Length of Diagonal

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The discussion revolves around the mathematical concepts of the diagonal of a square and the ambiguity in triangle solutions. The diagonal length is derived as \( \sqrt{2}a \), but concerns arise about the accuracy of drawn diagrams affecting problem-solving. Participants debate the relevance of drawing accuracy versus mathematical abstraction, particularly in the context of triangles and the Law of Sines versus the Law of Cosines. The thread also touches on the emotional toll of perceived misunderstandings in mathematical discussions and the desire for clarity in problem-solving methods. Ultimately, the conversation highlights the importance of precise representations in mathematics while acknowledging the challenges of interpretation.
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Homework Statement
I've drawn a sloppy mathematical diagram of a square of side length ##a## as I and almost everyone does. What is the length of the diagonal?
Relevant Equations
$$ a^2 + b^2 = c^2$$
1743091496618.png




$$ a^2 + a^2 = ?^2 $$

$$ ? = \sqrt{2}a $$


Now, do I expect an unreasonable solution that violates what the problem statement dictates because I was sloppy with my drawing of a square?

The bonus question is, what is the circumference of this circle of radius ##r##?

1743092310581.png


This will be my last thread; you all can finally be rid of me!
 
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The drawing accuracy is irrelevant if you are mentally dealing with an ideal square or circle.
In my case, I frequently replace mathematical ignorance with accurate drawings.
 
Lnewqban said:
The drawing accuracy is irrelevant if you are mentally dealing with an ideal square or circle.
In my case, I frequently replace mathematical ignorance with accurate drawings.
How about with uh...I don't know... triangles, would you say that should hold too? It is something fundamental to the mathematical abstraction of shapes...right?
 
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And in case this gets shut down like every other thread I participate in recently before its finished. I bid you all who were kind enough to give me a fair shake over the years a fond Fairwell. I've learned so much (maybe more than I ever did jamming it down the gullet in college), you get a chance to really sit a chew on an idea here and that was the main reason I participated. Problems come in, and you get a chance to actually think on it for a bit...what a refreshing change of pace from the way things are done.
 
If we are going to pontificate on inaccurate drawings, we can do better than non-square squares. How about "all triangles are isosceles"?
1743095492390.png

As I recall, a problem here is that ##B## is drawn between ##A## and ##R##. In fact, it should be on the far side of ##R## from ##A##
 
You see I can draw this triangle like this. It's nothing more than a typical mathematical construct.

Take for example this triangle

1743096349304.png


The objective is to find ##b## and ##\theta##. So most obviously you reach for the Law of Cosines for ##b##, and the law of sines for ##\theta##.

You will find that the angle ##\theta \approx 77^{\circ}, b \approx 28.78## ?

Plausible no?

After you have filled in all the blanks see if you have a solution to your problem. That is you have a triangle that is at least consistent with the fixed parameters in the problem in red.
 
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I'm going to save you all the suspense...you don't, I'm going to construct that triangle for you in CAD:

1743098346422.png


This is the triangle the solution gives you. You ask yourself the angle in the lower right is not 30 degrees, and the length of ##b## is not ##\approx 28.78## How strange.


Someone says, but you drew the mathematical construct triangle incorrectly. look the 56 is drawn shorter than 42. So against your will you reshape your construct.

1743098951604.png



That ought to do it! So optimistically you try again... Same result.

Someone then says, you drew the triangle...Its case of ambiguous triangle , and it should be obvious if you are just a little bit clever and you have to use this patchwork set of criteria to find out the actual solution is actually the angle outside the triangle that is the solution ( becuase the sine function isn't one to one). And sure enough there it is on the right!

1743099485093.png
 
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Even though I have a smiley face there I'm not really happy. I understand the work around, but why use it?

I can just apply the Law of Cosines here twice and go directly to the correct solution on the right. No side trip into LaLa land required. No worrying about whether or not I've drawn the triangle correctly, and/or whether or not you could draw another triangle.

