The acute angle between the lines given by $y-\sqrt{3} x+1=0$ and $\sqrt{3} y-x+7=0$ is
The acute angle between the lines given by $y-\sqrt{3} x+1=0$ and
$\sqrt{3} y-x+7=0$ is
$75^{\circ}$
$60^{\circ}$
$45^{\circ}$
$30^{\circ}$
Solution
Slope of given lines are $m_{1}=\sqrt{3}$, and $m_{2}=\frac{1}{\sqrt{3}}$
$\begin{array}{l}
\therefore \tan \theta=\left|\frac{\sqrt{3}-\frac{1}{\sqrt{3}}}{1+\sqrt{3} \times \frac{1}{\sqrt{3}}}\right|=\left|\frac{2}{\sqrt{3}}\right| \\
\tan \theta=\frac{1}{\sqrt{3}} \Rightarrow \theta=30^{\circ}
\end{array}$