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# Ratio of Areas watch

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1. Using the following diagram and various trig identities, I have worked out that Area = 2r^2 [1 - sin(x)]

But I don't know how to find sin(x) in terms of r or how to do it any other way. Any ideas?
2. i think you'll find it's actually cos(x)...
3. Length of side of square = 2r sin(x/2)

Area = length squared
=4 r^2 sin^2(x/2)

sinx = 1 - 2 sin^2(x/2)
=>2sin^2(x/2) = 1 - sin(x)
=>A = 2r^2 [1 - sin(x)]
4. o_0 one moment

yeah, r=5z/8

r2 = z2/4 +(z-r)2

this is from one of the little triangles, across horizontally from the centre to the edge of the square (z/2) , up, then down the radius
5. (Original post by e-unit)
sinx = 1 - 2 sin^2(x/2)
no, cosx = 1 - 2 sin^2(x/2)
6. (Original post by chewwy)
no, cosx = 1 - 2 sin^2(x/2)
SUGAR. I'm having a mid-life mathematical crisis.

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Updated: February 2, 2006
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