B How is it that law of sines does not work in this exercise?

[mcastillo356]

TL;DR Summary

It computes the supplementary

Exercise statement :

From a lighthouse you can see two boats. Ship $M$ is forty-two kilometers out to sea, and ship $N$ is at an unknown distance. If the distance between the ships is fifty-six kilometers and the angle formed in $M$ is thirty grades, how many kilometers are there from the lighthouse to $N$, and which is the angle formed in the position of the headlamp?

1) $$b^2=a^2+c^2-2ac\cos{(\theta)}=826.336\Rightarrow{b=\sqrt{826.336}}=28.746$$

2) $$\cfrac{28.74}{\sin{(30^o)}}=\cfrac{56}{\sin{(x)}}\Rightarrow{x=77}$$

As shown at the Geogebra file, the angle is $103^o$

Also, where is the guiding light at the sketch drawn?




Marcos

 

[jedishrfu]

You mentioned 30 grades and not degrees. They are not the same. There are 400 grads in a circle vs 360 degrees in a circle.

Could that be the issue?

Converting a grad to a degree:

$grads * (360/400) = degrees$

 

[Mark44]

You mentioned 30 grades and not degrees.

Clearly he meant "degrees" as the equations show the degree symbol. The Spanish equivalent of "degrees" is "grados," which he likely mistakenly wrote as "grades."

Could that be the issue?

No.

 

[FactChecker]

Your equation looks good, but there are two solutions for $x$. $\sin(77)=sin(103)$.
You should also check on @jedishrfu 's comment.

 

[Mark44]

TL;DR Summary: It computes the supplementary

From a lighthouse you can see two boats. Ship M is forty-two kilometers out to sea, and ship N is at an unknown distance. If the distance between the ships is fifty-six kilometers and the angle formed in M is thirty grades, how many kilometers are there from the lighthouse to N, and which is the angle formed in the position of the headlamp?

The plot you showed isn't very helpful, as it doesn't label the points where the ships are or where the lighthouse is (headlamp is a small light that you can wear on your head) or which angle is which. The Spanish equivalent of "lighthouse" is "faro." "Head lamp" is "lampara frontal.

Also, what does "the angle formed in M" mean? 30° relative to what?

In answer to your summary, the thing about the sine function is that both an angle and its supplement have the same sine value. IOW, $\sin(\theta) = \sin(180° - \theta)$ so using the Law of Sines can sometimes give misleading results if you aren't working from a reasonable drawing.

 

[mcastillo356]

$grads * (360/400) = degrees$

This is something I want to work too. I really don't know how to measure angles, not even the equivallences. Thanks.
I've taken a quick look to the responses, everyone amazingly clever. Now I'm going to sleep, @FactChecker might have given a clue. Tomorrow I will read everything more steadily.
Marcos

 

[FactChecker]

This is something I want to work too. I really don't know how to measure angles, not even the equivallences. Thanks.

Don't feel too bad. I never heard of 'grad' before as a measure of angle and had to look it up.

 

[Klystron]

Since it is not stated in the problem and the distances are relatively small, the computation ignores curvature of the Earth. I was thinking of this because of another thread requiring greater distances in which Law of Sines provided only an approximate solution due to Earth curvature.

 

[Mark44]

I never heard of 'grad' before as a measure of angle and had to look it up.

Some calculators allow you to work with angles in degrees, radians, or grads. Grads are used in Russia, or at least were in the past, along with possibly some other Eastern European countries.

 

[jtbell]

Gradian (Wikipedia) a.k.a. gon, grad, or grade.

I've never used them in calculations, myself, but my ancient HP-11C calculator can be set to work with them. Press g then GRD, then GRAD appears at the bottom of the display to remind you of the setting.

 

[mcastillo356]

The plot you showed isn't very helpful, as it doesn't label the points where the ships are or where the lighthouse is

(headlamp is a small light that you can wear on your head) or which angle is which. The Spanish equivalent of "lighthouse" is "faro." "Head lamp" is "lampara frontal.

