Looking for an elegant solution

[Kevin9679]

Hello,

I'm studying for my Surveying exam. From Barry Kavanaugh's textbook I have the following problem. I tried to solve the bearing portion first based on the fact that a five-sided traverse has 540d interior angles. Then I found out that I have two unknown interior angles, not one. Of course I could turn all the bearings into latitudes and departures and then do something really really complicated in which I'm sure I would make a mistake somewhere along the line. But I'm wondering if there is an elegant solution in which the problem can be solved in minutes by hand. Thanks in advance!

Course AB Distance = 537.144 Bearing = N37d10m49sE
Course BC Distance = 1109.301 Bearing = N79d29m49sE
Course CD Distance = ? Bearing = S18d56m31sW
Course DE Distance = 953.829 Bearing = ?
Course EA Distance = 483.669 Bearing = N62d58m31sW

Solve for Distance of CD and Bearing of DE.

 

[IRstuff]

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Only one unknown angle and one unknown distance, BUT E is SE of A, so DE must cross AB; it sounds like there's a typo in the problem A needs to be SW of E, not NW.

Brute force solution in Excel 1230.333 distance for CD, 37.83649386
degrees SW for EA

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[Kevin9679]

Am I mistaken or is there something funky about the graph? BC should be almost pointing east.

 

[Kevin9679]

Is there an elegant way to do this without Excel? For everyone's clarification, this is not coursework. It is self-study.

 

[IRstuff]

I definitely could have it backwards; 10 minutes worth of Google.

In any case, there is only one missing angle.

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[IRstuff]

Makes more sense now:

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[IRstuff]

There's a geometric solution. You know the distances EC and ED and the angle ECD, so SSA.

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[fel3]

This is actually just a right triangle solution wrapped in a fun little package.
[1] Run traverse E-A-B-C. Use arbitrary coordinates for E. You want the coordinates for C.
[2] Inverse C-E to find bearing and distance. You can use surveying methods or Pythagoras for distance and trig for bearing. The two methods are really the same thing, just handled a little differently.
[3] Line E-D is defined only by distance. Thus, Point D is somewhere along an arc of radius 953.829. For a unique solution, Line E-D must be perpendicular to Line C-D (here is the right triangle). Since Line C-D is defined by a bearing, you now know the bearing of Line E-D.
[4] To find the distance C-D, use Pythagoras since you already know distances C-E and E-D. I suggested getting the bearing from the inverse C-E so you can check your triangle solution using the bearings.

//EDIT// Measure once, cut twice. I plotted this in Autocad and the data provided has Line C-D crossing the arc of Line E-D at two locations. This means two solutions (i.e. no answer) or bad data as suggested above. //EDIT//

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[dylansdad]

Yes,
2 solutions.

 

[Kevin9679]

Ok. Wow that's complicated.

 

[fel3]

dylansdad...

That's what mine looked like. Reminds me of a lopsided ice cream cone.

============
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[IRstuff]

Going get that picture right eventually. I still think that Law of Sines can be applied, since you know two sides and a non-included angle

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