So then they say about a whole class of parameter sets that have this ambiguity. Triangles that look like this:

1743100143094.png


So instinctually again reach for the law of sines....Stop right there. Lets just instead solve the quadratic...


$$ a^2 = x^2 + c^2 - 2 c x \cos \theta $$

Low and behold, this ends very easily in:

## x = c \cos \theta \pm \sqrt{ a^2 - c^2 \sin ^2 \theta} ##. And we are done. Certain solutions will be ruled out as they always are. negative solutions for the distance and imaginary determinate.

Here is a plot in Desmos (set at the triangle that started all this)



"But we've been doing it that way for nearly 2000 years." Well, the general solution for the quadratic wasn't found until the 1500's.

Am I really that foolish for thinking "how we draw our mathematical construct actually shouldn't matter".

I also feel that in light of the presented solution, the law of signs is rendered a soft rule...that is fuzzy, and sometimes directly misleading in it solutions as was the case here.

End rant, feel free to terminate me now.
 
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I taught my students the law of cosines as an application of vector addition combined with the dot product. It is a foolproof method and avoids the confusion about the sign in front of the cosine term.

erobz_post.png
Three vectors adding up to zero form a closed triangle, see figure on the right. We are interested in the magnitude ##|\mathbf C|## and write $$\begin{align} & \mathbf A+\mathbf B+\mathbf C=0 \nonumber\\
& \mathbf C=-(\mathbf A+\mathbf B)\nonumber\\
& C^2=(\mathbf A+\mathbf B)\cdot(\mathbf A+\mathbf B)=A^2+B^2+2\mathbf A\cdot \mathbf B\nonumber\\
& C^2=A^2+B^2+2AB\cos\theta \nonumber \\
\end{align}$$Clearly, angle ##\theta## is the angle between vectors ##\mathbf A## and ##\mathbf B## when these vectors are put tail-to-tail. In this case ##\theta=150^{\circ}## as shown in the diagram.
 
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  • #10
I'm afraid I'm not sure how to take the sentiment of this thread.
Is @erobz really leaving - or asking to get booted? In which case, I'll consider not spending time on a thread that he may not even see? Or is he just goofing around?
 
  • #11
DaveC426913 said:
I'm afraid I'm not sure how to take the sentiment of this thread.
Is @erobz really leaving - or asking to get booted? In which case, I'll consider not spending time on a thread that he may not even see? Or is he just goofing around?
I've had three threads closed I've participated in the last month, I'm asking to be removed from membership. I said things like "The Law of Sines is lying to us", and that it is inferior to the "Law of Cosines" as it seems quite so in this problem, I was told I need to accurately represent the mathematical construct, and apply the ambiguous law of sines criterion to find out the solution is from an angle outside of the triangle that the solution gives.

I'm tired of wasting all our time with my "quackery".
 
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  • #12
erobz said:
Homework Statement: I've drawn a sloppy mathematical diagram of a square of side length ##a## as I and almost everyone does. What is the length of the diagonal?

Now, do I expect an unreasonable solution that violates what the problem statement dictates because I was sloppy with my drawing of a square?
The problem statement, as you've given it to us, would have the answer you've given us of ##a \sqrt{2}##. It's not a violation of the problem statement.

What concerns me is that I've never seen a homework statement that's anything like that. I'm used to a statement like "Find the length of the diagonal of a square with sides of length ##a##".

But if the homework statement specifies that you must include a diagram of the situation, then how sloppy of a drawing is allowed is something that would be set by your instructor. Certainly if the homework statement specified an accurate sketch, drawn to scale, then you would not be correct in expecting your solution to be reasonable with a sloppy drawing like the one you showed us.

But I can't answer your question as you've stated it because I have no way of knowing what your expectations are. That's something that only you can answer.

And by the way, how do you know that "almost everyone else" makes sloppy diagrams? If you have an instructor who insists on diagrams that are not sloppy, and is consistent about not allowing them, then you will find at least a significant number of students making neat diagrams, with all the other students receiving failing grades in the class.
 
