The slope of the line through the origin which makes an angle of $30^{\circ}$ with the positive direction of…
- $\frac{-2}{\sqrt{3}}$
- $-\sqrt{3}$
- $\frac{\sqrt{3}}{2}$
- $\frac{-1}{\sqrt{3}}$
Solution
Refer figure
Angle made by line $\mathrm{L}$ with positive direction of $\mathrm{X}$ axis is $\left(90^{\circ}+30^{\circ}\right)$ i.e. $120^{\circ}$.
$\therefore$ Slope of line $\mathrm{L}=\tan \left(120^{\circ}\right)=\tan \left(\pi-60^{\circ}\right)=-\tan 60^{\circ}=-\sqrt{3}$Asked in: MHT CET 2021 (20 Sep Shift 1)