What is the Cartesian form of #( 7 , ( 5pi)/3 ) #?
1 Answer
Feb 10, 2016
#(x,y) = (7/2,-{7sqrt3}/2)#
Explanation:
#(r,theta)=(7,{5pi}/3)#
#x = r cos theta#
# = 7 cos({5pi}/3)#
#= 7/2# y = r sin theta#
# = 7 sin({5pi}/3)#
#= -{7sqrt3}/2#
Therefore the Cartesian form is,
#(x,y) = (7/2,-{7sqrt3}/2)#
