more trigonometry help

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"HeyBub" wrote:

But not necessarily with the same degree of ease.
Lew
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HeyBub wrote:

So figure out the dimensions of a square with the same area as a circle or trisect an angle geometrically.
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J. Clarke wrote:

Hmm. Archimedes trisected an angle. You really should keep up.
You are right about "squaring the circle." Can't be done since pi is a trancendental number. Can't be done with algebra either, unless you simply express the area as an equation. In other words, it can't be "solved" for a value.
Possibly. I was drunk the day all that was covered.

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Yes, Archimedes trisected an angle, but his method requires putting marks on a straight edge, which advocates of pure geometry don't allow. Kerry
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ahead and built my bending form. Came out nicely. Thanks again for the help.
jc
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Kerry Montgomery wrote:

"Advocates of pure geometry"? The challenge issued in ancient times was to square the circle using Euclidean geometry. Euclidean geometry is a game with certain rules, one of which is no marks on the straightedge. If as a practical matter you need to square the circule then there are many ways to do it, however none of them answer the original challenge.
This isn't a matter of "advocacy". It's a matter of obeying the rules of the game. If you don't want to obey the rules of the game that's fine, but then you're playing som other game, not "Euclidean Geometry".
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Not *validly*.
You really should keep up.
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Doug Miller (alphageek at milmac dot com)
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And yes, I know of the marked straightedge method. It works, but purists don't like it.

a straight edge and compass.
Method: 1. Make a wheel the size of the circle to be squared. 2. Mark a point on the wheel. 3. Use wheel to measure distance on line equal to circumference. Finally, use the standard method of computing the square root of the length of the line segment you got in step 3.
See? It's easy provided you're not restricted to just a compass and straight edge.
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John Cochran wrote:

And what exactly does that have to do with geometry? Your method may be perfectly valid mathematically but that doesn't make it "geometry". And that assumes that it in fact squares the circle, which it does not--the challenge in squaring the circle is to find a square with the same area and I don't see how knowing the square root of the circumference helps you in that endeavor. It gives you s=SQRT(2*pi*r) and what you need is s=r*SQRT(pi).
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HeyBub wrote:

Yes, he did, but he wasn't using geometry. He was doing something which might to a layman look like geometry but he was not following the rules that define geometry.
You might find http://www.jimloy.com/geometry/trisect.htm to be of interest.

The equation is the general solution. Particular values are of little interest in mathematics. Squaring the circle was the first great triumph of analytic geometry.

Very likely.

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I think that Mr Cole and his fellow geometricians would have called Joe's 'apex' the 'sagitta'.
Jeff
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Joe wrote: | I need some trig help. I need to create a bending form but I need | the diameter (or radius, it doesn't matter) based on the following | | The diameter of a circle formed by an arc with a 1/8" apex over a 1 | 5/16" base length. Yeah, I know some of the terms are wrong, but | I'm a long way removed from Mr. Cole's high school trig class.
Sorry to have arrived so late. This gets asked enough that I put up a web page with diagram and step-by-step algebra at the link below.
-- Morris Dovey DeSoto Solar DeSoto, Iowa USA http://www.iedu.com/DeSoto/CNC/cove_geom.html
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Where A is the apex (or sagitta) and B is the chord (A = .125 and B 1.3125 in the question) then the Radius R is given by:
R =(A^2 + (B/2)^2)/(2A)
derived from:
R(1-cosT) = A RsinT = B/2
where T is the half angle of the arc.
Tom Veatch Wichita, KS USA
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