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Step-by-Step Solution
Step 1: Label the vertices
Let the square be OADB with vertex O at the origin (0,0). The side length of the square is 2, and the side through the origin makes a 30° angle with the positive x-axis.
Step 2: Find coordinates of vertex A
If the side from O to A makes an angle of 30°, and its length is 2, then:
$A = (2 \cos 30^\circ,\; 2 \sin 30^\circ)$
Recall:
$ \cos 30^\circ = \frac{\sqrt{3}}{2} \text{ and } \sin 30^\circ = \frac{1}{2}. $
Therefore,
$A = \Bigl(2 \times \frac{\sqrt{3}}{2},\; 2 \times \frac{1}{2}\Bigr) = \bigl(\sqrt{3},\; 1\bigr).$
Step 3: Find coordinates of vertex D
From O, the perpendicular side will be at a 90° angle from OA, so it will lie along the direction of $30^\circ + 90^\circ = 120^\circ$. Again, the side length is 2, so:
$D = (2 \cos 120^\circ,\; 2 \sin 120^\circ).$
Since
$ \cos 120^\circ = -\frac{1}{2}, \quad \sin 120^\circ = \frac{\sqrt{3}}{2}, $
we get
$D = \Bigl(2 \times -\frac{1}{2},\; 2 \times \frac{\sqrt{3}}{2}\Bigr) = (-1,\; \sqrt{3}).$
Step 4: Find coordinates of vertex B
Vertex B is the “top-right” vertex, which can be obtained by adding the vectors OA and OD. Equivalently, B is the point you reach if you move from A in the same direction you moved from O to D. Hence,
$B = A + (D - O) = \bigl(\sqrt{3},\,1\bigr) + \bigl(-1,\,\sqrt{3}\bigr).$
Therefore:
$B = \bigl(\sqrt{3} - 1,\; 1 + \sqrt{3}\bigr).$
Step 5: Sum of the x-coordinates of all vertices
The four vertices are:
$O = (0,\,0)$
$A = \bigl(\sqrt{3},\,1\bigr)$
$D = (-1,\,\sqrt{3}\bigr)$
$B = \bigl(\sqrt{3} - 1,\; 1 + \sqrt{3}\bigr)$
Sum of their x-coordinates = $0 + \sqrt{3} + (-1) + (\sqrt{3} - 1)$
$= \sqrt{3} + \sqrt{3} - 1 - 1$
$= 2\sqrt{3} - 2.$
Final Answer
$\displaystyle 2\sqrt{3} \;-\; 2$
