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Topic: TO DIVIDE AN ANGLE IN ANY NUMBER OF EQUAL PARTS
Replies: 23   Last Post: Mar 4, 2013 4:05 AM

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 Peter Scales Posts: 192 From: Australia Registered: 4/3/05
Re: TO DIVIDE AN ANGLE IN ANY NUMBER OF EQUAL PARTS
Posted: Jul 25, 2010 7:04 AM

SHYAMAL KUMAR DAS wrote:

> Respected Sirs,
>
> I would like to inform you that most probably I have
> invented a geometrical method / construction, by
> which any angle can be divided into any number of
> equal parts.
> So far I know, a few people had tried to trisect an
> angle, but failed to do so. My method can not only
> trisect but divide any angle into any no. of equal
> parts.
>
> Please be kind enough to look into my method.

Sir,

The method you propose does give a reasonable approximation to the 1/n th value of an acute angle for reasonably large n, but unfortunately it is not a valid method for exact angle division.

The flaw is that it equates the base (chord) of an isosceles triangle with the corresponding arc of a sector, which is simply not true.

This leads to an error in your construction. When you use a compass to lay off a length, it is a chord you lay off, not an arc. There is no unique arc associated with a chord, and in your example two different arcs occur: arc DM with r=1 and arc HI (=IJ etc) with r=5. The chords are identical, but the arcs are not.

Your comment on page 3 regarding 60degrees being an approximation for 1 radian arises from the same error. In an isosceles triangle, a side of unity subtends 60 degrees at the apex whereas an arc of unity subtends 1 radian = 57.296 degrees. This is not an approximation, but a different value from a different figure.

As an example, use your method to bisect a right angle:
BD=1, chordBE=sqrt2, BA=2, CF=FG=sqrt2
.: your approx to the bisected angle, theta, has
sin(theta/2)=1/4*sqrt2 .: theta/2=20.7deg theta=41.4deg
and the discrepancy is 90-2*41.4=7.2deg

It is even more obvious if you use your method to bisect a straight angle=180 BD=1 DE=2 and the resultant approx bisected angle is 60deg, not 90deg.

I hope this has been of some help to you.

Since you have access to the Internet you could Google angle division and explore the extensive existing literature.

Regards, Peter Scales.