MATH SOLVE

4 months ago

Q:
# HELP PLEASE! Can you please explain so I can understand how it was completed?Determine two pairs of polar coordinates for the point (3, -3) with 0°≤ θ < 360°. (1 point)(3 square root of 2 , 315°), (-3 square root of 2 , 135°)(3 square root of 2 , 225°), (-3 square root of 2 , 45°)(3 square root of 2 , 45°), (-3 square root of 2 , 225°)(3 square root of 2 , 135°), (-3 square root of 2 , 315°)

Accepted Solution

A:

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:[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= -45Then: r = 3\sqrt{2}[/tex]θ= -45 or 315Notice 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°)