  • #13
erobz said:
I'm asking to be removed from membership.
Then why don't you just stop writing posts? It seems that would achieve the same goal and would not depend on someone else answering your query.
 
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  • #14
Mister T said:
Then why don't you just stop writing posts? It seems that would achieve the same goal and would not depend on someone else answering your query.
I don't want the option to participate. I always will want to help if its open; it's an obsessive compulsion. I need to forget about problem solving amongst people of high levels of education that don't take my arguments seriously. Its unhealthy.
 
  • #15
erobz said:
I don't want the option to participate. I always will want to help if its open; it's an obsessive compulsion. I need to forget about problem solving amongst people of high levels of education that don't take my arguments seriously. Its unhealthy.
[ sidebar ]
(I hear you. Out of my own frustration, I took a two year break by requesting a ban on myself.)

But have you been infracted/warned along with these closed threads? If not, it's no harm, no foul, isn't it?
[ /sidebar ]
 
  • #16
DaveC426913 said:
[ sidebar ]
(I hear you. Out of my own frustration, I took a two year break by requesting a ban on myself.)

But have you been infracted/warned along with these closed threads? If not, it's no harm, no foul, isn't it?
[ /sidebar ]
No, but its clearly implied that I am the trouble maker, who perhaps hadn't been clearly breaking the rules. The end result is the same. The threads are shut down, and I am left to be a fool.
 
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  • #17
erobz said:
How about with uh...I don't know... triangles, would you say that should hold too? It is something fundamental to the mathematical abstraction of shapes...right?
Yes, it would.
I agree.

erobz said:
No, but its clearly implied that I am the trouble maker, who perhaps hadn't been clearly breaking the rules. The end result is the same. The threads are shut down, and I am left to be a fool.
I respectfully believe that most of that is only in your mind.
Please, stay with us, your posts are valuable, and for long time have taught me about things that I did not know or understand clearly.

:cool:
 
  • #18
erobz said:
... its clearly implied that I am the trouble maker, who perhaps hadn't been clearly breaking the rules.
That hardly seems a fair conclusion. PF has a mandate; you can't fault them for that. Comparatively, this is a strongly-moderated site. If you haven't got any warnings then they are not holding it against you. It's not personal unless it's personal.

erobz said:
The threads are shut down, and I am left to be a fool.
Consider the alternative: they must leave threads open that are against PF policy. That's basically special pleading. Is that what would be required for you to be able to stay? Is that reasonable?
 
  • #19
erobz said:
becuase the sine function isn't one to one
So you took the arc sine of a value and got the wrong result because the drawing convinced you that you were looking for a result in the first quadrant while the true result should have been in the second quadrant?

Yes, drawings can be confusing. They can draw one into making false assumptions. In this case, that a particular angle is acute rather than obtuse.

I am not sure what this has to do with the diagonal of a poorly drawn square or the circumference of a poorly drawn circle.
 
  • #20
Lnewqban said:
Yes, it would.
I agree.


I respectfully believe that most of that is only in your mind.
Please, stay with us, your posts are valuable, and for long time have taught me about things that I did not know or understand clearly.

:cool:
Well, I appreciate that. The respect/learning is mutual.
 
  • #21
DaveC426913 said:
That hardly seems a fair conclusion. PF has a mandate; you can't fault them for that. Comparatively, this is a strongly-moderated site. If you haven't got any warnings then they are not holding it against you. It's not personal unless it's personal.


Consider the alternative: they must leave threads open that are against PF policy. That's basically special pleading. Is that what would be required for you to be able to stay? Is that reasonable?
Ever been on the phone having a disagreement, and they hang up on you? It's personal. I have a significant portion of all "hang ups" this month... They might not be punishable offenses, but its surely personal.
 
  • #22
jbriggs444 said:
So you took the arc sine of a value and got the wrong result because the drawing convinced you that you were looking for a result in the first quadrant while the true result should have been in the second quadrant?

