What is the polar form of #( 36,48 )#?

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Answer 1 · 2016-09-10T09:31:18

Polar form of #(36,48)# is #(60,53.13^o)#

Polar coordinates #(r,theta)# are related to Cartesian coordinates #(x,y)# by the following:

#x=rcostheta#, #y=rsintheta# and #r^2=x^2+y^2#

Hence, let polar coordinates of #(36,48)# be #(r,theta)#

Hence #r=sqrt(36^2+48^2)=sqrt(1296+2304)=sqrt3600=60#

#costheta=36/60=0.60# and #sintheta=48/60=0.80#

and from tables #theta=53.13^o#

Hence, polar form of #(36,48)# is #(60,53.13^o)#