What is the distance between $(2, \frac{5 π}{6})$ $(2 ,(5 pi)/6 )$ and $(4, \frac{23 π}{12})$ $(4 , (23 pi )/12 )$ ?

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Answer 1 · 2016-09-06T14:22:19

$\sqrt{5 + \sqrt{2} (\sqrt{3} + 1)}$ $sqrt(5+sqrt 2(sqrt 3 +1))$

If O is the pole, A is $(2, \frac{5}{6} π)$ $(2, 5/6pi)$ and B is $(4, \frac{33}{12} π)$ $(4, 33/12pi)$ , then AB is

$\sqrt{O A^{2} + O B^{2} - 2 X O A X O B X cos ∠ A O B}$ $sqrt(OA^2+OB^2-2XOAXOBXcosangleAOB)$

$= \sqrt{2^{2} + 4^{2} - (2) (2) (4) cos (\frac{23}{12} π - \frac{5}{6} π)}$ $=sqrt(2^2+4^2-(2)(2)(4)cos(23/12pi-5/6pi)$

$= \sqrt{20 - 16 cos (\frac{13}{12} π)}$ $=sqrt(20-16 cos(13/12pi))$

$= 2 \sqrt{5 - 4 cos s (π + \frac{π}{12})}$ $=2sqrt (5-4coss(pi+pi/12)$

$= 2 \sqrt{5 + 4 {cos 15}^{o}}$ $=2 sqrt(5+4cos 15^o)$

Here, ${cos 15}^{o} = cos (45^{o} - 30^{o})$ $cos 15^o=cos(45^o-30^o)$

$= {cos 45}^{o} {cos 30}^{o} + {sin 45}^{o} {sin 30}^{o}$ $=cos 45^o cos 30^o + sin 45^o sin 30^o$

$= \frac{\sqrt{3} + 1}{2 \sqrt{2}}$ $=(sqrt 3+1)/(2sqrt 2)$ .

So, $A B = \sqrt{5 + 4 (\frac{\sqrt{3} + 1}{2 \sqrt{2}})}$ $AB = sqrt(5+4((sqrt3+1)/(2sqrt2))$

$= \sqrt{5 + \sqrt{2} (\sqrt{3} + 1)}$ $=sqrt(5+sqrt 2(sqrt 3 +1))$

What is the distance between (2,5π6)(2 ,(5 pi)/6 ) and (4,23π12)(4 , (23 pi )/12 )?