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

[Mark44]

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

In the post right after yours.

 

[erobz]

I had already found a way to make two positive solutions correcting the math. It happens when $c>a$.

The solutions I spot checked both correct.

 

[erobz]

In the post right after yours.

Here is your triangle(s).

I think this settles that the "Law of Cosines" dominates. As it doesn't appear to give any false solutions. What I mean is that it doesn't tell you that the angle outside of the triangle is an angle in the triangle.

If there are two triangles ( two positive solutions), then it gives both of the solutions, if there is one positive solution and one negative, toss the negative as we expect. two negative toss'em both. If there are no solutions the determinant will become negative. No triangles exist that fulfill the requirements. This happens just after 35 degrees in your example.

 

[Mark44]

I think this settles that the "Law of Cosines" dominates. As it doesn't appear to give any false solutions. What I mean is that it doesn't tell you that the angle outside of the triangle is an angle in the triangle.

It doesn't settle the comparison for me. For problems like the one I posed, the Law of Sines is simpler in that you don't need to solve a quadratic equation.

If there are two triangles ( two positive solutions), then it gives both of the solutions,

How do you figure? Your Desmos stuff shows just one solution for a, namely $\frac{\sqrt{10}}3$. There is another value for a, given the three parameters I showed.

With regard to your Desmos plot, what is the significance of the first line -- $y = 10^{125}(q - 15)$?
Also, what are lines 6 and 7 -- (15, 4.49...) and (15, 14.82...)? As far as I can tell, they have nothing to do with the question I asked.

 

[erobz]

If we had look at the blue triangle here as an example:

It doesn't settle the comparison for me. For problems like the one I posed, the Law of Sines is simpler in that you don't need to solve a quadratic equation.

Look at the blue triangle on the right?

lets say we know $a=3, c=5, \theta = 25^{\circ}$

The "Law of Sines" gives quite simply:

$$\frac{\sin(25^{\circ})}{3} = \frac{\sin \beta}{5} $$

$$\implies \sin \beta = \frac{5}{3}\sin(25^{\circ}) \approx 45^{\circ} $$

It gives you the corresponding angle in the red triangle!

A single solution???, Where is the other one (the blue triangle)...What if it was the one we are after? We don't even have an inkling it exists using the "Law of Sines"...

However, the "Law of Cosines" slaps you in the face and says " he buddy- wake up, just to let you know there are two solutions for this particular set of parameters and here they are, What's that? oh you want to vary the parameters...now there is just one again...you've gone to far - now there is none"

As we saw in the OP the "Law of Sines" says "hey here a solution, its wrong because none of the angles are what you required...but, if we just pretend that isn't concerning...the angle is the compliment of the angle you are searching for..."

With regard to your Desmos plot, what is the significance of the first line -- $y = 10^{125}(q - 15)$?

It’s a creative way to plot a ( very approximately) vertical line that I can shift around to get the solution at the angle. The program does the computation of the intersection point for you.

Also, what are lines 6 and 7 -- (15, 4.49...) and (15, 14.82...)? As far as I can tell, they have nothing to do with the question I asked.

They are the triangles that I tested would have two solutions, after I fixed the algebra.

EDIT: I see what you are saying about the points not corresponding to what you said. It wont make a difference to the argument. If you want to check it, just shift the vertical line over to 30. If you hover around the point of intersection it will pop up. in the right corner of the pop up you can add it to the plot. There is going to be two triangles just like in post 53 for your parameters when the angle is 30 deg.

 

[Mark44]

The "Law of Sines" gives quite simply:
$\frac{\sin(25^{\circ})}{3} = \frac{\sin \beta}{5}$
$\implies \sin \beta = \frac{5}{3}\sin(25^{\circ}) \approx 45^{\circ}$

No.
Your first equation results in $\sin(\beta) = \frac 5 3 \sin(25^\circ)$.

If you blindly use arcsin() on both sides, you get only one angle; namely $\beta \approx 44.78^\circ$.
OTOH, if you recognize that both an angle and its supplement have the same sine value, then you can state the other solution -- $\beta \approx 135.22^\circ$.

 

[Mark44]

However, the "Law of Cosines" slaps you in the face and says " he buddy- wake up, just to let you know there are two solutions for this particular set of parameters and here they are,

As you can see from my previous post, if you're only a little bit clever, you can see that the Law of Sines gives you both solutions.

As we saw in the OP the "Law of Sines" says "hey here a solution, its wrong because none of the angles are what you required...but, if we just pretend that isn't concerning...the angle is the compliment of the angle you are searching for..."

