MY 11 TRISECTION APPROXIMATION METHODS Pg. 50
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FIG. 10
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1.
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Draw any suitable straight line OBA and draw a unit circle GB
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with center pt O and a suitable radius OB = R = 1 unit, as shown
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in FIG. 10 above.
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| Pg. 51
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2.
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Draw any straight line OC so that ÐCOA = Æ with pt C lying on the
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circumference of the previous unity circle, see FIG. 10 above.
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3.
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Similarly draw a straight line OD so that the ÐDOA = 3Æ
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(measured counter clockwise) with pt D lying on the
circumference
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of the previous unity circle BCD, as shown in FIG. 10 above.
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4.
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Join pts C & D with a straight line.
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5.
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Draw a circle with center pt O and radius of length CD to intersect
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the line OD extended at pt E, as shown in FIG. 10 above.
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6.
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Repeat all steps 1 to 5 for every pt E' and ÐE'OA = 3Æ' and
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ÐC'OA = Æ' from 0o to 360o
to get the chart or complex figure
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GAHEO shown in FIG. 10 above.
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| NOTE:
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Line GOH is the Y-axis and line OBA is the X-axis.
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1.
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By going backwards from 3Æ' to get Æ' one can trisect all angles
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with little error.
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2.
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Assume in FIG. 10 that the angle to be trisected is ÐEOB = 3b.
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3.
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With center D and radius length OE draw an arc
to intersect the
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inner unity circle CGDRB at pt C.
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4.
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Join pts C & O with a straight line.
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5.
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Thus ÐCOB = b trisects ÐDOA = 3b as the required.
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1.
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In DCOD:
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Let ÐEOA = 3b and ÐCOE = Æ
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Since 2 sides OD = OC = R therefore DCOD is isosceles.
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length EO = DC = 2(OC)sin(ÐEOA - ÐCOA) =
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= 2RsinÐCOD = 2Rsin(3b - Æ)
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2.
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But length EO = 2Rsin2Æ, construction
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Thus (3b - Æ) = 2Æ or 3b = 3Æ and b = Æ.
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3.
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Thus ÐCOA = ÐEDA/3 = 3b/3 = b
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Thus ÐCOA = b trisects ÐEOA = 3b, as required.
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UPSIDE DOWN CAKE
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