Izzy Posted October 12, 2012 Report Share Posted October 12, 2012 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. 1 Quote Link to comment Share on other sites More sharing options...
curr3nt Posted October 12, 2012 Report Share Posted October 12, 2012 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... Quote Link to comment Share on other sites More sharing options...
phil1882 Posted October 12, 2012 Report Share Posted October 12, 2012 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. Quote Link to comment Share on other sites More sharing options...
curr3nt Posted October 12, 2012 Report Share Posted October 12, 2012 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. Quote Link to comment Share on other sites More sharing options...
Prof. Templeton Posted October 12, 2012 Report Share Posted October 12, 2012 h = R - sqrt(R2 - r2) You can easily find h if given R and r. 1 Quote Link to comment Share on other sites More sharing options...
Izzy Posted October 12, 2012 Author Report Share Posted October 12, 2012 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? Quote Link to comment Share on other sites More sharing options...
curr3nt Posted October 12, 2012 Report Share Posted October 12, 2012 >< 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. 1 Quote Link to comment Share on other sites More sharing options...
Izzy Posted October 12, 2012 Author Report Share Posted October 12, 2012 Wow, I can't believe I overlooked that. Thanks guys. Quote Link to comment Share on other sites More sharing options...
Prof. Templeton Posted October 12, 2012 Report Share Posted October 12, 2012 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. Quote Link to comment Share on other sites More sharing options...
curr3nt Posted October 13, 2012 Report Share Posted October 13, 2012 yes but I didn't explain it well the first time... or really at all hence the >< Quote Link to comment Share on other sites More sharing options...
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