What is the distance between #(2 ,(5 pi)/4 )# and #(8 , (11 pi )/12 )#?

1 Answer
Shwetank Mauria
Mar 13, 2016

#7.211#

Explanation:

#(2,(5pi)/4)# in rectangular coordinates would be #(2cos((5pi)/4), 2sin((5pi)/4))#. As #sin((5pi)/4)=-sqrt2/2# and #cos((5pi)/4)=-sqrt2/2#

This is #(-sqrt2,-sqrt2)# or #(-1.4142,-1.4142)#.

#(8,(11pi)/12)# in rectangular coordinates would be #(8cos((11pi)/12), 8sin((11pi)/12))#. As #sin((11pi)/12)=0.2588# and #cos((5pi)/4)=-0.9659#

This is #(-7.7272,2.0704)#.

Hence distance between two is #sqrt{((-7.7272-(-1.4142))^2+((2.0704-(-1.4142))^2#

= #sqrt{(6.313)^2+(3.4846)^2}=sqrt51.9964=7.211#

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