Question

Find the angles between the pairs of straight lines:
$x-y\sqrt{3}=5$ and $\sqrt{3}x+y=7$

Answer 1

Given pairs of straight lines are

$x-y\sqrt{3}=5$ and $\sqrt{3}x+y=7$

Slopes are $m_1=\frac{1}{\sqrt{3}}$ are $m_2=-\sqrt{3}$

Then,

$\tan \theta =\frac{-\sqrt{3}-\frac{1}{\sqrt{3}}}{1+(-\sqrt{3}\times \frac{1}{\sqrt{3}})}$

$\tan \theta =\frac{-\sqrt{3}-\frac{1}{\sqrt{3}}}{1-1}$

$\Rightarrow \tan \theta =\infty$

$\Rightarrow \theta =\frac{\pi }{2}={90}^0$ is the angle.