What is the distance between #(-6 , (5 pi)/8 )# and #(3 , (5 pi )/3 )#?

1 Answer
Jim G.
Jan 13, 2016

≈3.1 ( 1 decimal place )

Explanation:

Use the Polar coordinate form of the distance formula which is :

#d^2 = r_1^2 + r_2^2 - (2 xxr_1 xx r_2 xx cos (theta_2 - theta_1 )) #

In this question let # (r_1 , theta_1 ) = (-6 , (5pi)/8 ) # and

#( r_2 , theta_ 2) = (3 , (5pi)/3 ) #

substitute values into formula :

# d^2 = (- 6 )^2 + 3^2 - ( 2 xx(-6) xx 3 xx cos ((5pi)/3 -(5pi)/8) #

#rArr d^2 = 36 + 9 - ( - 36 xx cos((25pi)/24) = 45 - ( 35.69)#

#d^2 = 9.31 rArr d = sqrt9.31 ≈3.1 #

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