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(a) \[2\sqrt{3}-1\]

(b) \[2\sqrt{3}-2\]

(c) \[\sqrt{3}-1\]

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To solve this question we will first of all make the figure of given condition,

Draw the square as OABC, O being at origin.

A = \[\left( {{x}_{1}},{{y}_{1}} \right)\]

B = \[\left( {{x}_{2}},{{y}_{2}} \right)\]

C = \[\left( {{x}_{3}},{{y}_{3}} \right)\]

Join the diagonal OB.

We are given that, \[\angle AOX={{30}^{\circ }}\].

Now consider the triangle drawn as AOX below,

Given that, OA = 2 and \[\angle AOX={{30}^{\circ }}\].

Let OX = \[{{x}_{1}}\] then by using property of \[\cos \theta \] we have,

\[{{x}_{1}}=2\cos {{30}^{\circ }}\]

Also, as \[\cos {{30}^{\circ }}=\dfrac{\sqrt{3}}{2}\].

\[\Rightarrow {{x}_{1}}=2\left( \dfrac{\sqrt{3}}{2} \right)=\sqrt{3}\] - (1)

Again let \[AX={{y}_{1}}\], then by property of \[\sin \theta \] we have, \[{{y}_{1}}=2\sin {{30}^{\circ }}\].

As, \[\sin {{30}^{\circ }}=\dfrac{1}{2}\].

\[\Rightarrow {{y}_{1}}=2\left( \dfrac{1}{2} \right)=1\]

Therefore, \[\left( {{x}_{1}},{{y}_{1}} \right)=\left( \sqrt{3},1 \right)\].

Similarly, we have to find \[{{x}_{2}}\] & \[{{x}_{3}}\].

Because \[\Delta BAO\] is a right angled triangle. So, we have applying Pythagoras theorem which says that –

“In a right angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides”.

Then \[\Rightarrow {{\left( BO \right)}^{2}}={{2}^{2}}+{{2}^{2}}\]

\[\Rightarrow {{\left( BO \right)}^{2}}=8\]

Taking square roots both sides we have,

\[BO=2\sqrt{2}\]

Hence diagonal, \[BO=2\sqrt{2}\].

We have angle of square measures \[{{90}^{\circ }}\] and diagonals of square bisects the angles.

\[\Rightarrow \angle BOA=\dfrac{1}{2}\left( {{90}^{\circ }} \right)={{45}^{\circ }}\] also \[\angle COB={{45}^{\circ }}\] (same reason).

Then, \[\angle BOX={{45}^{\circ }}+{{30}^{\circ }}={{75}^{\circ }}\].

Consider \[\Delta OAX\] then given OB = \[2\sqrt{2}\] and \[\angle BOX={{75}^{\circ }}\].

Then, \[{{x}_{2}}=2\sqrt{2}\cos {{75}^{\circ }}\]

\[{{x}_{2}}=2\sqrt{2}\left( \dfrac{\sqrt{3}-1}{2\sqrt{2}} \right)\]

\[{{x}_{2}}=\sqrt{3}-1\] - (2)

This so as, \[\cos {{75}^{\circ }}=\dfrac{\sqrt{3}-1}{2\sqrt{2}}\].

Again, \[\angle COX=\angle COB+\angle BOX\]

\[\angle COX={{45}^{\circ }}+{{75}^{\circ }}={{120}^{\circ }}\]

Then consider triangle COX, where OC = 2 & \[\angle COX={{120}^{\circ }}\].

Then, \[{{x}_{3}}=2\cos {{120}^{\circ }}\].

\[{{x}_{3}}=2\left( \dfrac{-1}{2} \right)\]

\[{{x}_{3}}=-1\] - (3)

As, \[\cos {{120}^{\circ }}=\dfrac{-1}{2}\].

Now finally adding (1), (2) and (3) we get,

\[\begin{align}

& {{x}_{1}}+{{x}_{2}}+{{x}_{3}}=\sqrt{3}+\left( \sqrt{3}-1 \right)+\left( -1 \right) \\

& {{x}_{1}}+{{x}_{2}}+{{x}_{3}}=2\sqrt{3}-2 \\

\end{align}\]

Hence the sum of co – ordinates is \[2\sqrt{3}-2\].