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Q.
The slope of a line, which makes an angle of $30^{\circ}$ with the positive direction of $Y$-axis measured anti-clockwise, is
Straight Lines
Solution:
Given, $\angle Y P Q=30^{\circ}$
To find slope of line $A Q$.
Here, $ \angle Y P Q=\angle O P A $ (vertically opposite angles)
$\because \angle O P A+\angle P O A+\angle P A O=180^{\circ}$
$\left(\because\right.$ sum of all angles of a triangle is $\left.180^{\circ}\right)$
$\Rightarrow 30^{\circ}+90^{\circ}+\angle P A O=180^{\circ}$
$\Rightarrow \angle P A O=180^{\circ}-120^{\circ}=60^{\circ}$
$\Rightarrow \angle P A X=180^{\circ}-60^{\circ}=120^{\circ}$
$\therefore$ Slope of line $A Q=m=\tan 120^{\circ}$
$\left[\because m=\tan \theta \tan \left(180^{\circ}-\theta\right) =-\tan \theta\right]$
$=\tan \left(180^{\circ}-60^{\circ}\right)$
$=-\tan 60^{\circ}=-\sqrt{3}$
