TRISECTION OF ANY ANGLE
Author : Shyamal Kumar Das E-mail ID : das.shyamal1c@gmail.com
Given : < ABC of any value.
Required : To divide < ABC into 3 equal parts i.e. trisection of < ABC.
Construction : The given angle is to be bisected once, twice, trice and so on
to get each bisected part equals or closes to 13.5 degree.
(why 13.5 ? please see the explanation beneath)
The angle in the above figure has been bisected twice for
demonstration purpose. So, < IBC is 1/4th of the given <ABC.
From BC, 3 equal parts ( suitable and convenient ) BD, DE,
and EF are cut off. With center B and radius BD, an arc is
drawn which intersects BI at G. Similarly, with center B and
radius BF, another arc FH is drawn and extended, which
intersects BA at H.
,
From bigger arc FH, taking DG as a chord, 4 equal parts i.e.
4 graduations FJ, JK, KL, LM are cut off.
( why 4 equal parts ? please see the explanation )
If BM is joined and extended to N,
then, < NBC is 1/3rd of the given <ABC.
Again, from arc FH, taking FM as a chord, another 2 portions
namely, MP and PH are obtained. BP is joined and extended to Q.
Hence, < ABQ = < QBN = < NBC = 1/3rd of <ABC
Explanation: 1/3rd of 13.5 degree = 4.5 degree
To get best and accurate result, θ /e*n should be equal to
4.5degree ( Magic value ), where, θ is the given angle, ‘e’ is the no of
bisected portions, i.e. 2, 4, 8, or 16 (here, e = 4 )and ‘n’ is the no. of
reqd. divisions (here, n =3). This Magic value, as calculation reveals
is that angle when subtended at the centre by an arc , the chord
length and arc length are almost equal. Also, note that, no. of
graduations taken at a time= no of bisected portions i.e., 2,4,8 or 16
(here, no. of graduations is 4) of the given angle.
Proof :Let < ABC = Ѳ, therefore, < IBC = Ѳ/4 and BD = r
then, BF =3 × BD = 3r
Arc DG = rѲ/4 and arc FL = 3rѲ/4 = 3 times arc DG
Hence, FJ = JK = KL =rѲ/4 = LM
So, FM = FJ+ JK+ KL+ LM = 4 × rѲ/4 =rѲ
therefore, <NBC = rѲ/3r = Ѳ/3 =1/3rd of < ABC
Calculation : When the angle subtended at the center =4.5 degree
= π/40 radian,
the difference between arc length and chord length =
π/40 × R – 2R × sin 2.25 = R(0.07854 -2× 0.03926) =
R( 0.07854 – 0.07852) = 0.00002 R
Hence, it is revealed that when angle subtended at the
center by an arc is 4.5 degree, the chord length is almost
equal to the arc length.
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