An Alternative Way of Finding the Angle in a Rectangular to Polar Conversion
Welcome to a special Monday edition of Eddie’s Math and Calculator Blog.
The Traditional Method
Often we are required to find polar coordinates of a given point (x,y). Finding the radius, r, is fairly simple:
r = √(x^2 + y^2)
When talking about complex numbers, r represents the absolute value of x + yi where i = √-1.
Finding the angle, θ, often uses the formula:
θ = atan(y/x)
In complex numbers, θ represents the argument (arg) function.
On a scientific calculator the range of the arctangent function is ( -90°, 90° ). (open interval). In finding the true angle, adjustments will be required:
Let a = atan(y/x). Then:
Quadrant I (x and y are both positive): θ = a
Quadrant II (x is negative, y is positive): θ = a + 180°
Quadrant III (x and y are both negative): θ = a - 180°
Quadrant IV (x is positive, y is negative): θ = a
If you are working with radian angle measures, know that 90° = π/2, and 180° = π.
This does not take into consideration situations where either x or y is 0:
If x > 0 and y = 0: θ = 0°
If x = 0 and y > 0: θ = 90°
If x < 0 and y = 0: θ = 180°
If x = 0 and y < 0: θ = -90°
Is there a shorter way to calculate θ?
The Vector Method
Consider the point (x, y) as a vector [x, y]. Now draw another vector [x, 0]. In a regular Cartesian coordinate system, angles are measured from the x-axis counter clockwise.
Let a and b represent two vectors. Then the angle between two vectors are:
cos θ = (a ● b) / ( ||a|| ||b|| ) = dot(a,b) / ( norm(a) * norm(b) )
with:
dot(a,b) = a1 * b1 + a2 * b2
norm(a) = √(a1^2 + a2^2)
norm(b) = √(b1^2 + b2^2)
Let a = [x, y] and b = [x, 0]. Then:
cos θ = (x^2) / (√(x^2 + y^2) * √(x^2))
cos θ = (x^2) / (√(x^2 + y^2) * x)
cos θ = x / √(x^2 + y^2)
θ = acos( x / √(x^2 + y^2) )
The range of the arccosine function of a calculator is [ 0°, 180° ].
If y < 0, the angle would be measured clockwise, and therefore I would make the adjustment:
θ = -acos( x / √(x^2 + y^2) )
In summary:
If y ≥ 0, θ = acos( x / √(x^2 + y^2) )
If y < 0, then θ = -acos( x / √(x^2 + y^2) )
Examples:
Find the angle, in degrees, in a rectangular to polar conversions:
Quadrant I (2, 4): y ≥ 0: θ = acos( 2 / √(2^2 + 4^2) ) ≈ 63.43494882°
Quadrant II (-2, 4): y ≥ 0: θ = acos( -2 / √((-2)^2 + 4^2) ) ≈ 116.5650512°
Quadrant II (-2, -4): y < 0: θ = -acos( (-2) / √((-2)^2 + (-4)^2) ) ≈ -116.5650512°
Quadrant IV (2, -4): y < 0: θ = -acos( 2) / √(2^2 + (-4)^2) ) ≈ -63.43494882°
Vector Method for Navigation
In navigation, angles start from true North (up) and rotate clockwise towards East (right). Angles are measured from 0° to 360°.
Use the vectors [0, N] and [E, N], then the angle between these vectors are:
If E ≥ 0, θ = acos( N / √(E^2 + N^2))
If E < 0, θ = 360° - acos( N / √(E^2 + N^2))
Note: This is the first blog entry that I have typed on Google Docs. I have been using Windows app Wordpad for the last three years. I am testing Google apps as I am considering buying a Chromebook.
Eddie
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