Question

HELP PLEASE! Can you please explain so I can understand how it was completed?

Answer 1

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:

• The radial coordinate which is often denoted by r.

• The angular coordinate by θ.

So we need to find r and θ. So we know that:

$r=\sqrt{x^{2}+y^{2}}$       (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:

$r=\sqrt{x^{2}+y^{2}}=\sqrt{3^{2}+(-3)^{2}} = 3\sqrt{2}$

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°)