What is the Cartesian form of #(-24,(17pi)/6))#?

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Answer 1 · 2016-01-17T09:14:47

Cartesian Form #(12 sqrt(3), -12)#

Given Polar Form #(r, theta)#=#(-24, 17 pi/6)#
Where #r=-24# and #theta=(17 pi)/6#
#x=r *cos theta#
#x=-24*cos ((17 pi)/6)#

#x=-24*(-sqrt(3)/2)#

#x=12 sqrt(3)#

#y=r*sin theta#

#y=-24 *sin((17 pi)/6)#

#y= -24*(1/2)#
#y=-12#

therefore
#(x, y)=(12 sqrt(3), -12)#