What is the distance between $(4, \frac{5 π}{6})$ $(4 ,( 5 pi)/6 )$ and $(1, \frac{- 3 π}{2})$ $(1 , ( -3 pi )/2 )$ ?

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What is the distance between $(4, \frac{5 π}{6})$ $(4 ,( 5 pi)/6 )$ and $(1, \frac{- 3 π}{2})$ $(1 , ( -3 pi )/2 )$ ?

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Answer 1 · 2016-01-06T19:25:40

$3.605$ $3.605$ units

Explanation:

The distance formula for polar coordinates is

$d = \sqrt{r_{1}^{2} + r_{2}^{2} - 2 r_{1} r_{2} cos (θ_{1} - θ_{2})}$ $d=sqrt(r_1^2+r_2^2-2r_1r_2Cos(theta_1-theta_2)$
Where $d$ $d$ is the distance between the two points, $r_{1}$ $r_1$ , and $θ_{1}$ $theta_1$ are the polar coordinates of one point and $r_{2}$ $r_2$ and $θ_{2}$ $theta_2$ are the polar coordinates of another point.
Let $(r_{1}, θ_{1})$ $(r_1,theta_1)$ represent $(4, \frac{5 π}{6})$ $(4,(5pi)/6)$ and $(r_{2}, θ_{2})$ $(r_2,theta_2)$ represent $(1, \frac{- 3 π}{2})$ $(1,(-3pi)/2)$ .
$\Rightarrow d = \sqrt{4^{2} + 1^{2} - 2 \cdot 4 \cdot 1 cos (\frac{5 π}{6} - \frac{- 3 π}{2})}$ $implies d=sqrt(4^2+1^2-2*4*1Cos((5pi)/6-(-3pi)/2)$
$\Rightarrow d = \sqrt{16 + 1 - 8 cos (\frac{5 π}{6} + \frac{3 π}{2})}$ $implies d=sqrt(16+1-8Cos((5pi)/6+(3pi)/2)$

$\Rightarrow d = \sqrt{17 - 8 cos (\frac{14 π}{6})}$ $implies d=sqrt(17-8Cos((14pi)/6)$
$\Rightarrow d = \sqrt{17 - 8 \cdot 0.5} = \sqrt{17 - 4} = \sqrt{13} = 3.605$ $implies d=sqrt(17-8*0.5)=sqrt(17-4)=sqrt(13)=3.605$ units
$\Rightarrow d = 3.605$ $implies d=3.605$ units (approx)
Hence the distance between the given points is $3.605$ $3.605$ units.

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What is the distance between $(4, \frac{5 π}{6})$ $(4 ,( 5 pi)/6 )$ and $(1, \frac{- 3 π}{2})$ $(1 , ( -3 pi )/2 )$ ?

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