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Find Theta

Started by Izzy, Oct 12 2012 06:42 AM

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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.