How do you convert #(8.3, 4.2)# into polar forms?

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Answer 1 · 2016-01-29T13:46:08

#(9.302,0.149pi)#

suppose,
#x=8.3#
#y=4.2#
now, we know,
for polar form,
#r^2=x^2+y^2#

#or,r=root(2)(x^2+y^2)#

now, by putting the value of x and y in the equation, we get

#r=root(2)(8.3^2+4.2^2#

#=root(2)(86.53)#

#=9.302#

again
we know,
#theta=tan^(-1)(y/x)#

by putting the value of x and y on the above equation, we get,

#theta=tan^(-1)(4.2/8.3)#

#=0.149pi#

so, the polar form is ( #9.302,0.149pi#)