Yes, drawings can be confusing. They can draw one into making false assumptions. In this case, that a particular angle is acute rather than obtuse.

jbriggs444 said:
I am not sure what this has to do with the diagonal of a poorly drawn square or the circumference of a poorly drawn circle.


There was argument that "had I drawn it accurately I would have got the correct solution". I feel I have proven that false. Drawing the triangle accurately has no effect.

You'll see the Law of Sines in a text book as:

$$ \boxed{ \begin{array} ~\text{The Law of Sines} \\ \text{Suppose that a triangle has sides of length a,b,and c} \\ \text{with corresponding opposite angles A,B,and C, as shown, Then} \\ \frac{\sin A}{a} = \frac{\sin B}{b}=\frac{\sin C}{C} \end{array} } $$

Then it will give a couple of examples, and a few pages later ##but##, and it will list all the ways it fails to give a proper result and then show you what to do about it.

The entirety of caveats and criterion are inseparable from the "Law of Sines". There is a mental picture set in your head for the "Law", and it does not include all the "buts". In fact it turns out that the "Law of Sines" to be a Law, it is truly inseparable from its list of "buts", or the math can "lie to you". Giving you false solutions. And I say it is a false solution, because what is pumped out here does not represent the triangle you asked it to find using your parameters...it just decides to give you a solution in which the fixed parameters, and a calculated value are altered. In goes these numbers and out comes something else!

This is not true for the "Law of Cosine". If you apply it here, twice ...instead of once, there is no false solution. You go directly to the actual triangle in question.

If you have a triangle like post 8., just solve the quadratic. I don't find it "silly" (as I was told it was) to say the Law of sines deserves a lower status, seeing how I'm not forced to use the "Law of Sines" at all, and have consistent outcomes every time.

Another member (in the thread that was cut short) said that I have a point in this looks very much like a problem where we expect that the application of "Law of Sines" will give an expected result, and it doesn't. They asked for comment twice, and wanted someone/counter arguer to address a specific case that he had in mind, near almost right triangles. Instead, their comments were ignored, not only once, but twice. Then the thread was shut down by said counter arguer.

Thats what all this business with "mathematical constructs" comes from.
 
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  • #23
erobz said:
it will list all the ways it fails to give a proper result
The law of sines did not give an improper result. It gave a proper result: the sine of the angle you were after.

It is not the fault of the law of sines if one then applies a particular partial inverse of the sine function without paying attention to the fact that it is not the only partial inverse.

This is much ado about nothing, in my opinion.
 
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  • #24
erobz said:
Ever been on the phone having a disagreement, and they hang up on you? It's personal. I have a significant portion of all "hang ups" this month... They might not be punishable offenses, but it’s surely personal.
My suggestion is to just take a break for a while, cool down then come back. I’ve had frustrating moments occasionally too but putting it in perspective, it’s not a big deal and the benefits of the dialogs are far greater than the occasional moment of grief.
 
  • #25
bob012345 said:
My suggestion is to just take a break for a while, cool down then come back. I’ve had frustrating moments occasionally too but putting it in perspective, it’s not a big deal and the benefits of the dialogs are far greater than the occasional moment of grief.
I concur.

And it would be a shame to lose you. i find your posts interesting and insightful.
 
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  • #26
jbriggs444 said:
The law of sines did not give an improper result. It gave a proper result: the sine of the angle you were after.

It is not the fault of the law of sines if one then applies a particular partial inverse of the sine function without paying attention to the fact that it is not the only partial inverse.

This is much ado about nothing, in my opinion.
It may be much ado about nothing.

I think it fails in the sense that the application of the mathematics fails to give any indication of a second definitive solution, there are not always 2, and sometimes none. And because it gives no such indication you will sometimes fail to realize it that you had better check your solution makes sense by fully analyzing the resulting triangle against your parameters if you are using it. I've done it so many times...and it never bit me until now.

This is why I think the Law of Cosines clearly stands apart. It tells you upfront, as a direct output. Unique, ambiguous, or no solution are all treated equally.

