What is the distance between #(3 , (3 pi)/4 )# and #(9, pi )#?

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Answer 1 · 2016-03-14T06:20:37

#14.739#

#(r,theta)# in polar coordinates is #(rcostheta,rsintheta)# in rectangular coordinates.

Hence #(3,(3pi)/4)# is #(3cos((3pi)/4),3sin((3pi)/4))# or #(3*(-1)/sqrt2,3*(-1)/sqrt2)# or #((-3sqrt2)/2,-(3sqrt2)/2)#

and #9.pi)# is #(9cospi,9sinpi)# or #(9xx1,9xx0)# or #(9,0)#

The distance between #(9,0)# and #((-3sqrt2)/2,-(3sqrt2)/2)# is

#sqrt((9-((-3sqrt2)/2))^2+(0-((-3sqrt2)/2))^2)# or

#sqrt((9+(3sqrt2)/2)^2+((3sqrt2)/2)^2)# or

#sqrt(81+27sqrt2+9/2+9/2)# or

#sqrt(90+27sqrt2)# or #3sqrt(10+3sqrt2)# or

= #3sqrt(10+14.142)=3sqrt24.142=3xx4.913=14.739#