For the triangle you posted, the two solutions are:
1. a = 3, c = 5, $\theta = 25^\circ$ -- (the given information)
$\beta \approx 44.78^\circ, \gamma \approx 110.22^\circ$
I leave finding the third side as an exercise for you in using the Law of Sines.

2. a = 3, c = 5, $\theta = 25^\circ$ -- (the given information)
$\beta \approx 135.22^\circ, \gamma \approx 19.78^\circ$
As before, I leave finding the third side as an exercise for you in using the Law of Sines.

On a minor note, you can avoid using Greek letters for the names of angles by using capital letters for the angles, and lowercase letters for the sides. The usual way of doing things is that angle C is across from side c, and so on.

 

[erobz]

In the OP the solution directly given by the Law of Sines is:

Its a perfectly valid solution...until you realize the angle in the corner is inexplicably shifted to 56 degrees...when it needs to be 30.

The "Law of Cosine" handles it with ease (as I have already shown in post 19), and any other problem you throw at it. I’ve just eliminated the guesswork in falsifying a direct solution…by being a little bit clever.

 

[Mark44]

In the OP the solution directly given by the Law of Sines is:

You are completely missing my point. The Law of Sines gives you the sine of an angle. It's up to you to figure out which angle has this number as its sine, and also matches the diagram you've drawn for the problem. If you are so foolish as to think that there is only one angle with this sine (i.e., by taking the inverse sine of that number), then you will of course get only one angle that could very well be wrong for the problem you're trying to solve.
A trig problem where you are given two sides and an angle opposite one of them is well known to be ambiguous. Such problems are presented in any trig textbook
These are points I've made several times. How many more times do I have to repeat them before they sink in?

 

[erobz]

A trig problem where you are given two sides and an angle opposite one of them is well known to be ambiguous. Such problems are presented in any trig textbook

But they aren't ambiguous. I just solved them all unambiguously. How many times do I have to make that point before it sinks in?

And it solves more than just ambiguity of multiple triangle owning the fixed parameter $\theta$. It also throws out the solution that it is more than ambiguous, for Pete sake...the law of sines outright changed the fixed parameter $\theta$ in the OP...it will do that to me no more! If you choose to be tricked that’s your business.

 

[Mark44]

But they aren't ambiguous. I just solved them all unambiguously.

For the problem I posed, you showed one solution in post #53. Where was the other solution?

How many times do I have to make that point before it sinks in?

Only as many times until your point becomes true...

the law of sines outright changed the fixed parameter θ ...it will do that to me no more!

I have no idea what you're talking about regarding a fixed parameter. For the triangle at the right of your drawing in post #55, I showed both solutions in post #57. Both solutions used the same given information, the parameters of the problem, so what are you talking about when you say that the "law of sines changed the fixed parameter"?

 

[erobz]

I have no idea what you're talking about regarding a fixed parameter. For the triangle at the right of your drawing in post #55, I showed both solutions in post #57. Both solutions used the same given information, the parameters of the problem, so what are you talking about when you say that the "law of sines changed the fixed parameter"?

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?


View attachment 358853

Marcos

The law of sines tells us as a "solution" the angle $MLN$ is 77 degrees. But the resulting triangle ( though legitimate as a triangle) has no angle that is 30 degrees in it! Ill post it again...

The fixed parameter is the angle we have to work with $\theta = 30 deg$. The true solution must have it, and it does. The law of Sines gives us a solution that changes the angle from 30 degrees to 56 degrees without telling us! It is lying through its teeth to present us with a solution. (notice that $b$ has changed too - it is no longer 28 ish its 47 ish. If you hold the angle at 30 and try to construct the solution, the distance between the ships has changed, and $b$ as well.

Let me guess, with the Law of sines you always must take the angle outside of the triangle that the solution comes from...Are you really going to go down that path. The only way to rule out the non-sense is to check everything - every time. There is no reason to expect the criteria can't be met in general when the math is leading you on until it's all said and done.

I've never had a "true" solution alter a constant just to present us a solution. How many times have you fully checked a solution and found out one of the constants in it has inexplicably changed...and the math is correct? My guess is never.

It's the result of a patchwork. Pay no attention to the triangle, take the sine of the angle outside of it... has to be.

 

[FactChecker]

For the problem I posed, you showed one solution in post #53. Where was the other solution?

Only as many times until your point becomes true...

I have no idea what you're talking about regarding a fixed parameter. For the triangle at the right of your drawing in post #55, I showed both solutions in post #57. Both solutions used the same given information, the parameters of the problem, so what are you talking about when you say that the "law of sines changed the fixed parameter"?

