What is the distance between #(3 , (3 pi)/8 )# and #(9, pi )#?

2 Answers
A. S. Adikesavan
Mar 28, 2016

#sqrt(90-54cos(5pi/8))# = #sqrt110.665=10.52# nearly.

Explanation:

The position vectors to the points are of lengths a = 3 and b = 9. The angle in-between is C = #5pi/8#.
Use the formula c = #sqrt(a^2+b^2-2ab cos C)#

Bio
Mar 28, 2016

#sqrt{90 + 54cos({3pi}/8)} ~~ 10.520#

Explanation:

To convert the polar coordinates to Cartesian coordinates, we use

#x = r cos(theta)#
#y = r sin(theta)#

The cartesian coordinate of #(3,{3pi}/8)# is #(3/2sqrt(2-sqrt2),3/2sqrt(2+sqrt2))#. Use the half angle formula to get the values.

The cartesian coordinate of #(9,pi)# is #(-9,0)#.

We can use the Pythagoras Theorem to find the distance between the 2 points

#d = sqrt{(-9 - 3cos({3pi}/8))^2 + (0 - 3sin({3pi}/8))^2}#

#= sqrt{(81 + 9cos^2({3pi}/8) + 54cos({3pi}/8)) + 9sin^2({3pi}/8)}#

#= sqrt{90 + 54cos({3pi}/8)}#

#= 3sqrt{3sqrt{2-sqrt2}+10}#

#~~ 10.520#

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