Question

# 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) $\sqrt{3}-1$

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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”.

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 $2\sqrt{3}-2$.
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.