Some well rated translators misleaded me.

Also, what does "the angle formed in M" mean? 30° relative to what?

Relative to... Lighthouse is on the left side of this late sketch.
Attempt :
It's something like this: the rules are two: Earth is plane. I can imagine @Klystron, and everyone open-mouthed; and is static, not moving or changing.

In answer to your summary, the thing about the sine function is that both an angle and its supplement have the same sine value. IOW, $\sin(\theta) = \sin(180° - \theta)$ so using the Law of Sines can sometimes give misleading results if you aren't working from a reasonable drawing.

$\sin{77^o}=\sin{180^o-77^o}$ Why the second one? The exercise is got not frame of reference, since it is static (lighthouse should be, but ships don't move); this is, I choose $M$ boat, because this way everything fits.

 

[FactChecker]

View attachment 358864
$\sin{77^o}=\sin{180^o-77^o}$ Why the second one? The exercise is got not frame of reference, since it is static (lighthouse should be, but ships don't move); this is, I choose $M$ boat, because this way everything fits.

When your equation has multiple solutions, you have to see which solutions fit the original problem.

 

[erobz]

IOW, $\sin(\theta) = \sin(180° - \theta)$ so using the Law of Sines can sometimes give misleading results if you aren't working from a reasonable drawing.

What is unreasonable about the drawing though? It looks like every application that I've ever seen. I was unaware that it could do this.

 

[Mark44]

What is unreasonable about the drawing though?

There are no indications of where the two ships are, where the lighthouse is, angle measures, or side lengths. I suppose one might take some time to reason where all these things are, but I didn't feel like taking that time.

 

[FactChecker]

What is unreasonable about the drawing though? It looks like every application that I've ever seen. I was unaware that it could do this.

I didn't like that none of the labels matched the description or the equation. In fact, no labels in the description, equation, or diagram match any other. I didn't even try to figure it out. It's not my job.

 

[erobz]

I see that in an my pre-calc text, the author does indeed talk about funny things happening in the Law of Sines. I certainly didn't remember it, and I can't believe I haven't stumbled on to it. It seems a little bit too "iffy" to be given the status of "Law". I thought we would expect that you can't mess it up, by drawing it so long as you made sensible a triangle.

 

[Mark44]

It's something like this: the rules are two: Earth is plane. I can imagine @Klystron, and everyone open-mouthed; and is static, not moving or changing.

I suppose you mean that the problem assumptions are that the Earth can be considered to be a plane, and that the positions of the ships are static. The first is reasonable if the distances are not too large, but for the distances involved, any calculations will be off a bit.

$\sin{77^o}=\sin{180^o-77^o}$

Looking at the LaTeX you wrote I can see that you meant $\sin(180° - 77°)$, but as you wrote it, with braces instead of parentheses, it looks like ##$\sin(180°) - 77°$.

Why the second one?

I'm not sure I understand your question. My point was that if you have an equation $\sin(x) = <value>$, you can't just blindly take the arcsin of <value> to get an angle, since both x and 180° - x have the same sine values. (Not to mention that there are an infinite number of values x for which sin(x) = <value>.)

 

[erobz]

The Law of Cosines applied twice gives the correct solution if you take the positive root for the distance between the lighthouse to the ship at $N$. Which is consistent with the definition of distance. It gives the false solution of 77° if you say the distance is $- \sqrt{~~~}$.

I vote to demote the Law of Sines to a "soft rule".

