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A square of each side 2 lies above the x – axis and has one vertex at origin. If one of the side passing through the origin, if one of the side passing through the origin makes an angle \[{{30}^{\circ }}\] with positive direction of x – axis then the sum of x co – oridinates of vertices of the square is
(a) \[2\sqrt{3}-1\]
(b) \[2\sqrt{3}-2\]
(c) \[\sqrt{3}-1\]

Last updated date: 27th Feb 2024
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IVSAT 2024
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Hint: To solve this question first of all draw the figure of the given situation. Locate the co – ordinates of vertices of square ABCD, and then by determining the angle which each vertex makes from the x – axis we can easily determine the x – coordinates. Also we will use Pythagora's theorem to solve 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”.

Complete step-by-step answer:
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,
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,
  & {{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 \\
Hence the sum of co – ordinates is \[2\sqrt{3}-2\].
So, the correct answer is “Option B”.

Note: The possibility of error in this question can not determine the angles with which vertex O, A, B, C are attached to the x – axis. Rather going to directly determine \[{{x}_{1}},{{x}_{2}}\] and \[{{x}_{3}}\]. This can complicate the question which can lead to incorrect solutions, also the student might get confused with the horizontal and vertical distance as x and y co – ordinates.