How can I find the length of a bisected shape with trigonometry?

[BicycleTree]

http://img62.imageshack.us/my.php?image=geometry9cn.png

 

[AKG]

\sqrt{2.3925} - 1.35 \approx 0.197

 

[BicycleTree]

Yes, that appears to be the correct answer. I am interested how you got it. The expression I got was
sin((60-2*(90-arccos(.45)))/2) * 3^.5 * 2

 

[AKG]

2(sin(120)/sin(30))sin((60 - Y)/2)
= 2(sin(120)/sin(30))(sin(30)cos(Y/2) - cos(30)sin(Y/2))
= 2(3^0.5)(0.5(1 - 0.45²)^0.5 - 0.5(3^0.5)(0.45))
= (3(1 - 0.45²))^0.5 - 1.35
= (2.3925)^0.5 - 1.35

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.

 

[NateTG]

Hmm, if you bisect the figure, then you have an isoscoles trianlge at the top so the interior angle of the top triangle is
2\sin^{-1}(.45)
Moreover, it's not hard to see that the length from the center intersection to the edge of the red segment has length
\sqrt{3}

That means that you can go with:
2\left(\sqrt{3} \sin\left(30-\sin^{-1}(.45)\right)\right)

Which can easily be simplified using angle addition forumlas.