A square of each side 2, lies above the x-axis and has one vertex at the origin. If one of the sides passing through the origin makes an angle `30^(@)` with the positive direction of the x-axis, then the sum of the x-coordinates of the vertices of the square is:

Options

A. `sqrt(3)-2`
B. `2sqrt(3)-1`
C. `sqrt(3)-1`
D. `2sqrt(3)-2`

Answer

To solve the problem step by step, we will first visualize the square and then calculate the coordinates of its vertices based on the given conditions.

### Step 1: Visualize the Square
We know that one vertex of the square is at the origin (0, 0) and one side of the square makes an angle of 30 degrees with the positive x-axis.

### Step 2: Determine the Coordinates of the Vertices
1. **Vertex A (Origin)**: This is at (0, 0).
2. **Vertex B**: Since one side makes an angle of 30 degrees with the x-axis and has a length of 2, we can find the coordinates of vertex B using trigonometry:
- \( x_B = 2 \cos(30^\circ) \)
- \( y_B = 2 \sin(30^\circ) \)
- Using the values \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \) and \( \sin(30^\circ) = \frac{1}{2} \):
- \( x_B = 2 \cdot \frac{\sqrt{3}}{2} = \sqrt{3} \)
- \( y_B = 2 \cdot \frac{1}{2} = 1 \)
- So, the coordinates of B are \( B(\sqrt{3}, 1) \).

3. **Vertex C**: The next vertex C will be perpendicular to AB. The angle between AB and the line BC is 90 degrees. Therefore, the coordinates of C can be calculated by rotating point B by 90 degrees:
- The x-coordinate of C will be the same as B's y-coordinate and the y-coordinate will be the negative of B's x-coordinate:
- \( x_C = y_B = 1 \)
- \( y_C = -x_B = -\sqrt{3} \)
- So, the coordinates of C are \( C(1, -\sqrt{3}) \).

4. **Vertex D**: Finally, the coordinates of vertex D can be calculated by moving vertically from the origin:
- Since the square has a side length of 2, and we have already moved 1 unit up to B, we need to move 1 unit down from C to get D:
- \( x_D = 0 \)
- \( y_D = 2 \) (the height of the square)
- So, the coordinates of D are \( D(0, 2) \).

### Step 3: Sum of the x-coordinates
Now we have the coordinates of all vertices:
- A(0, 0)
- B(\(\sqrt{3}, 1\))
- C(1, -\(\sqrt{3}\))
- D(0, 2)

Now we can sum the x-coordinates:
\[ \text{Sum} = x_A + x_B + x_C + x_D = 0 + \sqrt{3} + 1 + 0 = \sqrt{3} + 1 \]

### Final Answer
The sum of the x-coordinates of the vertices of the square is \( \sqrt{3} + 1 \).