Step 1: Understanding the Question:
We need to determine the mathematical slope of a straight line passing through the origin that is oriented at an angle of $30^\circ$ relative to the positive y-axis, measured in the counter-clockwise (anticlockwise) direction.
Step 2: Key Formula or Approach:
The slope $m$ of a straight line is defined by the tangent of its inclination angle $\theta$, where $\theta$ is the angle that the line makes with the
positive direction of the x-axis measured anticlockwise:
$$ m = \tan\theta $$
Step 3: Detailed Explanation:
Let's trace out the angles geometrically starting from the standard coordinate layout:
• The positive direction of the y-axis is already located at an angle of $90^\circ$ relative to the positive direction of the x-axis.
• The problem states that our line is rotated an additional $30^\circ$ anticlockwise away from this positive y-axis.
Therefore, the total inclination angle $\theta$ of the line relative to the positive x-axis is:
$$ \theta = 90^\circ + 30^\circ = 120^\circ $$
Now, we calculate the slope by taking the tangent of this total angle:
$$ m = \tan(120^\circ) $$
Using the trigonometric reduction identity $\tan(180^\circ - \phi) = -\tan\phi$:
$$ m = \tan(180^\circ - 60^\circ) = -\tan(60^\circ) $$
Substituting the exact standard value $\tan(60^\circ) = \sqrt{3}$:
$$ m = -\sqrt{3} $$
Step 4: Final Answer:
The slope of the line is $-\sqrt{3}$, which corresponds to option (B).