\right|=\left|\frac{-2}{2 \times \sqrt{3}}\right|$
$\tan \theta=\frac{1}{\sqrt{3}}$
$\theta=30^{\circ}$
Thus, the angle between the given lines is either $30^{\circ}$ or $180^{\circ}-30^{\circ}=150^{\circ}$.