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Find Theta


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9 replies to this topic

#1 Izzy

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Posted 12 October 2012 - 06:42 AM

Well, this isn't really homework. What I'm actually trying to do is find h, and I have a way to do that if I know theta. So, I'd appreciate help with finding theta, but if you can do h, that works too.

(This is actually a really simplified version of the problem. In reality, the circle is a sphere, and h is the distance of a line that intersects perpendicularly with the center of a small circle within the sphere, whose center lines up with a point on the surface of the sphere. Didn't know how to draw that, so help me find h or theta pl0x?)

All I've managed to establish is that the angle to the left of theta will be 135, regardless of h.


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#2 curr3nt

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Posted 12 October 2012 - 02:28 PM

Well likely doesn't help much but per your picture h = r - x. Since it's a 45-45-90 triangle those two sides are the same length.

So now we need x...
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#3 phil1882

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Posted 12 October 2012 - 02:32 PM

assuming R is the radius, h = r-x by similar triangles. further, theta = tan-1((r-x)/r).
so we just need x; using the chord property: two crossing chords AB, CD intersecting at point E gives the relationship
AE*EB = CE*ED; we have...
x*(2*r-x) =sqrt(2)*(r-x)*(sqrt(2)*R -sqrt(2)*(r-x))

x*(2*r-x) =(r-x)*(2*R -2*(r-x)) which should be solvable if you know R and r.
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#4 curr3nt

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Posted 12 October 2012 - 02:53 PM

I think I have this right... x = r - R + sqrt(R^2 - r^2)

Lets call the distance from R to r, d.

d = sqrt(R^2 - r^2)

And we know R = d + r - x we can solve for x.
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#5 Prof. Templeton

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Posted 12 October 2012 - 02:53 PM

h = R - sqrt(R2 - r2)

You can easily find h if given R and r.
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#6 Izzy

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Posted 12 October 2012 - 06:20 PM

h = R - sqrt(R2 - r2)

You can easily find h if given R and r.

Since I don't actually know x, this response has been the most useful. Thanks Prof. Can you explain it, though?
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#7 curr3nt

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Posted 12 October 2012 - 07:36 PM

><

Spoiler for I will try again...

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#8 Izzy

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Posted 12 October 2012 - 11:52 PM

Wow, I can't believe I overlooked that. Thanks guys. :)
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#9 Prof. Templeton

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Posted 13 October 2012 - 12:09 AM

Since I don't actually know x, this response has been the most useful. Thanks Prof. Can you explain it, though?


curr3nt explained it. It's the Pythagorean theorem between R, r and R-h.
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#10 curr3nt

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Posted 13 October 2012 - 01:33 AM

yes but I didn't explain it well the first time... or really at all hence the ><
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