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 Measure that angle IV (Posted on 2004-11-02)
Triangle ABC is isosceles with AB=AC. Point D is on side AB such that angle BCD is 70 degrees. Point E is on side AC such that angle EBC is 60 degrees. Angle ABE equals 20 degrees, and angle DCE equals 10 degrees.

 See The Solution Submitted by Brian Smith Rating: 3.6667 (9 votes)

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 Iterative Solution | Comment 2 of 21 |

The base angles of the isosceles triangle are 80 degrees each (B and C are 60+20 and 70+10, respectively).  Angle BEC is 40 degrees (as EBC and ECB add up to 140) and angle BDC is 30 degrees (as DCB and DBC add up to 150).

In the quadrilateral EDBC, the angles at B and C add up to twice 80, that is, 160, leaving 200 for the angles at D and E.  Subtracting the 70 that CDB and BEC add up to, leaves 130 for EDC and DEB.

Call mEDC = d and mDEB = e.

Then d+e=130.

Triangles BDC and BEC share a side BC, whose length we'll call z.

Call side EC x, and side DB y. By the law of sines in their respective triangles,

x/sin60 = z/sin40, or x = z sin 60 / sin 40
y/sin70 = z/sin30, or y = z sin 70 / sin 30

Then, using w to represent the length of ED, and applying the law of sines to the top right and top left triangles:

x/sin(d) = w/sin 10, or x = w sin(d) / sin 10
y/sin(e) = w/sin 20, or y = w sin(e) / sin 20

Equating the ratios of x to y:

sin(d) sin 20 / (sin(e) sin 10) = sin 60 sin 30 / (sin 40 sin 70)

or

sin(d)/sin(e) = sin 10 sin 30 sin 60 / (sin 20 sin 40 sin 70)

Start with d and e estimated as equal halves of 130 degrees, and then iterate using this last formula to evaluate d from e, and then use d+e=130 to evaluate e from d.  This converges rapidly on d = 20, e = 110.  This could be done with a calculator, putting the result of sin 10 sin 30 sin 60 / (sin 20 sin 40 sin 70) into memory, and entering 65 sin x mem = asin - 130 = +/- and repeat sin x mem = asin - 130 = +/- until convergence is reached. Depending on accuracy, 10 to 18 iterations will do. Here's a convergence done via a computer program:

1             19.26082801420013           110.7391719857999
2             19.90037747996541           110.0996225200346
3             19.98677161255989           110.0132283874401
4             19.99824703218261           110.0017529678174
5             19.99976776717145           110.0002322328286
6             19.99996923494301           110.000030765057
7             19.99999592441669           110.0000040755833
8             19.99999946008978           110.0000005399102
9             19.99999992847575           110.0000000715243
10            19.99999999052487           110.0000000094751
11            19.99999999874479           110.0000000012552
12            19.99999999983372           110.0000000001663
13            19.99999999997797           110.000000000022
14            19.99999999999708           110.0000000000029
15            19.99999999999961           110.0000000000004
16            19.99999999999995           110.0000000000001
17            19.99999999999999           110
18            20            110
19            20            110

making angle EDC 20 degrees.

DECLARE FUNCTION asin# (x#)
DEFDBL A-Z
pi = ATN(1) * 4
dr = pi / 180
frac = SIN(10 * dr) * SIN(30 * dr) * SIN(60 * dr) / (SIN(20 * dr) * SIN(40 * dr) * SIN(70 * dr))

d = 130 / 2: e = d

FOR i = 1 TO 40
d = asin(frac * SIN(e * dr)) / dr
e = 130 - d
PRINT i, d, e
NEXT

FUNCTION asin (x)
asin = ATN(x / SQR(1 - x * x))
END FUNCTION

 Posted by Charlie on 2004-11-02 18:36:54

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