I think I follow your logic, after adding some labels to your last diagram. I would never have come up with it.
I think the values in the problem are not self consistent to sufficient accuracy for the answers to agree any better.?ÿ Consider the locus for point E as determined from ABE as you change the angle at A. Then from CDE as you change the angle at D. The intersection of those loci is at a very shallow angle, so that a small change in any measurement has a big effect on the northing of E and thus the center angle.
I got 217.748 from my program, but I was holding all three angles tightly.
?ÿ(and my eraser use several times) I'll bet there aren't too many of us left here that know what that know what an eraser actually is, besides "backspace". ?????ÿ
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@bill93?ÿ
It's really cool logic, but it's not sound. The problem is that when we rotate those triangles, their bases may overlap, meet, or have a gap. The value of x will be half of the base of the created triangle, but the base of the created triangle probably won't be equal to the sum of the two given lengths.
If the logic were sound, then every overlapping triangle problem presented like this would have a missing segment equal to the average of the two given ones. It was just happenstance that my answer was a close to Dave's as it was.
Sorry for the misinformation; it just looked so good that it "had to be right." Such things should always stew overnight.
Here are a couple of examples created in Geogebra by stretching and shrinking objects:
I had tentatively come to that conclusion this morning, but wasn't sure enough to post.
The other approach I want to try is based on the fact an angle inscribed in a circle, with 2 fixed points, can have the vertex anywhere on the circle. Thus the given data defines 2 circles that may move apart or together until their intersection E is where the middle angle results.
