# CONVERTING POLAR COORDINATES TO RECTANGULAR COORDINATES

Find the rectangular coordinates of the point with polar coordinates.

Problem 1 :

(4, 4π/3)

Solution :

Polar coordinates will be in the form (r, θ)

Rectangular coordinates in the form of

(x, y) ==> (r cos θ, y sin θ)

 r cos θ = 4 cos 4π/3= 4 cos (π + π/3)Angle lies in 3rd quadrant= -4 cos (π/3)= -4(1/2)= -2 r sin θ = 4 sin 4π/3= 4 sin (π + π/3)Angle lies in 3rd quadrant= -4 sin (π/3)= -4(√3/2)= -2√3

So, the rectangular coordinate is (-2, -2√3).

Problem 2 :

(3, 315°)

Solution :

Polar coordinates will be in the form (r, θ)

Rectangular coordinates in the form of

(x, y) ==> (r cos θ, y sin θ)

 r cos θ = 3 cos 315= 3 cos (270+45)Angle lies in 4th quadrant= 3 sin (45)= 3(√2/2) = 3√2/2 r sin θ = 3 sin 315= 3 sin (270 + 45)Angle lies in 4th quadrant= -3 sin (45)= -3(√2/2)= -3√2/2

So, the rectangular coordinate is (3√2/2, -3√2/2).

Problem 3 :

(-5, -π/6)

Solution :

Polar coordinates will be in the form (r, θ)

Rectangular coordinates in the form of

(x, y) ==> (r cos θ, y sin θ)

 r cos θ = -5 cos (-π/6)= -5 cos (π/6)Angle lies in 1st quadrant= -5 cos (π/6)= -5√3/2 r sin θ = -5 sin (-π/6)= 5 sin (π/6)Angle lies in 1st quadrant= 5 (1/2)= 5/2

So, the rectangular coordinate is (-5√3/2, 5/2).

Problem 4 :

(6, π)

Solution :

Polar coordinates will be in the form (r, θ)

Rectangular coordinates in the form of

(x, y) ==> (r cos θ, y sin θ)

 r cos θ = 6 cos (π)= 6 cos π= 6(-1)= -6 r sin θ = 6 sin (π)= 6 sin π= 6 (0)= 0

So, the rectangular coordinate is (-6, 0).

Problem 5 :

(1/2, 3π/4)

Solution :

Polar coordinates will be in the form (r, θ)

Rectangular coordinates in the form of

(x, y) ==> (r cos θ, y sin θ)

 r cos θ = (1/2) cos (3π/4)= (1/2) cos (π - π/4)Angle lies in 2nd quadrant= (1/2) cos (π/4)= (1/2) (-√2/2)= -√2/4 r sin θ = (1/2) sin (3π/4)= (1/2) sin (π - π/4)Angle lies in 2nd quadrant= (1/2) sin (π/4)= (1/2) (√2/2)= √2/4

So, the rectangular coordinate is (-√2/4√2/4).

Problem 6 :

(-2, π/6)

Solution :

Polar coordinates will be in the form (r, θ)

Rectangular coordinates in the form of

(x, y) ==> (r cos θ, y sin θ)

 r cos θ = -2 cos (π/6)Angle lies in 1st quadrant= -2 (√3/2)= -√3 r sin θ = -2 sin (π/6)Angle lies in 1st quadrant= -2 (1/2)= -2

So, the rectangular coordinate is (-√3, -2).

Problem 7 :

(10, π/3)

Solution :

Polar coordinates will be in the form (r, θ)

Rectangular coordinates in the form of

(x, y) ==> (r cos θ, y sin θ)

 r cos θ = 10 cos (π/3)Angle lies in 1st quadrant= 10 (1/2)= 5 r sin θ = 10 sin (π/3)Angle lies in 1st quadrant= 10 (√3/2)= 5√3

So, the rectangular coordinate is (5, 5√3).

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