A square,of each side 2,lies above the x-axis | vedclass

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(B) Let the vertices of the square be $O(0, 0)$,$A$,$B$,and $C$. The side $OA$ makes an angle of $30^o$ with the positive $x-$axis. Since the side len

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$2\sqrt{3} - 2$

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(B) Let the vertices of the square be $O(0, 0)$,$A$,$B$,and $C$. The side $OA$ makes an angle of $30^o$ with the positive $x-$axis. Since the side length is $2$,the coordinates of $A$ are $(2 \cos 30^o, 2 \sin 30^o) = (2 \cdot \frac{\sqrt{3}}{2}, 2 \cdot \frac{1}{2}) = (\sqrt{3}, 1)$.The side $OC$ is perpendicular to $OA$ and lies at an angle of $30^o + 90^o = 120^o$ with the positive $x-$axis. The coordinates of $C$ are $(2 \cos 120^o, 2 \sin 120^o) = (2 \cdot -\frac{1}{2}, 2 \cdot \frac{\sqrt{3}}{2}) = (-1, \sqrt{3})$.The vertex $B$ is the sum of vectors $\vec{OA}$ and $\vec{OC}$,so $B = (\sqrt{3} - 1, 1 + \sqrt{3})$.The $x-$coordinates of the vertices are $0$,$\sqrt{3}$,$-1$,and $\sqrt{3} - 1$.The sum of the $x-$coordinates is $0 + \sqrt{3} - 1 + \sqrt{3} - 1 = 2\sqrt{3} - 2$.

$A$ 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 $30^o$ with the positive direction of the $x-$axis,then the sum of the $x-$coordinates of the vertices of the square is

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