Help with Trig (Updated drawing)

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Can you post a screen shot or a picture of the actual math problem?
See post 52 or 56. This is the drawing that my son presented me....it's from his head for some (yet unknown) rocket project of his......and his math teacher couldn't solve it.
 
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One of the things most of you guys are missing, is that you can solve systems of equations, if you have as many equations as you have variables.
If I say "a + b = 7", you could look at that and say, "You can't solve it - a and b could be anything!" - and you'd be right. But if I also say "a -b = 3" you could also look at that by itself and say "You still can't solve it, a and b could be anything!". But if you take those pieces of information together, it's easy to solve, since you have 2 unknowns and 2 equations. The only answer is a=5, b=2.

As far as I know, it's not possible to use trig or geometry figure out any individual angle or length in the presented problem. But the angles and lengths are all properly constrained (build the model with drinking straws and pins, to prove it to yourself) and so by simultaneously solving a sufficient number of known equations, we can solve for everything. I presented 4 equations (if any of them are incorrect, tell me why you think so!) in 4 variables, therefore the problem is solvable.
 
One other thing, all of the drawings shown by rocketjet are incorrect. They do not capture all the constraints. The 3" and 11.5" line segments MUST lie along a straight 14.5" line, and the point that is 3" along this line MUST lie along the 34" long horizontal line. None of his drawings include both of these facts. If you add these constraints, you will see that all of the angles and dimensions are fixed, and thus solvable.
 
The large triangle is a not really a given either Fattbank. We don't know for positive that x is a bisector. We all assume it is which gives us the 3 pieces of info needed to solve it.
Yes, you do need to assume that 'x' is a bisector. That goes along with the assumption that the two sides are symmetric, an assumption I stated.

After that we have to assume either the 7.5 or 3 is parallel to the 45's....only then do we have an angle for the smaller triangles.
No, as you can see it was solved without that assumption.
 
Excellent Kelly. I understand your explanation. Even though I don't know exactly the system of equations you followed to reach your answer, (you definitely damaged my brain), I bow to your expertise here...your answer of 33.2 IS correct.
 
So we're going with 33.2?

I soooooooooooooo wanna do the 'show your work' thing so that I could repeat the solution given any variations that my son might want to make.......but I think it would hurt my brain!
 
So, I'll admit... after I set down the 4 equations, I tried to solve with pencil and paper... and gave up. Had better things to do with my time. I used an online, simultaneous equations solver to get the numerical answer.
Banzai, here's some other numbers, if it helps verify the answer:
Side a: 9.2
side b: 7.8
side d: 8.45
angle y: 47.3°
length of the bisector of the 45/45/34 triangle: 41.7
length of x: 41.7 - 8.45 = 33.2
 
A couple of notes in response to Kelly:
1. None of the constraints you mentioned were a part of OP's post. Before OP confirmed those constraints, there are indeed multiple solutions. But those constraints make your system of equations valid, except:
2. Your equation for the law of cosines is a little off. It should be 7.5^2 = 3^2 + a^2 - 2*3*a*cos(y) <--you forgot the "a" in the cosine term.
3. Why not use cos(y) = b/11.5? That makes it a little easier to get the solution.
 
A couple of notes in response to Kelly:
1. None of the constraints you mentioned were a part of OP's post. Before OP confirmed those constraints, there are indeed multiple solutions. But those constraints make your system of equations valid, except:
2. Your equation for the law of cosines is a little off. It should be 7.5^2 = 3^2 + a^2 - 2*3*a*cos(y) <--you forgot the "a" in the cosine term.
3. Why not use cos(y) = b/11.5? That makes it a little easier to get the solution.

Correct on all counts. The assumptions I made were strongly implied (to me) by the drawing, but not expicitly stated at first. On #2, that was a transcription error - I solved the problem correctly, but erred when writing on TRF (just fixed it). Thanks for this correction! And #3, true also. Sometimes when a brute force method (automated solver) is available the obvious/easier stuff gets overlooked.
 
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There is a solution for any X in this range:

Sqrt(1736) - sqrt(195) <= X <= sqrt(1736) + sqrt(195)

Can you show a solution for X= sqrt(1736) + sqrt(195) = 55.6, that doesn't involve sqrt(-1) or some other monkey business?
;)
 
X= 55.6 is the solution when the angles between the 7.5 segment and the 3 segment are 180 degrees, and the angle between the 11.5 segments are 180 degrees. In other words, the figure looks like a kite, with length 45 on each bottom side and 22 on each top side. This is one of the valid solutions before the other constraints were set. In fact, kbrocket’s range of solutions is what I also had before the other facts were known.
 
Ah, OK. I've never consider a solution where the 3 segment didn't terminate on the 17 (or 34) segment, as shown.
 
Fattbank...33.2125 is the correct answer. Tell me, did you get this answer using the altitude?

I got the answer using cad. Our answer is correct (if calculated down to 4 decimal places), but I had to move circles and lines around until I came up with the 3 dimension.
Yes.
 
Yesterday I was away putting pointy things up in the sky. During a break I typed in the X range answer. This only pertains to the original post. I hadn't reviewed the discussion to realize that more information had been given. I was just typing out what happened in my head after reading the post the night before.

Once constraints got added the problem became more complicated to solve for multiple unknowns with relationships.

I have attached a python program that numerically solves this problem by starting with a vector of potential X values. It uses a few math/trig relationships to compute what is the length of the "7.5 side" that I call b. The final solution is the X value that makes side b actually equal to 7.5. This method may be one of the least complicated and most understandable ways of solving the relationships for X.

Here is a plot of a range of X values and the predicted length of the 7.5 side.
1601224753462.png

Here is a plot of the fin shape with the final solution of X = 33.212. I added a few symbols that you can match up with the python. The vertical and horizontal axes should have equal scales so this should be an undistorted depiction.

1601225375197.png

I had to make the change the file extension from .py to .txt to get it to upload. I could have zipped it but thought this was better.

I used Python because it is fairly well known and free and perhaps something a son might tinker with.
 

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  • fin_calc.txt
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