 

[erobz]

With the Law of Cosines we find that:

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

$$ \implies b = \sqrt{a^2 + c^2 - 2 a c \cos \theta} $$

Now apply again:

$$ c^2 = a^2 + b^2 - 2 a b \cos \beta $$

$$ \implies \cos \beta = \frac{b^2 + a^2 - c^2}{2ab} $$

Substitute in for $b^2$ , and $b$:

$$ \implies \cos \beta = \frac{ [a^2 \cancel{+ c^2} - 2 a c \cos \theta]+ a^2 \cancel{- c^2}}{2a\sqrt{a^2 + c^2 - 2 a c \cos \theta}} $$

$$ \cos \beta = \frac{ 2a^2 - 2 a c \cos \theta }{2a \sqrt{a^2 + c^2 - 2 a c \cos \theta} } $$

$$ \cos \beta = \frac{\cancel{2a}( a- c \cos \theta) }{\cancel{2a} \sqrt{a^2 + c^2 - 2 a c \cos \theta} } $$



$$ \boxed{ \beta = \arccos \left( \frac{ a- c \cos \theta }{ \sqrt{a^2 + c^2 - 2 a c \cos \theta} } \right) } \tag{General Solution}$$

for $a = 42, c= 56, \theta = 30^{\circ}$

$$ \beta = \arccos \left( \frac{ 42- 56 \cos 30^{\circ} }{ \sqrt{42^2 + 56^2 - 2 \cdot 42\cdot 56 \cos 30^{\circ}} } \right) \approx 103^{\circ} $$

A single correct solution...

EDITED.

 

[FactChecker]

Much better.
One problem here is that the choice of $103^\circ$ is due to the specific distances given in the problem. With other distances, $77^\circ$ would be correct. So it is important to show those distance numbers in the diagram.
Making a clear presentation is a learned skill. It's a thankless job.

 

[erobz]

Much better.
One problem here is that the choice of $103^\circ$ is due to the specific distances given in the problem. With other distances, $77^\circ$ would be correct. So it is important to show those distance numbers in the diagram.
Making a clear presentation is a learned skill. It's a thankless job.

I don't follow. 77 is a false answer for the parameters given in the problem statement, found by way of the application of a flawed “law”. Saying that the answer could be 69 for some mix of parameters would be just as true. The algebraic result is a general one. I'll plug in numbers on another line if that is what you are getting at.

 

[erobz]

Notice this is what you get if you construct the solution given by the "Law of Sines"

Our parameter for $\theta$ is nowhere to be found. Its application has re-written the problem statement...

 

[FactChecker]

I don't follow. 77 is a false answer for the parameters given in the problem statement,

Yes, but none of those parameter values are given in the diagram, so the diagram and following algebra would be appropriate for both answers, $77^\circ$ and $103^\circ$. If you want a diagram that is only appropriate for $103^\circ$ you must put the parameter values on the diagram.

 

[erobz]

If you want a diagram that is only appropriate for $103^\circ$ you must put the parameter values on the diagram.

I want a diagram that is as general as it can be. It's understood that $L$ is the lighthouse, $M$ is ship a vertex, $N$ is another ship the remaining vertex, and the angle formed within $M$ ( the vertex) is $\theta$. There is no solution in which $\beta$, is $77°$ given the parameters in the problem statement and applying the Law of Cosines won't produce it. The fact that someone could change up all the values crafting a new problem is irrelevant. I've EDITED the solution earlier to include a general solution in the box, the parameters for the problem substituted on the next line.

There is nothing left to discuss, unless it's about demoting the "Law of Sines".

 

[FactChecker]

I want a diagram that is as general as it can be.

This was not a general math question; it was a question about why $77^\circ$ was the wrong answer. When solving a particular problem, the diagram is most helpful if it displays the problem in question, not a general situation.

 

[FactChecker]

I want a diagram that is as general as it can be.

This was not a general math question; it was a question about why $77^\circ$ was the wrong answer. When solving a particular problem, the diagram is most helpful if it displays the problem in question, not a general situation.

There is nothing left to discuss, unless it's about demoting the "Law of Sines".

You have a good point the the law of sines is trickier than some think, but the law of sines is still perfectly valid. We understand that the sine function is not one-to-one.

 

[erobz]

This was not a general math question; it was a question about why $77^\circ$ was the wrong answer. When solving a particular problem, the diagram is most helpful if it displays the problem in question, not a general situation.

I feel it is a problem about why the law of sines does this as a properly applied law.

