equation from graph from problem

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antiantianti

Joined Aug 4, 2016
45
Hi
I saw in a book that they got the equation A cos a = B cos b from the graph but I didnt understand how they figured it out from the graph
Thank you
 

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MrAl

Joined Jun 17, 2014
11,389
Hi
I saw in a book that they got the equation A cos a = B cos b from the graph but I didnt understand how they figured it out from the graph
Thank you
Hi,

Refer to the new drawing in the attachment.

It looks like they drew 'A' correctly but then drew the horizontal axis incorrectly.
Figure 1 is the original, figure 2 shows how 'A' is drawn with the arrowhead position more clear, and then figure 3 shows how we would extend the horizontal axis (in thicker red) because of the way 'A' is drawn with the position of the arrowhead.

Thus, the horizontal axis is longer than shown in your diagram. When it is drawn correctly, then you get the relationship you seek with the cos() functions. You actually do get this relationship, but the original drawing is misleading because of the way they drew the 'x' axis.

This is a guess, but is quite likely this is what happened. The horizontal line was drawn too short so it made it look like there was no relationship.

Actually if this was drawn the way we usually draw things, there would be no illusion because the horizontal axis would extend far to the right, in fact to infinity.
That would mean the relationship would hold by just drawing A, a, B, and b
(the correct axes would be already drawn before that).

Another way of looking at this is that A and B are drawn correctly and if we draw the axes as we normally do, there's no problem.
 

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Last edited:

WBahn

Joined Mar 31, 2012
29,979
The horizontal axis is infinitely long -- how long it is drawn is irrelevant. It is merely a reference to indicate that the angle 'a' is measured relative to the horizontal.

The key thing that is implied, but not explicit, in the graph is that the x-extent of both vector A and B are the same.

A better way to depict it would have been

upload_2017-4-2_22-30-39.png
 

MrAl

Joined Jun 17, 2014
11,389
I agree that's probably the point. Without the explicit verticals added by @WBahn, it's just a guess.
Hi,

Really? I think the drawings are subjective except for the fact that vectors are more general than that.
See attached drawing, and also note that the equation itself imposes certain implications about the drawing too (such as a single vertical at the end of each independently positioned "C").
 

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djsfantasi

Joined Apr 11, 2010
9,156
...Really? I think the drawings are subjective except for the fact that vectors are more general than that.
Huh? The drawing is very subjective. What if the vertex of angle a is on a different vertical line the the vertex of angle b. Or easier to imagine, the head of line A had a different x value than that line B?
It could be interpreted that way without the extra lines @WBahn added. In that case \(C_a\) would not be equal to \(C_b\). And the equation would be false.
 

MrAl

Joined Jun 17, 2014
11,389
Huh? The drawing is very subjective. What if the vertex of angle a is on a different vertical line the the vertex of angle b. Or easier to imagine, the head of line A had a different x value than that line B?
It could be interpreted that way without the extra lines @WBahn added. In that case \(C_a\) would not be equal to \(C_b\). And the equation would be false.
Hello,

In drawings you have to 'interpret' the meaning sometimes and so that makes it subjective.
It seems you may have interpreted my drawing unfavorably so you could up short.
In my drawing, if you take it exactly as is, you see that both lines labeled 'C' are the same length that's why they are both labeled "C" and not "Ca" and "Cb" which for some reason you decided to change, and they are approximately the same length as drawn also.
If you want to make up some other way to interpret the drawing that's up to you, but the intent of the author was that both lines "C" are the same length.
You can perhaps make up your own drawing and call each line or angle anything you want.
 

djsfantasi

Joined Apr 11, 2010
9,156
I am saying that there is nothing in the drawing that justifies your conclusion that \(C_a\) equals \(C_b\). IMHO.

And hence, how do you know the author's intent? I propose you are reading into the diagram just as I have, only we have come to different conclusions. Both of our interpretations are thus equally valid. Except for some reason you have decided differently.

Wayneh agreed with me. Without additional information added to the picture, it's just a guess.
 

MrAl

Joined Jun 17, 2014
11,389
I am saying that there is nothing in the drawing that justifies your conclusion that \(C_a\) equals \(C_b\). IMHO.

And hence, how do you know the author's intent? I propose you are reading into the diagram just as I have, only we have come to different conclusions. Both of our interpretations are thus equally valid. Except for some reason you have decided differently.

Wayneh agreed with me. Without additional information added to the picture, it's just a guess.

Hi again,

Oh ok, no problem, perhaps i misunderstood you.
All is well in AAC :)
 
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