What is the Cartesian form of #( 12 , (23pi)/3 ) #?

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Answer 1 · 2016-12-27T20:08:39

#(6,-10.4)#

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in polar forms, the first coordinate is always the hypotenuse.

#r=12#

#theta=(23pi)/3#

#x# and #y# can be solved as the adjacent and opposite, respectively.

using #sin# and #cos# ratios:

#sin (23pi)/3 = O /H = y/12#

#y = 12 * sin((23pi)/3)#

#= -10.4 (3s.f.)#

#cos (23pi)/3 = A/H = x/12#

#x = 12 * cos(23pi)/3#

#= 6#

cartesian coordinates: #(6, -10.4)#