You have a good point the law of sines is trickier than some think, but the law of sines is still perfectly valid. We understand that the sine function is not one-to-one.

As far as I’m concerned from now on the law of sines becomes inseparable from the “law of potentially lying to you”. Having to check the criteria that it will produce a valid solution, where it doesn’t self manipulate the parameters in its application renders its use no more efficient than the alternative at best (its main claim to fame).

“an oath to a liar is no oath at all”

 

[FactChecker]

I feel it is a problem about why the law of sines does this as a properly applied law.

The law of sines, as stated, is correct. Its application may be a little tricky.

As far as I’m concerned from now on the law of sines becomes inseparable from the “law of potentially lying to you”. Having to check the criteria that it will produce a valid solution, where it doesn’t self manipulate the parameters in its application renders its use no more efficient than the alternative at best (its main claim to fame).

Mathematics is full of multiple "solutions" that must be checked against the original problem statement to see which are correct. This is nothing new.

 

[Mark44]

I feel it is a problem about why the law of sines does this as a properly applied law.

As far as I’m concerned from now on the law of sines becomes inseparable from the “law of potentially lying to you”.

The fact that there can be answers that aren't applicable to a specific problem is less about the Law of Sines and more about the fact that the sine of an angle is equal to the sine of the angle's supplement.

If you're given an angle and the two sides adjacent to it, you can always use the Law of Cosines to find the other angles and sides of a triangle. However, if you are given an angle and two sides, with one of the sides adjacent to the angle and the other opposite the angle, there are two possible triangles that can be formed.

In the drawing below you're given angle B and sides a and b of the two triangles. If you use the Law of Sines to find angle A, you will need to choose whether the acute angle (as in the triangle on the right) or the obtuse angle (as in the triangle on the left) is the appropriate solution. This is why having a drawing with the sides and angles labeled is important.

 

[erobz]

The law of sines, as stated, is correct. Its application may be a little tricky.

Mathematics is full of multiple "solutions" that must be checked against the original problem statement to see which are correct. This is nothing new.

This scenario is worse. It gives a single solution that is incorrect - sometimes. The fact that we can patchwork it to an angle that doesn't exist in our triangle is of no comfort (to me). It gives an incorrect result. period

 

[erobz]

The fact that there can be answers that aren't applicable to a specific problem is less about the Law of Sines and more about the fact that the sine of an angle is equal to the sine of the angle's supplement.

If you're given an angle and the two sides adjacent to it, you can always use the Law of Cosines to find the other angles and sides of a triangle. However, if you are given an angle and two sides, with one of the sides adjacent to the angle and the other opposite the angle, there are two possible triangles that can be formed.

In the drawing below you're given angle B and sides a and b of the two triangles. If you use the Law of Sines to find angle A, you will need to choose whether the acute angle (as in the triangle on the right) or the obtuse angle (as in the triangle on the left) is the appropriate solution. This is why having a drawing with the sides and angles labeled is important.
View attachment 358911

Yeah, there is ambiguity. But I have never had the math alter a fixed parameter before, to give me an unrelated solution of its choosing. It's clearly lesser in the hierarchy than the "Law of Cosines". There is no good reason to use a law that can unpredictably lie to you. If this were but a step in a larger problem and you accept it (it was a perfectly fine triangle after all), it could be a huge waste of time.

 

[Mark44]

This scenario is worse. It gives a single solution that is incorrect.

Likely this is due to blindly using arcsine to find an angle. This is akin to using the square root to solve the equation $x^2 = 4$ and getting 2 as an answer. Of course, 2 is a solution to this equation, but so is -2.

If the question is, find the solution(s) of $\frac{x^2 - 4}{x - 2} = 0$ and you multiply both sides by x - 2, getting $x^2 - 4 = 0$ or $x^2 = 4$, and then blindly take the square root of both sides to get x = 2, that solution is incorrect because it is not a solution of the equation we started with.

Sometimes in mathematics you need to have your wits about you...

