What is the Cartesian form of #(-23,(5pi)/16))#?

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Answer 1 · 2017-04-08T20:33:26

(x,y)
(#23sqrt3/2#, #-23/2# )

Polar and cartesian points are related through:

#x=rcos(theta)#
#y=rsintheta#

#x=rcos(theta)#
#x=-23(cos((5pi)/6))#
# cos ((5pi)/6)= -sqrt(3)/2#

#-sqrt 3/2(-23)= 23sqrt3/2#
#x=23sqrt3/2#

#y=rsintheta#
#x=23sqrt3/2#

#sin((5pi)/6)=1/2#

#y=-23/2#

(x,y)
(#23sqrt3/2#, #-23/2# )