What is the distance between #(3 ,(5 pi)/6 )# and #(-2 , (5 pi )/12 )#?

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Answer 1 · 2017-11-09T17:08:20

#208.028#

( 3 .d.p. )

First you need to convert the polar coordinates into Cartesian coordinates. This can be done using the following:

#x= rcos(theta)#

#y= rsin(theta)#

#:.#

#x= 3cos((5pi)/6)=-2.598# ( 3 .d.p.)

#y=3sin((5pi)/6)=1.500#

#color(blue)((-2.598 , 1.500)#

#x=-2cos((5pi)/12)=-0.518#

#=-2sin((5pi)/12)=-1.932#

#color(blue)(( -0.518 , -1.932 )#

Using the distance formula:

#d = sqrt((x_2-x_1)^2 + (y_2-y_1)^2#

#d= sqrt((-2.598-(-0.518))^2+(1.500-(-1.932))^2#

#d= sqrt((-208)^2+(3.432)^2)=208.0283121=208.028#

( 3 .d.p. )