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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 30⁰ with the positive direction of the x-axis, then the sum of the x-coordinates of the vertices of the square

Option 1)
$2\sqrt{3} -1$

Option 2)
$2\sqrt{3} -2$

Option 3)
$\sqrt{3}-2$

Option 4)
$\sqrt{3}-1$

Answers (1)

best_answer

As we learnt in

Parametric form -

$x=x_{1}+r\cos \Theta$

$y=y_{1}+r\sin \Theta$

- wherein

Where $\Theta$ is the inclination of the line and $r$ is the distance between $(x,y)$ and $(x_{1},y_{1})$

x-coordinates of A is $2cos30^{\circ}$

$=2\times \frac{\sqrt{3}}{2}=\sqrt{3}$

angle of OB with positive x-axis is $30^{\circ}+45^{\circ}=75^{\circ}$

Hence x - coordinate of B is $2\sqrt{2}\cos75^{\circ}$

$=2\sqrt{2}\left ( \frac{\sqrt{3}-1}{2\sqrt{2}} \right ) =(\sqrt{3}-1)$

x-cordinate of C is -2sin $30^{\circ}$ =-1

Hence Sum= $(\sqrt{3}+\sqrt{3}-1+(-1))$

$=2\sqrt{3}-2$

Option 1)

$2\sqrt{3} -1$

This option is correct

Option 2)

$2\sqrt{3} -2$

This option is incorrect

Option 3)

$\sqrt{3}-2$

This option is incorrect

Option 4)

$\sqrt{3}-1$

This option is incorrect

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