What is the polar form of #( 216,-6 )#?

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Answer 1 · 2017-12-21T15:36:47

#( 6sqrt(1296) , 0.028 )#

Using:

Cartesian #( x , y)-># Polar#-> ( r , theta )#

#r = sqrt(x^2+y^2)#

#theta=arctan(y/x)#

#r=sqrt((216)^2+(-6)^2)=sqrt(46692)=6sqrt(1296)#

#theta = arctan(-6/216)=arctan(-1/36)=0.028# ( 3 .d.p)

#:.#

#( 6sqrt(1296) , 0.028 )#