Only if you know distance AD.
Well hell, even a 72 year old idiot (just released) can figure out AD = AB + BC + CD ? (and what's up with the angle's?) ?????ÿ
I think there is probably only one possible point for the common vertex of the angles that will allow that combination of given angles and distances.?ÿ
Finding it will be tricky, but I expect computer work will yield it faster than trig equations.
?ÿLooks like fun for after the meeting I'm now headed to.
5 knowns
5 unknowns
Lotsa equations that may be created
Line AE is common to three triangles
Line BE is common to three triangles
Line CE is common to three triangles
Line DE is common to three triangles
Throw in a right angle just for fun
Back from the meeting, I put the constraints and initial guesses into my least squares program and it readily converged to a solution.Will post numbers tomorrow.
I'll maybe play around with equations, too.
@bill93?ÿ ?ÿWhy post a computer solution? Let all work on it long hand first, say 1 month; that's what Dave Lindell's?ÿ problems
are all about, checking your Math skills, not your button pushing skills.
After solving Dave Lindell's above problem long hand I then put it into MicroSurvey STAR*NET Version 11,0,6,2263
Got the same answer (long hand vs computer) but a BIG difference in time to solve it. It took me something like
1 hour (and my eraser use several times) compared to 1 second on the computer.
Thanks Dave; another great problem.
JOHN NOLTON
Thanks Dave; another great problem.
John, they may be great for you but they seem to "overclock" my brain. ?????ÿ
I've started a couple different ways that got messy in a hurry. Haven't found any magic bullet approach.
@flga-2-2?ÿ ?ÿThey do the same to me. Many of Dave Lindell's problems I do not solve.
I've started a couple different ways that got messy in a hurry. Haven't found any magic bullet approach.
If you know the Law of Sines, simultaneous equations, and the quadratic equation you are all set to go!
Law of Sines, simultaneous equations, and the quadratic equation
Whut dat? ?????ÿ
@dave-lindell?ÿ ?ÿThat's what I used Dave. Since it's such an "innocent " looking problem I am going to spend more time on
it and see if it can be solved any other way.?ÿ
Thanks again Dave.
JOHN NOLTON
This is an overlapping triangle problem. Teachers and good geometry students love to solve them and test makers who want to prove how smart they are love to put them on standardized tests. I have to say that this is the first one I've seen that involved trig, but you really don't need trig to solve it.?ÿ
It just takes a translation and a rotation as shown below. I know, I know, I got 214 and Dave got 217 but reconciling the two might be a good Sunday project. I apologize for the crude drawings, but I can assure you that my students suffered through worse.
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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.