What is the distance between $(3, \frac{3 π}{8})$ $(3 , (3 pi)/8 )$ and $(9, π)$ $(9, pi )$ ?

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What is the distance between $(3, \frac{3 π}{8})$ $(3 , (3 pi)/8 )$ and $(9, π)$ $(9, pi )$ ?

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Answer 1 · 2016-03-28T06:53:35

$\sqrt{90 - 54 cos (5 \frac{π}{8})}$ $sqrt(90-54cos(5pi/8))$ = $\sqrt{110.665} = 10.52$ $sqrt110.665=10.52$ nearly.

Explanation:

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

Answer 2 · 2016-03-28T07:06:46

$\sqrt{90 + 54 cos (\frac{3 π}{8})} \approx 10.520$ $sqrt{90 + 54cos({3pi}/8)} ~~ 10.520$

Explanation:

To convert the polar coordinates to Cartesian coordinates, we use

$x = r cos (θ)$ $x = r cos(theta)$
$y = r sin (θ)$ $y = r sin(theta)$

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

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

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

$d = \sqrt{{(- 9 - 3 cos (\frac{3 π}{8}))}^{2} + {(0 - 3 sin (\frac{3 π}{8}))}^{2}}$ $d = sqrt{(-9 - 3cos({3pi}/8))^2 + (0 - 3sin({3pi}/8))^2}$

$= \sqrt{(81 + 9 {cos}^{2} (\frac{3 π}{8}) + 54 cos (\frac{3 π}{8})) + 9 {sin}^{2} (\frac{3 π}{8})}$ $= sqrt{(81 + 9cos^2({3pi}/8) + 54cos({3pi}/8)) + 9sin^2({3pi}/8)}$

$= \sqrt{90 + 54 cos (\frac{3 π}{8})}$ $= sqrt{90 + 54cos({3pi}/8)}$

$= 3 \sqrt{3 \sqrt{2 - \sqrt{2}} + 10}$ $= 3sqrt{3sqrt{2-sqrt2}+10}$

$\approx 10.520$ $~~ 10.520$

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What is the distance between $(3, \frac{3 π}{8})$ $(3 , (3 pi)/8 )$ and $(9, π)$ $(9, pi )$ ?

2 Answers

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