Find the inclination of a line whose slope is
(i) $\sqrt{3}$
(ii) $\frac{1}{\sqrt{3}}$
(iii) 1
(iv) $-1$
(v) $-\sqrt{3}$
We know that the slope of a given line is given by
Slope $=\tan \theta$ Where $\theta$ angle of inclination
(i) $\tan \theta=\sqrt{3}$
$\Rightarrow \theta=\tan ^{-1}(\sqrt{3})$
$\Rightarrow \theta=60^{\circ}$
(ii) $\tan \theta=\frac{1}{\sqrt{3}}$
$\Rightarrow \theta=\tan ^{-1}\left(\frac{1}{\sqrt{3}}\right)$
$\Rightarrow \theta=30^{\circ}$
(iii) $\tan \theta=1$
$\Rightarrow \theta=\tan ^{-1}(1)$
$\Rightarrow \theta=45^{\circ}$
(iv) $\tan \theta=-1$
$\Rightarrow \theta=\tan ^{-1}(-1)$
$\Rightarrow \theta=-45^{\circ}=315^{\circ}$
(v) $\tan \theta=-\sqrt{3}$
$\Rightarrow \theta=\tan ^{-1}(-\sqrt{3})$
$\Rightarrow \theta=-60^{\circ}=300^{\circ}$