The position vector of point P (20, 10) is rotated anti-clockwise in X-Y plane by an angle 6 =30° such that point P occupies position Q, as shown in the figure. The coordinates (x, y) of Q are 

The correct option is (12.32, 18.66).

Step-by-Step Explanation:

When a point $P(x,y)$ is rotated counter-clockwise (anti-clockwise) about the origin by an angle $\theta$ in the 2D Cartesian plane, its new coordinates $Q(x^{\text{'}},y^{\text{'}})$ are determined using the standard rotation transformation equations:

$$x^{\text{'}}=x\cos \left(\theta \right)-y\sin \left(\theta \right)$$

$$y^{\text{'}}=x\sin \left(\theta \right)+y\cos \left(\theta \right)$$

From the problem statement and the provided figure:
- The initial point is $P(20,10)$, so we have $x=20$ and $y=10$.
- The angle of anti-clockwise rotation is $\theta ={30}^{ˆ}$.

Let us calculate the trigonometric values for $\theta ={30}^{ˆ}$:
- $\sin \left({30}^{ˆ}\right)=0.5$
- $\cos \left({30}^{ˆ}\right)=\frac{\sqrt{3}}{2}\approx 0.8660$

1. Calculating the new x-coordinate ($x^{\text{'}}$):

$$x^{\text{'}}=20·\cos \left({30}^{ˆ}\right)-10·\sin \left({30}^{ˆ}\right)$$

$$x^{\text{'}}=20·0.8660-10·0.5$$

$$x^{\text{'}}=17.32-5.00=12.32$$

2. Calculating the new y-coordinate ($y^{\text{'}}$):

$$y^{\text{'}}=20·\sin \left({30}^{ˆ}\right)+10·\cos \left({30}^{ˆ}\right)$$

$$y^{\text{'}}=20·0.5+10·0.8660$$

$$y^{\text{'}}=10.00+8.66=18.66$$

Thus, the coordinates $(x,y)$ of point Q after rotation are (12.32, 18.66).