(-2,6) is placed in the Cartesian coordinate system. convert to polar coordiantes in degrees
r^2 = x^2+y^2 = 40
r = √40
tanθ = y/x = 6/-2 = -3
Note that the point is in QII, so θ = π-arctan(3) ≈ 1.892
So, (-2,6) = (2√10, 1.892)
To convert coordinates from Cartesian to polar form, we can use the following equations:
r = √(x^2 + y^2)
θ = arctan(y/x)
Given the point (-2,6), we can substitute the values into these equations to find the polar coordinates.
First, let's find the value of r:
r = √((-2)^2 + 6^2)
r = √(4 + 36)
r = √40
r = 2√10
Next, let's find the value of θ:
θ = arctan(6/-2)
θ = arctan(-3)
θ ≈ -56.31 degrees
Therefore, the polar coordinates of the point (-2,6) in degrees are approximately (2√10, -56.31°).
To convert a point from Cartesian coordinates to polar coordinates, we can use the following formulas:
r = sqrt(x^2 + y^2)
θ = arctan(y / x)
where (x, y) is the Cartesian coordinate and (r, θ) is the polar coordinate in degrees.
Given the point (-2, 6) in Cartesian coordinates, we can plug the values into the formulas to convert it to polar coordinates:
r = sqrt((-2)^2 + 6^2) = sqrt(4 + 36) = sqrt(40) = 2√10
θ = arctan(6 / -2) = arctan(-3) ≈ -71.57°
Now we have the polar coordinates in terms of r and θ. But if you prefer them in standard form, we can write it as:
(2√10, -71.57°)
So, the polar coordinates of the point (-2, 6) in degrees are approximately (2√10, -71.57°).