A square , of each side 2, lies above the x-axis and has one vertes 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 verties of the square is

Let side of the square be r = 2θ=300, O0, 0OA¯, OB¯, OC¯ making an anglesθ1=300, θ2=750, θ3=1200 with X−axis Sum of X−cordinates = x1+OAcosθ1+x2+OBcosθ2+x3+OCcosθ3⇒ 2cos3i+22cos750+2cos1200⇒ 232+223−122+2−12⇒ 3+3−1−1⇒ 23−2

A square , of each side 2, lies above the x-axis and has one vertes 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 verties of the square is

Let side of the square be r = 2θ=300, O0, 0OA¯, OB¯, OC¯ making an anglesθ1=300, θ2=750, θ3=1200 with X−axis Sum of X−cordinates = x1+OAcosθ1+x2+OBcosθ2+x3+OCcosθ3⇒ 2cos3i+22cos750+2cos1200⇒ 232+223−122+2−12⇒ 3+3−1−1⇒ 23−2

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A square , of each side 2, lies above the x-axis and has one vertes 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 verties of the square is

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$2 \sqrt{3} - 1$

$2 \sqrt{3} - 2$

$\sqrt{3} - 2$

$\sqrt{3} - 1$

answer is B.

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Detailed Solution

Let side of the square be r = 2

$\begin{array}{l} θ = 30^{0}, O (0, 0) \\ \bar{O A}, \bar{O B}, \bar{O C} m a k i n g a n a n g l e s \\ θ_{1} = 30^{0}, θ_{2} = 75^{0}, θ_{3} = 120^{0} w i t h X - a x i s \end{array}$

$\begin{array}{l} S u m o f X - c o r d i n a t e s = (x_{1} + O A \cos θ_{1}) + (x_{2} + O B \cos θ_{2}) + (x_{3} + O C \cos θ_{3}) \\ \Rightarrow 2 \cos 3 i + 2 \sqrt{2} \cos 75^{0} + 2 \cos 120^{0} \\ \Rightarrow 2 (\frac{\sqrt{3}}{2}) + 2 \sqrt{2} (\frac{\sqrt{3} - 1}{2 \sqrt{2}}) + 2 (\frac{- 1}{2}) \\ \Rightarrow \sqrt{3} + \sqrt{3} - 1 - 1 \Rightarrow 2 \sqrt{3} - 2 \end{array}$