# CONVERTING POLAR COORDINATES TO RECTANGULAR COORDINATES

Let P be the point have the polar coordinates (r,  θ) and its rectangular coordinates will be (x, y).

Then,

x  =  r cosθ and y  =  r sin θ

r2  =  x2 + y2

tan θ  =  y/x

Example 1 :

Convert the given polar coordinates to rectangular coordinates.

(a)  (-4, 2π/3)    (b)  (√3, π/6)

Solution :

From the point (-4, 2π/3), r is -4 and θ is 2π/3

 x  =  r cos θx  =  -4 cos 2π/3x  =  -4(-1/2)x  =  2 y  =  r sin θy  =  -4 sin 2π/3y  =  -4(√3/2)y  =  -2√3

So, the required rectangular co ordinate is (2, -2√3).

(b)  (√3, π/6)

From the point (√3, π/6), r is √3 and θ is π/6

 x  =  r cos θx  =  √3 cos π/6x  =  √3 (√3/2)x  =  3/2 y  =  r sin θy  =  √3 sin π/6y  =  √3 (1/2)y  =  √3/2

So, the required rectangular co ordinate is (3/2√3/2).

Example 2 :

Convert the given polar coordinates to rectangular coordinates.

(a)  (2, π/4)       (b)  (-3, 5π/6)

(c)  (5, 10π/3)    (d)  (47, 17π/2)

Solution :

(a)  (2, π/4)

From the point (2, π/4), r is 2 and θ is π/4

 x  =  r cos θx  =  2 cos π/4x  =  2(1/√2)x  =  2/√2 y  =  r sin θy  =  2 sin π/4y  =  2(1/√2)y  =  2/√2

So, the required rectangular co ordinate is (2/2, 2/2).

(b)  (-3, 5π/6)

From the point (-3, 5π/6), r is -3 and θ is 5π/6

 x  =  r cos θx  =  -3 cos 5π/6x  =  -3(√3/2)x  =  -3√3/2 y  =  r sin θy  =  -3 cos 5π/6y  =  -3(1/2)y  =  -3/2

So, the required rectangular co ordinate is

(-3√3/2, -3/2).

(c)  (5, 10π/3)

From the point (5, 10π/3), r is 5 and θ is 10π/3

 x  =  r cos θx  =  5 cos 10π/3x  =  5(-√3/2)x  =  -5√3/2 y  =  r sin θy  =  5 cos 10π/3y  =  5(-1/2)y  =  -5/2

So, the required rectangular co ordinate is

(-5√3/2, -5/2).

(d)  (47, 17π/2)

From the point (47, 17π/2), r is 47 and θ is 17π/2

 x  =  r cos θx  =  47 cos 17π/2x  =  47(0)x  =  0 y  =  r sin θx  =  47 sin 17π/2x  =  47(1)x  =  47

So, the required rectangular co ordinate is (0, 47). Apart from the stuff given above, if you need any other stuff in math, please use our google custom search here.

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