But anyhow. I'll let it go.

Thanks all for the words of encouragement, and sound advice. Maybe I'll talk to in the future. Take care.
 
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  • #27
erobz said:
... you had better check your solution makes sense ...
This is the take away IMO. So many students don't do what I call the sanity check.

"Setting aside the exact numbers; did I get a sensical outcome?"

If you are unable to decide if it's a sensible outcome then you have not understood the problem, and are just pushing numbers around.
 
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  • #28
Emphasis added...
jbriggs444 said:
The law of sines did not give an improper result. It gave a proper result: the sine of the angle you were after.
Seems like I've seen this before.
jbriggs444 said:
It is not the fault of the law of sines if one then applies a particular partial inverse of the sine function without paying attention to the fact that it is not the only partial inverse.

This is much ado about nothing, in my opinion.
Amen!
 
  • #29
erobz said:
This is why I think the Law of Cosines clearly stands apart.
It stands apart by comparison with the law of sines in that the cosine is single-valued in the interval ##0 < \theta <\pi##while the sine is double-valued. If you are given two sides of a triangle and the cosine of the angle between them, you can find the third side. However, if you are given two sides and the sine of the angle between them, there are two triangles that can be found because ##~\sin(\pi-\theta)=\sin\theta.## In Euclid's time people used a ruler and a compass to find (i.e. draw) a specific triangle. Needed for the construction were two sides and the angle between them or one side and the angles at its two ends or all three sides.
 
  • #30
The Law of Sines is often used when the angles are known and the length of a side is to be determined. In that case, there is only one solution.
If it is used to determine the angle, multiple angles are suggested and you have to decide which is correct. A drawing, with the known side lengths and angles can be helpful. Usually, the drawing does not have to be very accurate -- just enough to determine which angle is valid. It's not that big a deal.

PS. If you teach math, you will constantly need to remind students to check their answers in the original problem statement. It is fundamental.
 
  • #31
This is a common theme with teachers to remind students to never trust a diagram and eyeball an answer from it even though in the real world one might do that. But students being students will try to optimize their energy in such a pursuit.

I have been known to do that and then I learned to listen for clues from the teacher like this looks like a great quiz question, or subtlety when I was a student I got tricked by this. It's helped me on three separate occasions in college and a few times in high school too.
 
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  • #32
erobz said:
You see I can draw this triangle like this. It's nothing more than a typical mathematical construct.

Take for example this triangle

1743096349304-png.png


The objective is to find ##b## and ##\theta##. So most obviously you reach for the Law of Cosines for ##b##, and the law of sines for ##\theta##.
. . .
No, that's not so obvious.

You're right about first determining ##b##, and the Law of Cosines will do that for you.
But, if next, you're going to use the Law of Sines (followed by application of the arcsine function) to determine one of the remaining angles, you would be well advised to pick an acute angle. That angle is the unmarked angle in your figure, the angle opposite the side of length 42 . This angle is guaranteed to be less than ##90^\circ##. The largest angle is opposite the longest side, and no more than one of the angles of a triangle can be obtuse.

In short: Avoid using the arcsine to solve for the largest angle.
 
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  • #33
There was someone on PF who posted a trick interview question with an isoscelles triangle with each being 10m and the base being 20m asking for the angles.

It came with a triangle diagram labeled as such causing some confusion since given the side values its becomes clear its not a triangle when two sides sum equals the third side.

Who was that masked poster?
 
  • #34
erobz said:
I don't want the option to participate.
Yet you continue to participate. Grow up.
 
  • #35
You said @erobz,
"I've drawn a sloppy mathematical diagram of a square of side length a as I and almost everyone does".

No I dont. To draw a neat diagram is essential for thought, unless the problem is (trivially) simple. I go above and beyond to do it, or else I can't think. I suppose it will be less inportant for others. Yet, a diagram inaccurately drawn can make you suspect your own thinking. You could be correct and believe you aren't.

I insist on all students drawing accurate diagrams to questions, when lengths and angles are given. They lose a little mark if they do not.
 
