Inverse tangent

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

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 (in message ):

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 (in message ):

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

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

AAR

...

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

Jerry

On Mon, 13 Aug 2007 15:32:39 +0100, Jerry Avins wrote (in message ):

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

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