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- Thread starter BicycleTree
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AKG

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[itex]\sqrt{2.3925} - 1.35 \approx 0.197[/itex]

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sin((60-2*(90-arccos(.45)))/2) * 3^.5 * 2

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AKG

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= 2(sin(120)/sin(30))(sin(30)cos(Y/2) - cos(30)sin(Y/2))

= 2(3

= (3(1 - 0.45²))

= (2.3925)

Y is the small angle in the triangle the contains the blue line (the bottom angle). I didn't want to use any inverse trigonometric functions, so I used Pythogras' Theorem, Sine Law, and the angle addition formula for sine.

- #5

NateTG

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[tex]2\sin^{-1}(.45)[/tex]

Moreover, it's not hard to see that the lenght from the center intersection to the edge of the red segment has length

[tex]\sqrt{3}[/tex]

That means that you can go with:

[tex]2\left(\sqrt{3} \sin\left(30-\sin^{-1}(.45)\right)\right)[/tex]

Which can easily be simplified using angle addition forumlas.

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