 

[erobz]

However, if you are given an angle and two sides, with one of the sides adjacent to the angle and the other opposite the angle, there are two possible triangles that can be formed.


View attachment 358911

And I suspect the Law of Cosines will give you both.

 

[Mark44]

And I suspect the Law of Cosines will give you both.

No. Using the Law of Cosines you need two sides and the angle included between them.
$c^2 = a^2 + b^2 - 2ab\cos(C)$. Here C is the angle between sides a and b. c is the side opposite angle C.

 

[erobz]

No. Using the Law of Cosines you need two sides and the angle included between them.
$c^2 = a^2 + b^2 - 2ab\cos(C)$. Here C is the angle between sides a and b. c is the side opposite angle C.

I can repeatedly apply the Law of Cosines and find the solutions. You have to solve a quadratic for the unknown side (yielding two solutions - from which we could eliminate the violator if it exist - negative distance). The rest is substitution into the developed equations. No invocation of Law of Sines. Take it for a spin.

 

[Mark44]

I can repeatedly apply the Law of Cosines and find the solutions. You have to solve a quadratic for the unknown side (yielding two solutions - from which we could eliminate the violator if it exist - negative distance).

I don't think it's as simple as you describe. If you look at my drawing in post #29, the two solutions you would get for the missing side would both be positive. In that case you're in the same boat as if you had used the Law of Sines. IOW, you would have to choose which of the potential solutions matched the problem you're trying to solve.

 

[erobz]

I don't think it's as simple as you describe. If you look at my drawing in post #29, the two solutions you would get for the missing side would both be positive. In that case you're in the same boat as if you had used the Law of Sines.

I don’t think that will be a problem. I said “if “ there is a solution that violates by giving a negative distance... From what I see there may be two positive valid solutions from this approach.

 

[Mark44]

I don’t think that will be a problem. I said “if “ there is a solution that violates by giving a negative distance... From what I see there may be two positive valid solutions from this approach.

And in all likelihood, it will be the case that both solutions for the missing side are positive.

 

[erobz]

And in all likelihood, it will be the case that both solutions for the missing side are positive.

Doesn't look like it.

 

[Mark44]

doesn't look like it.

For ambiguous cases such as the triangles in the drawing of post #29, both solutions will be positive. Using the Law of Cosines you will still need to determine which of the two solutions matches your problem.

 

[erobz]

For ambiguous cases such as the triangles in the drawing of post #29, both solutions will be positive. Using the Law of Cosines you will still need to determine which of the two solutions matches your problem.

If true, I would still rather that be the case than have it outright give only the false solution. The Law of Cosines is superior.

 

[Mark44]

The Law of Cosines is superior.

Each of these tools is useful. To say that one is superior to the other is just silly. It's a little like the old saying, "If the only tool you have is a hammer, everything starts to look like a nail."

 

[erobz]

To say that one is superior to the other is just silly. ..."If the only tool you have is a hammer, everything starts to look like a nail."

I disagree. Which is why I said a very good saying "an oath to a liar is no oath at all".

 

[FactChecker]

Yeah, there is ambiguity. But I have never had the math alter a fixed parameter before, to give me an unrelated solution of its choosing. ... There is no good reason to use a law that can unpredictably lie to you.

If you are asked to solve $\sin(x)=0$ for $x$ and you say that $x=0$ with no other conditions or explanation, then you are just plain wrong. The correct answer is $x=l\pi, l\in \mathbb{I}$. None of those values are "lies".

 

[erobz]

If you are asked to solve $\sin(x)=0$ for $x$ and you say that $x=0$ with no other conditions or explanation, then you are just plain wrong. The correct answer is $x=l\pi, l\in \mathbb{I}$. None of those values are "lies".

The problem we are talking about the only solution found is a "lie"! It changes the parameter $\theta$ under our noses. This Is the lie it tries to tell us. The patchwork that tries to say the angle outside of the triangle is the relevant one for the triangle is ridiculous!

