If the axes are rotated through an angle `60^(0)`, fine the coordinates of the point `(6sqrt3,4)` in the new system.

Answer

To find the coordinates of the point \( (6\sqrt{3}, 4) \) after the axes have been rotated through an angle of \( 60^\circ \), we will use the transformation formulas for coordinates after rotation.

### Step 1: Identify the original coordinates and the angle of rotation
The original coordinates of the point are:
- \( x = 6\sqrt{3} \)
- \( y = 4 \)

The angle of rotation \( \theta \) is \( 60^\circ \).

### Step 2: Use the rotation transformation formulas
The transformation formulas for rotating the axes are:
\[ x' = x \cos \theta + y \sin \theta \]
\[ y' = -x \sin \theta + y \cos \theta \]

### Step 3: Calculate \( \cos 60^\circ \) and \( \sin 60^\circ \)
From trigonometric values:
- \( \cos 60^\circ = \frac{1}{2} \)
- \( \sin 60^\circ = \frac{\sqrt{3}}{2} \)

### Step 4: Substitute the values into the formulas
Substituting \( x \), \( y \), \( \cos 60^\circ \), and \( \sin 60^\circ \) into the transformation formulas:

#### For \( x' \):
\[ x' = (6\sqrt{3}) \cdot \frac{1}{2} + 4 \cdot \frac{\sqrt{3}}{2} \]
\[ x' = 3\sqrt{3} + 2\sqrt{3} \]
\[ x' = 5\sqrt{3} \]

#### For \( y' \):
\[ y' = - (6\sqrt{3}) \cdot \frac{\sqrt{3}}{2} + 4 \cdot \frac{1}{2} \]
\[ y' = - 9 + 2 \]
\[ y' = -7 \]

### Step 5: Write the new coordinates
The new coordinates after the rotation are:
\[ (x', y') = (5\sqrt{3}, -7) \]

### Final Answer
The coordinates of the point \( (6\sqrt{3}, 4) \) in the new system after rotation through \( 60^\circ \) are \( (5\sqrt{3}, -7) \).

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