A square, of each side 2 , lies above the x -axis and has one...

For $A,$

\dfrac{x}{\cos 30 °} = \dfrac{y}{\sin 30 °} = 2

$\to x = \dfrac{2\sqrt{3}}{2} = \sqrt{3}$ and $y = 1$

For $C,$

\dfrac{x}{\cos 120 °} = \dfrac{y}{\sin 120 °} = 2

$x = - 1, y = \sqrt{3}$

For $B,$

\dfrac{x}{\cos 75 °} = \dfrac{y}{\sin 75 °} = 2\sqrt{2}

$x = \sqrt{3} - 1$, $y = \sqrt{3} + 1$

Sum of the $x$-coordinates $= 0 + \sqrt{3} + \sqrt{3} - 1 + \left(- 1\right)$

$= 2\sqrt{3} - 2$