The slope of the line through the origin which makes an angle of $30^\circ$ with the positive direction of Y-axis measured anticlockwise is :
Show Hint
Draw a quick mental sketch of the coordinate axes! A line tilted $30^\circ$ into the second quadrant is clearly sloping downwards from left to right, meaning its slope must be negative. Since it is steeper than a $45^\circ$ line, its absolute value must be greater than 1, which instantly points you directly to $-\sqrt{3}$ over $-\frac{1}{\sqrt{3}}$!
Step 1: Recall what slope means.
The slope is the tangent of the angle the line makes with the positive x-axis, measured anticlockwise.
Step 2: Find that angle.
The positive y-axis is already at $90^\circ$ from the x-axis. The line is a further $30^\circ$ past it, anticlockwise. So the total angle is $90^\circ+30^\circ=120^\circ$.
Step 3: Take the tangent.
\[ m=\tan120^\circ=-\tan60^\circ=-\sqrt3 \]
\[ \boxed{-\sqrt3} \]