Solveeit Logo
Login

Question

Question: Find the rectangular coordinate of the point \(\left( {5,300} \right).\)...

Find the rectangular coordinate of the point (5,300).\left( {5,300} \right).

Explanation

Solution

To convert Polar coordinates(r,θ)\left( {r,\theta } \right)to rectangular coordinates (x,y)\left( {x,y} \right),
we have the equation:
x=rcos⁡θ y=rsin⁡θ  x = r\cos \theta \\\ y = r\sin \theta \\\
So by using the above equation and substituting the needed values we can find the rectangular coordinate corresponding to polar coordinate (5,300).\left( {5,300} \right).

Complete step by step solution:
Given
(5,300).....................(i)\left( {5,300} \right).....................\left( i \right)
We know that rectangle coordinates are the Cartesian coordinates seen in the Cartesian plane which is represented by (x,y)\left( {x,y} \right)and polar coordinates give the position of a point in a plane by using the lengthrrand the angle made to the fixed point θ\theta , and is represented by (r,θ).\left( {r,\theta } \right).
We know that (i) which is a polar coordinate is to be converted to a rectangular coordinate.
For that we can use the formula:
x=rcos⁡θ.................(ii) y=rsin⁡θ..................(iii)  x = r\cos \theta .................\left( {ii} \right) \\\ y = r\sin \theta ..................\left( {iii} \right) \\\
So by substituting the values of r  and  θr\,\,{\text{and}}\,\,\theta in the equation (ii) and (iii) we can find the
values of x  and  y.x\,\,{\text{and}}\,\,y.
Now we know that on comparing (i) we can write:
r=5r = 5
θ=300=(2π−π3)\theta = 300 = \left( {2\pi - \dfrac{\pi }{3}} \right)
i.e. changing θ\theta from degrees to radians.
Now substituting the values of r  and  θr\,\,{\text{and}}\,\,\theta in the equation (ii) and (iii), we get:
⇒x=rcos⁡θ=5×cos⁡(2π−π3)          x=5×12  \Rightarrow x = r\cos \theta = 5 \times \cos \left( {2\pi - \dfrac{\pi }{3}} \right) \\\ \,\,\,\,\,\,\,\,\,x = 5 \times \dfrac{1}{2} \\\
⇒x=52.......................(iv)\Rightarrow x = \dfrac{5}{2}.......................\left( {iv} \right)
Now for finding y:
⇒y=rsin⁡θ            =5×sin⁡(2π−π3)            =5×−32 ⇒y=−532..................(v)  \Rightarrow y = r\sin \theta \\\ \,\,\,\,\,\,\,\,\,\,\, = 5 \times \sin \left( {2\pi - \dfrac{\pi }{3}} \right) \\\ \,\,\,\,\,\,\,\,\,\,\, = 5 \times - \dfrac{{\sqrt 3 }}{2} \\\ \Rightarrow y = - \dfrac{{5\sqrt 3 }}{2}..................\left( v \right) \\\
So from (iv) and (v) we have got the values of x=52  and  y=−532x = \dfrac{5}{2}\,\,{\text{and}}\,\,y = - \dfrac{{5\sqrt 3 }}{2}, which are our rectangular coordinates.
Therefore the rectangular coordinate of the point (5,300)\left( {5,300} \right) is (52,−532)\left( {\dfrac{5}{2}, - \dfrac{{5\sqrt 3 }}{2}} \right).

Note: We know that to convert a polar coordinate(r,θ)\left( {r,\theta } \right)to a rectangular coordinate(x,y)\left( {x,y} \right), we can use the formula:
x=rcos⁡θ y=rsin⁡θ  x = r\cos \theta \\\ y = r\sin \theta \\\
In a similar manner convert rectangular coordinate(x,y)\left( {x,y} \right)to a polar coordinate (r,θ)\left( {r,\theta } \right), we can use the formula:
r=(x2+y2) θ=tan⁡−1(yx)  r = \sqrt {\left( {{x^2} + {y^2}} \right)} \\\ \theta = {\tan ^{ - 1}}\left( {\dfrac{y}{x}} \right) \\\
Also while choosing θ\theta it’s better to choose it in radians since when θ\theta is in radians the calculations become much easier.