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Question: How do you convert the polar point \( (6,{120^ \circ }) \) into rectangular form ?...

How do you convert the polar point (6,120∘)(6,{120^ \circ }) into rectangular form ?

Explanation

Solution

Hint : In order to solve this question , we need to first understand the mathematical terms what are polar coordinates and what are rectangular coordinates and what is their significance . If we talk about the polar coordinates , then Polar Coordinates (p,θ)(p,\theta ) is actually a 2D coordinate system in which every point on the plane is found by a distance pp from a reference point and an angle i . e . θ\theta from a reference direction where pp is the radial coordinate and θ\theta is known as the angular coordinate.
If we take our question then the given polar coordinate is (6,120∘)(6,{120^ \circ }) .
Radial coordinate = p / r=6p\,/\,r = 6
Angular coordinate =θ=120∘= \theta = {120^ \circ }

Complete step-by-step answer :
So , Lets convert the polar coordinate into rectangular coordinates .
Now to transformation by which we can find our rectangular coordinates (x,y)\left( {x,y} \right) is
x=rcos⁡θ y=rsin⁡θ   x = r\cos \theta \\\ y = r\sin \theta \;
In our case r=6 and θ=120∘=2π3r = 6\,and\,\theta = {120^ \circ } = \dfrac{{2\pi }}{3}
x=6cos⁡(2π3) =6cos⁡(π−π3)   x = 6\cos \left( {\dfrac{{2\pi }}{3}} \right) \\\ = 6\cos \left( {\pi - \dfrac{\pi }{3}} \right) \;
Using Allied angle in trigonometry cos⁡(π−θ)=−cos⁡θ\cos \left( {\pi - \theta } \right) = - \cos \theta
=−6cos⁡(π3) =−6(12) =−3   = - 6\cos \left( {\dfrac{\pi }{3}} \right) \\\ = - 6\left( {\dfrac{1}{2}} \right) \\\ = - 3 \; using trigonometric value of cos⁡(π3)=12\cos \left( {\dfrac{\pi }{3}} \right) = \dfrac{1}{2}
y=6sin⁡(2π3) =6sin⁡(π−π3)   y = 6\sin \left( {\dfrac{{2\pi }}{3}} \right) \\\ = 6\sin \left( {\pi - \dfrac{\pi }{3}} \right) \;
Using Allied angle in trigonometry sin⁡(π−θ)=sin⁡θ\sin \left( {\pi - \theta } \right) = \sin \theta
=6sin⁡(π3) =6(32) =33   = 6\sin \left( {\dfrac{\pi }{3}} \right) \\\ = 6\left( {\dfrac{{\sqrt 3 }}{2}} \right) \\\ = 3\sqrt 3 \; using trigonometric value of sin⁡(π4)=12\sin \left( {\dfrac{\pi }{4}} \right) = \dfrac{1}{{\sqrt 2 }}
Therefore, polar coordinates (6,120∘)(6,{120^ \circ }) in rectangular coordinates are (−3,33)\left( { - 3,3\sqrt 3 } \right) .
So, the correct answer is “(−3,33)\left( { - 3,3\sqrt 3 } \right) ”.

Note : 1.One must be careful while taking values from the trigonometric table and cross-check at least once
2.To verify your answer, plot both polar and rectangular coordinates on the cartesian plane and if both are the same then the answer is correct and if they both do not represent the same point then your transformation is wrong ,check your solution.
We use the following formulaTrigonometric Form or Polar form
sin⁡(π+θ)=−sin⁡θ\sin \left( {\pi + \theta } \right) = - \sin \theta
cos⁡(π+θ)=−cos⁡θ\cos \left( {\pi + \theta } \right) = - \cos \theta