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# Find Theta

## 10 posts in this topic

Posted · Report post

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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Posted · Report post

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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Posted · Report post

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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Posted · Report post

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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Posted · Report post

h = R - sqrt(R2 - r2)

You can easily find h if given R and r.

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Posted · Report post

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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Posted · Report post

><

From center to where r meets the circle is R making a triangle of r, d and R. This gives d = sqrt(R^2 - r^2) using a^2 + b^2 = c^2.

From center to the top is R. This is split by h and d. So R = h + d

Substitute d and you get R = h + sqrt(R^2 - r^2) or h = R - sqrt(R^2 - r^2).

I had just solved for x instead of h in my other post.

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Posted · Report post

Wow, I can't believe I overlooked that. Thanks guys.

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Posted · Report post

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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Posted · Report post

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

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