Answer:
(3 square root of 2 , 135°), (-3 square root of 2 , 315°)
Step-by-step explanation:
Hello!
We need to determine two pairs of polar coordinates for the point (3, -3) with 0°≤ θ < 360°.
We know that the polar coordinate system is a two-dimensional coordinate. The two dimensions are:
So we need to find r and θ. So we know that:
[tex]r=\sqrt{x^{2}+y^{2}}[/tex] (1)
x = rcos(θ) (2)
x = rsin(θ) (3)
From the statement we know that (x, y) = (3, -3).
Using the equation (1) we find that:
[tex]r=\sqrt{x^{2}+y^{2}}=\sqrt{3^{2}+(-3)^{2}} = 3\sqrt{2}[/tex]
Using the equations (2) and (3) we find that:
3 = rcos(θ)
-3 = rsin(θ)
Solving the system of equations:
θ= -45
Then:
r = 3\sqrt{2}[/tex]
θ= -45 or 315
Notice that there are two feasible angles, they both have a tangent of -1. The X will take the positive value, and Y the negative one.
So, the solution is:
(3 square root of 2 , 135°), (-3 square root of 2 , 315°)
