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) $\sqrt{3}-1$

Answer 1

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,
$BO=2\sqrt{2}$
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,

& {{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{aligned}$$ Hence the sum of co – ordinates is $$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.