Question

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

Answer 1

$y-\sqrt{3}x=5$

$y=\sqrt{3}x+5$

The above equation is of the form $y=mx=c$ whereas, $m$ is the slope.

Therefore,

$m_1=\sqrt{3}$

Similarly,

$\sqrt{3}y-x+6=0$

$\sqrt{3}y=x-6$

$y=\frac{x}{\sqrt{3}}-2\sqrt{3}$

$\Rightarrow m_2=\frac{1}{\sqrt{3}}$

As we know that,

$\tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$

$\Rightarrow \tan \theta =\left|\frac{\sqrt{3}-\frac{1}{\sqrt{3}}}{1+\sqrt{3}\left(\frac{1}{\sqrt{3}}\right)}\right|$

$\Rightarrow \tan \theta =\frac{3-1}{2\sqrt{3}}=\frac{1}{\sqrt{3}}$

$\Rightarrow \theta ={\tan }^{-1}\left(\frac{1}{\sqrt{3}}\right)$

$\Rightarrow \theta =30°$

Thus the acute angle between the given lines is $30°$

Also,

Obtuse angle between the lines $=180°-30°=150°$