Question

Find the angle between the lines $y-\sqrt{3}x-5=0$ and $\sqrt{3}y-x+6=0$.

Answer 1

We have,

$\begin{array}{c}y=\sqrt{3x}-5=0m_1=\sqrt{3}\\ \sqrt{3y}=x-6m_2=\frac{1}{\sqrt{3}}\\ \tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|=\frac{1}{\sqrt{3}}\\ \therefore \theta ={30}^0\end{array}$

$\therefore$ Angle b/w llines is ${30}^0$.