  • #36
You can at least make the 42 side shorter than the 56 side can't you? That doesn't seem to be expecting too much.
 
  • #37
Let's debate on principal. Visually examine the triangle...Is it drawn incorrectly?

1744032067175.png
 
  • #38
brotherbobby said:
You said @erobz,
"I've drawn a sloppy mathematical diagram of a square of side length a as I and almost everyone does".

No I dont. To draw a neat diagram is essential for thought, unless the problem is (trivially) simple. I go above and beyond to do it, or else I can't think. I suppose it will be less inportant for others. Yet, a diagram inaccurately drawn can make you suspect your own thinking. You could be correct and believe you aren't.

I insist on all students drawing accurate diagrams to questions, when lengths and angles are given. They lose a little mark if they do not.
But it shouldn't be essential for the mathematics. This isn't about "marks"
 
  • #39
erobz said:
Let's debate on principal. Visually examine the triangle...Is it drawn incorrectly?

View attachment 359560
I'd say an inaccurate diagram serves the purpose of teaching students to "do the math" and not trust diagrams.
But in a real world scenario, an accurate-(ish) diagram is a good form of sanity check.

Somewhere out there on teh webz is a diagram of a triangle that is actually a degenerate straight line (altitude=0) but this fact is obscured by a diagram of an innocent-looking typical triangle.
 
  • #40
erobz said:
Let's debate on principal. Visually examine the triangle...Is it drawn incorrectly?
It's an order of magnitude better than the one you drew in post #6.
erobz said:
But it shouldn't be essential for the mathematics.
A human brain has two hemispheres. As I understand things, one side is more visually oriented and the other side is more analytically oriented. An inaccurate drawing is a hindrance in doing the mathematics, something that others in this thread have noted.
 
  • #41
Mark44 said:
It's an order of magnitude better than the one you drew in post #6.
You haven't answered. Is it incorrectly drawn?
 
  • #42
erobz said:
You haven't answered. Is it incorrectly drawn?
It looks fine. My comment was more about your drawing of post #6, which was obviously drawn incorrectly.
 
  • #43
Mark44 said:
It looks fine.
stay on post 37 for a moment... But its clearly incorrectly drawn - not fine. So I disagree. An accurate diagram would capture both triangles that satisfy the constraints.
 
  • #44
erobz said:
stay on post 37 for a moment... But its clearly incorrectly drawn - not fine. So I disagree. An accurate diagram would capture both triangles that satisfy the constraints.
I see nothing wrong with the triangle you drew -- you drew one triangle, and the sides you labeled are not clearly out of whack, unlike your figure of post #6.
The fact that there might be two distinct triangles with sides and one angle as shown is a consequence of the information given with the picture; namely, two sides and a non-included angle. It is very well-known that such triangles are not unique.

You're beating a dead horse here. Please stop.
 
  • #45
Mark44 said:
I see nothing wrong with the triangle you drew -- you drew one triangle, and the sides you labeled are not clearly out of whack, unlike your figure of post #6.
The fact that there might be two distinct triangles with sides and one angle as shown is a consequence of the information given with the picture; namely, two sides and a non-included angle. It is very well-known that such triangles are not unique.
So you check the ambiguity criteria's every time you use the law of sines to see which triangle you need...what happened; you said, "it was fine"? You had no idea from the diagram that it wasn't unique. The Law of cosines did.
 
  • #46
erobz said:
So you check the ambiguity criteria's every time you use the law of sines to see which triangle you need...what happened; you said, "it was fine"?
I have retracted part of what I wrote, both in my post and in what you quoted.
First off, the drawing is fine. Second, no, I don't check ambiguity criteria (BTW criteria already is the plural of criterion).

I would use the Law of Sines to find the angle across from the ##36.3^\circ## angle and would then find the remaining angle. This is the same advice @SammyS gave some time ago.

Since you are bringing nothing new to this thread, I'm labeling you done in it, again.
 

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