MY 11 TRISECTION APPROXIMATION METHODS Pg. 54
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TRISECTION USING M/3 & M/PI
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FIG. 11
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1
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Draw any suitable straight line AOB with line AO = 2 units, as
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shown in FIG. 11, above.
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2.
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From pt O draw a straight line OC = 1 unit with ÐCOB = 3Æ, as
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shown in FIG. 11 above.
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| Pg. 55
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3.
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Mark off a distance DO = 1/2
unit to the left of pt O on line ADO.
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4.
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Join pts D & C and pts A & C
with a straight line.
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NOTE:
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DAOC & DCOD
are similar and alike by construction:
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i.
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DAOC/DCOD = AO/OC = OC/DO = 2/1 = 1/.5 = 2
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ii.
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ÐCAD = ÐDCO = b and ÐAOC = ÐCOD = 2i - 3Æ.
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5.
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Draw the right bisector of line AEC at pt E to meet the line AFD
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at pt F, as shown in FIG. 11 above.
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6.
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Join pts F & E
with a straight line.
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7.
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Let FD = m and divide by 3 to get m/3, as explained by Dabe,
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as shown
in top right inset of FIG. 11 above.
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| NOTE:
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To sivide FD into 3 equal parts: first mark off 3 units at 'right angles'
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| Pg. 56
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to line FD, bc = cd = de = 1 unit, then join pts F or a & e. Then
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using parallel lines, to line Fe, draw through pts d & c to get pts f
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& G, respectfully, on line afGb. Now length Ff = af = fG = Gb =
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= GD = FD/3 = m/3 as needed.
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8.
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Mark off a distance OG = 1/2 + m/3 = b to the left of pt O on line
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AFGDO in DACO, reproduced in DGCO, as shown in top left inset
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of FIG. 11 above.
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9.
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Join pts G & C with a
straight line.
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10.
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Thus in DGCO, ÐCGO = Æ trisects ÐCOB = 3Æ.
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1.
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Let AC = a thus AE = AC/2 = EC = GC = a/2
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NOTE: ratio DACO/DFCO = AO/OC = OC/DO = 2, by construction.
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| Pg. 57
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2.
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Let AF = n, FD = m, GC = e, and GO = b = m/3 + 1/2.
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3.
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AC2 = AG2 + OC2 + 2*AO*OC*cosÐCOB
[COSINE LAW]
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a2 = 22 + 12 + 2*2*1*cosÐCOB = 5 + 4cos3Æ
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a = (5 + 4cos3Æ)½
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4.
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cosb
= (a/2)/n
[from CRC TABLES]
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OC2 = AO2 + AC2 - 2*AO*AC*cosÐCAO
[COSINE LAW]
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12 = 22 + n2 - 2*2*n*cosb
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cosb = (a/2)/n = a/2n = (3 + a2)/(4a) = (8 + 4cos3Æ)/(4a)
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thus n = 2n2/(3 + n2) = 2(5 + 4cos3Æ)/(8 + 4cos3Æ) = .5n/cosb
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and m = AO - AF - DO = 2.0 - n - 0.5
= 1.5 - .5asecb
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m = 1.5 - 2a2/(3 + a2) = 1.5 - 2(5 + 4cos3Æ)/(8 + 4cos3Æ)
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5.
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in DGCO:
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Let GO = DO + GO = b = 1/2 + m/3
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| Pg. 58
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GC2 = OC2 + GO2 + 2*OC*GO*cosÐCOB
[COSINE LAW]
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e2 = 1 + b2 + 2bcos3Æ
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e = (1 + b2 + 2bcos3Æ)½
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6.
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sinÐGCO/b = sina/b = sin3Æ/e
[from CRC TABLES]
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sina/b = sin3Æ/(1 + b2 + 2bcos3Æ)½
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sina = b*sin3Æ/(1 + b2 + 2bcos3Æ)½
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a = sin-1[b*sin3Æ/(1 + b2 + 2bcos3Æ)½]
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Thus ÐCGO = a » Æ trisects ÐCOB = 3Æ with a small error, see
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TABLE 3, below.
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CALCULATIONS FOR ERROR TABLE 3:
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In DGCO:
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ÐGCO = a » Æ = sin-1[bsin3Æ/(1 + b2 + 2bcos3Æ)½]
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% ERROR = (a - Æ)Æ*100% giving the ERROR TABLE 3 below.
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CALCULATIONS FOR ERROR TABLE 4: In DGCO:
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sin2Æ/1 = sin2Æ/b
[SINE LAW]
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b = sinÆ/sin2Æ = .5/cosÆ = .5secÆ
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GD = GO - DO = b - .5 = .5/cosÆ - .5 = .5(secÆ - 1)
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thus the division factor = FD/GD = 2m/(secÆ - 1)
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giving TABLE 4 for the 1/TRUE FACTOR for each angle below.
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