When the axes are rotated through an angle `60^(0)` the point P is changed as (3,4). Find original coordinates of P.

Answer

To find the original coordinates of point P before the axes were rotated by an angle of \(60^\circ\), we can use the transformation formulas for rotation of axes. Let's denote the original coordinates of point P as \((x, y)\) and the new coordinates after rotation as \((x', y') = (3, 4)\).

### Step-by-Step Solution:

1. **Identify the Rotation Angle**:
The angle of rotation is given as \( \theta = 60^\circ \).

2. **Use the Transformation Formulas**:
The transformation formulas for rotating the axes are:
\[ x = x' \cos \theta - y' \sin \theta \]
\[ y = x' \sin \theta + y' \cos \theta \]

3. **Substitute the Values**:
Substitute \(x' = 3\), \(y' = 4\), and \(\theta = 60^\circ\) into the formulas. We need the values of \(\cos 60^\circ\) and \(\sin 60^\circ\):
\[ \cos 60^\circ = \frac{1}{2}, \quad \sin 60^\circ = \frac{\sqrt{3}}{2} \]

4. **Calculate the Original x-coordinate**:
Using the formula for \(x\):
\[ x = 3 \cdot \cos 60^\circ - 4 \cdot \sin 60^\circ \]
\[ x = 3 \cdot \frac{1}{2} - 4 \cdot \frac{\sqrt{3}}{2} \]
\[ x = \frac{3}{2} - 2\sqrt{3} \]

5. **Calculate the Original y-coordinate**:
Using the formula for \(y\):
\[ y = 3 \cdot \sin 60^\circ + 4 \cdot \cos 60^\circ \]
\[ y = 3 \cdot \frac{\sqrt{3}}{2} + 4 \cdot \frac{1}{2} \]
\[ y = \frac{3\sqrt{3}}{2} + 2 \]

6. **Final Result**:
The original coordinates of point P before the rotation are:
\[ (x, y) = \left(\frac{3}{2} - 2\sqrt{3}, \frac{3\sqrt{3}}{2} + 2\right) \]