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

[mcastillo356]

TL;DR

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