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# Find Theta

### #1

Posted 12 October 2012 - 06:42 AM

(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.

### #2

Posted 12 October 2012 - 02:28 PM

So now we need x...

### #3

Posted 12 October 2012 - 02:32 PM

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.

### #4

Posted 12 October 2012 - 02:53 PM

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.

### #5

Posted 12 October 2012 - 02:53 PM

^{2}- r

^{2)}

You can easily find h if given R and r.

### #6

Posted 12 October 2012 - 06:20 PM

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

^{2}- r^{2)}

You can easily find h if given R and r.

### #7

Posted 12 October 2012 - 07:36 PM

### #8

Posted 12 October 2012 - 11:52 PM

### #9

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.

### #10

Posted 13 October 2012 - 01:33 AM

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