Inverse tangent

On Thu, 9 Aug 2007 12:19:12 +0100, Boen S. Liong wrote


Let Z be the base length of both triangles, then :

tan(w3) = (x1 - x2).cos(w1) / { Sqrt(z^2 + y2^2).cos(w1-w2) }

AAR

On Thu, 9 Aug 2007 12:19:12 +0100, Boen S. Liong wrote


Sorry, previous post should read :

Let Z be the base length of both triangles, then :

tan(w3) = (x1 - x2).cos(w1) / { Sqrt(z^2 + x2^2).cos(w1-w2) }

AAR

On Sun, 12 Aug 2007 18:03:39 +0100, AAR wrote


Or, in terms of x1, x2 and Z only :

tan(w3) = Z(x1-x2) / (Z^2 + x1x2)


AAR

AAR wrote:

   ...


Can you tie that back to arctangent? That was the original question.

Jerry
--
Engineering is the art of making what you want from things you can get.
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On Mon, 13 Aug 2007 15:32:39 +0100, Jerry Avins wrote


The original question was "What is tan(w3) in terms of x1, x2, z1, z2 ?"  So
it seems to me that I've given what was asked for.  

Note: there's no need for two z's because the triangles have a common base
length which I call Z.

AAR

AAR wrote:

The subject of this thread: Inverse tangent.
The original query begins:

Let w1 = tan-1(x) i.e. tan(x) = w1

  w2 = tan-1(y) i.e. tan(y) = w2

I guess I was misled.

Jerry
--
Engineering is the art of making what you want from things you can get.
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