How do you find the polar coordinate of #(2sqrt3, 2)#?

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Answer 1 · 2018-05-05T18:46:16

#(4, (pi)/6)# (radians) or #(4, 60^@)# (degrees)

Rectangular #-># Polar: #(x, y) -> (r, theta)#

Find #r# (radius) using #r = sqrt(x^2 + y^2)#
Find #theta# by finding the reference angle: #tantheta = y/x# and use this to find the angle in the correct quadrant

#r = sqrt((2sqrt3)^2 + (2)^2)#

#r = sqrt((4*3)+4)#

#r = sqrt(12+4)#

#r = sqrt(16)#

#r = 4#

Now we find the value of #theta# using #tantheta = y/x#.

#tantheta = 2/(2sqrt3)#

#tantheta = 1/sqrt3#

#theta = tan^-1(1/sqrt3)#

#theta = (pi)/6# or #(7pi)/6#

To determine which one it is, we have to look at our coordinate #(2sqrt3, 2)#. First, let's graph it:
enter image source here

As you can see, it is in the first quadrant. Our #theta# has to match that quadrant, meaning that #theta = (pi)/6#.

From #r# and #theta#, we can write our polar coordinate:
#(4, (pi)/6)# (radians) or #(4, 60^@)# (degrees)

Hope this helps!