What will be the co ordinates of the point in original position if its co ordinates after rotation of axes through an angle `60^(@)` be `(2,-sqrt(3))`?

Answer

To find the coordinates of the point in its original position after a rotation of axes through an angle of \(60^\circ\), we can use the formulas for the transformation of coordinates due to rotation.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- The coordinates after rotation are \( (x', y') = (2, -\sqrt{3}) \).
- The angle of rotation \( \theta = 60^\circ \).

2. **Use the Rotation Formulas:**
The formulas to convert the coordinates from the rotated system back to the original system are:
\[ x = x' \cos \theta + y' \sin \theta \]
\[ y = -x' \sin \theta + y' \cos \theta \]

3. **Calculate \( \cos 60^\circ \) and \( \sin 60^\circ \):**
- \( \cos 60^\circ = \frac{1}{2} \)
- \( \sin 60^\circ = \frac{\sqrt{3}}{2} \)

4. **Substitute the Values into the Formulas:**
- For \( x \):
\[ x = 2 \cdot \frac{1}{2} + (-\sqrt{3}) \cdot \frac{\sqrt{3}}{2} \]
Simplifying this:
\[ x = 1 - \frac{3}{2} = 1 - 1.5 = -\frac{1}{2} \]

- For \( y \):
\[ y = -2 \cdot \frac{\sqrt{3}}{2} + (-\sqrt{3}) \cdot \frac{1}{2} \]
Simplifying this:
\[ y = -\sqrt{3} - \frac{\sqrt{3}}{2} = -\frac{2\sqrt{3}}{2} - \frac{\sqrt{3}}{2} = -\frac{3\sqrt{3}}{2} \]

5. **Final Coordinates:**
The coordinates of the point in its original position are:
\[ \left(-\frac{1}{2}, -\frac{3\sqrt{3}}{2}\right) \]

### Summary:
The coordinates of the point in its original position are \( \left(-\frac{1}{2}, -\frac{3\sqrt{3}}{2}\right) \).