The law of Cosines gives us both the lie and the truth (when it ambiguous). The only reason the OP could tell something had misfired was because they drew it out! If the equation correctly spits out two solutions (what I propose) the OP would have immediately been curious. No drawing required. For every one that figures it out, there are a 1000 that will forever have no idea why their parameter somehow changed before there very eyes. It's a parameter for pete sake. Why would anyone expect it to be the variable they didn't agree to?

The equation is outright lying without warning! If it looks like crap, smells like crap, and tastes like crap...Its crap!

 

[erobz]

I take back what I said, they are both behaving in unsensible ways. They apparently give you answers that aren't exactly what you want!

This is the figure I work from.

Here are the solutions for some set of parameters:



One triangle, the one with a positive solution- the 20 degree angle - nowhere to be found.

The other triangle, corresponding to the absolute value of the negative root, gives an angle of 20 degrees in the triangle (perhaps by dumb luck), but it's the wrong angle!

 

[erobz]

The angle of 20 being in one of the triangle was a coincidence. If you set $q= 120$ you get these two triangles.

It gives you two triangles it can make with the side lengths 7 and 4, and completely ignores the parameter $\theta$ you tried to suggest. I find this a bit fascinating...

 

[erobz]

Its the gift that keeps on giving, if I flip $c= 4$, $a=7$ and keep $q = 120$ I get more triangles!

How is it picking these triangles? Somehow these 4 triangles are associated with the parameter 120°

We know there are an infinite number of triangles that have one side length of 4 and one of 7. Its plucked 4 of them from the abyss.

So its decidedly worse than the law of Sines, but was somehow better for the OP. Pardon me while I go jump off a bridge...

 

[erobz]

I think something in Desmos is messing up computationally, or how I've entered it there.
EDIT: I have an algebra mistake in the Desmos calculator.



My TI 89 for the parameters $a=7,c=4,\theta = 120^{\circ}$

$$ 7^2 = x^2 + 4^2 -2 \cdot 4 \cdot x \cdot \cos ( 120^{\circ} )$$

$$ 49 = x^2 + 16 -8 ~x~\left( \frac{-1}{2} \right) $$

$$ x^2 + 4x -33 = 0 $$

$$ x = \frac{-4 \pm \sqrt{16+4 \cdot 33}}{2} = -2 \pm \frac{1}{2}\sqrt{148} \approx [4.083, -8.083] $$

The positive root draws the triangle I expect.

Does someone have a case where the result where will give two positive roots like @Mark44 suggests?

 

[Mark44]

The problem we are talking about the only solution found is a "lie"! It changes the parameter θ under our noses. This Is the lie it tries to tell us.

The law of Cosines gives us both the lie and the truth (when it ambiguous).

You keep saying that the solutions found are lies. What are you talking about? It is very well known that given only some information about a triangle, it might not be possible to determine a unique triangle.

Here's an example of what I'm talking about.

The given information is that side c = 10, angle B = 30°, and side b = $\frac {10}{\sqrt 3}$ are given. (I used Paint to create the drawing, so I couldn't include the value for b in the drawing, as Paint's text feature is very limited.)
Use the Law of Cosines to find the length of side a (the unlabeled side at the base of the triangle). Hint: there are two solutions, both of which are positive.
Bonus question: Is one of the solutions a "lie"?

I take back what I said, they are both behaving in unsensible ways. They apparently give you answers that aren't exactly what you want!

I'm not sure what exactly you are taking back. If you believe that both (i.e., Law of Sines and Law of Cosines) are behaving insensibly, then that means that you don't understand how some sets of triangle parameters lead to solutions that are unique while other sets of parameters can lead to two different triangles or possibly none at all. For some reason you seem to be adamant in thinking that a drawing of the triangle is unimportant or even unnecessary, but that's the only way you can decide which of two possible triangles is the one you're after.

Regarding the plot you showed in post #46 that is labeled "Law of Cosines," I have no idea what the two blobs are supposed to convey.