The OP question was answered satisfactorily after a few posts and the last 40 to 50 posts have been about this issue. There has been no progress.

 

[Charles Link]

I think @erobz does make an interesting point though. On this one, a good close to scale drawing tells you which is the correct answer, but what if the angle turns out to be very near 90 degrees? The law of sines will give two results, and you won't be able to tell which is the correct one without going to the law of cosines.

(i.e.You'll at least need to check whether $c^2> a^2+b^2$ or if $c^2<a^2+b^2$)

 

[Mark44]

The law of sines tells us as a "solution" the angle MLN is 77 degrees. But the resulting triangle ( though legitimate as a triangle) has no angle that is 30 degrees in it! Ill post it again...

This wikipedia article explains the ambiguous case pretty well, which can occur when you are given two sides and a non-included angle - https://en.wikipedia.org/wiki/Law_of_sines - see the section titled "The ambiguous case of triangle solution".

Let me guess, with the Law of sines you always must take the angle outside of the triangle that the solution comes from...

No.
Here are the conditions (from the site I linked to:

Given a general triangle, the following conditions would need to be fulfilled for the case to be ambiguous:

The only information known about the triangle is the angle α and the sides a and c.
The angle α is acute (i.e., α < 90°).
The side a is shorter than the side c (i.e., a < c).
The side a is longer than the altitude h from angle β, where h = c sin α (i.e., a > h).

The figure in the article has labels for all sides and all angles.

 

[erobz]

This wikipedia article explains the ambiguous case pretty well, which can occur when you are given two sides and a non-included angle - https://en.wikipedia.org/wiki/Law_of_sines - see the section titled "The ambiguous case of triangle solution".

No.
Here are the conditions (from the site I linked to:

The figure in the article has labels for all sides and all angles.

What I am talking about is not ambiguity. The triangle that the solution directly gives to the OP Does Not Exist!

Try to construct this solution for the OP - check to see that everything holds true mathematically. The parameters in red are fixed by the problem statement. You are going to find that it can't be done! It's not an ambiguous result, it's a "no result"!

Ambiguity is when there are two valid results for the same constraints. These triangles (below) are ambiguous. The OP is not ambiguous.

 

[Mark44]

What I am talking about is not ambiguity. The triangle that the solution directly gives to the OP Does Not Exist!

Try to construct this solution for the OP - check to see that everything holds true mathematically. The parameters in red are fixed by the problem statement. You are going to find that it can't be done! It's not an ambiguous result, it's a "no result"!

Your drawing is not to scale. The side you labeled as 42 in your triangle appears to be a lot longer than the side labeled 56! Obviously it's important to work with drawings as realistic as possible. Here's a more realistic drawing based on the problem in the OP.

Note that this triangle, with its given data of an angle and two sides that don't include the angle, satisfies the conditions listed in the Wikipedia article I cited. We are very much in the situation of ambiguous triangles. Your major complaint about finding a solution that is a triangle that does not include the given angle ($30^\circ$ in this case) is valid but doesn't take into account the ambiguity involved here. If the given angle had been angle LNM rather than angle LMN, that angle would have been part of both triangles.

In my drawing, L and L' are the two possible locations of the lighthouse. I used the Law of Cosines to find LN, the distance from the lighthouse to ship N, with $LN \approx 28.74$ km.
I then used the Law of Sines to find $\theta$ (angle LNM), with $\theta \approx 46.94^\circ$.

Finding the third angle is tricky, as we're dealing with ambiguous triangles. Let $\alpha$ = angle MLN (the obtuse angle), and let $\alpha'$ = angle ML'L (the acute angle). Per the same wikipedia article, $\alpha' = \sin^{-1}\left(\frac{56\sin(\theta)}{42} \right)$ and $\alpha = 180 - \alpha'$.

My results for these are $\alpha' \approx 76.95^\circ$ and $\alpha \approx 103.05^\circ$.

 

[erobz]

Your drawing is not to scale. The side you labeled as 42 in your triangle appears to be a lot longer than the side labeled 56! Obviously it's important to work with drawings as realistic as possible.

You are making my point. Using the Law of Cosines (as I have clearly demonstrated), it is not important to work with realistic drawings of triangles. The actual solution(s) show up no matter what I did on the paper...other than the mathematics. If you want to draw triangles to scale, check the four criteria are satisfied, and then rule out a potentially bonkers (false) solutions all the time... be my guest; don't solve a quadratic!

 

[erobz]

My results for these are $\alpha' \approx 76.95^\circ$ and $\alpha \approx 103.05^\circ$.

And the first is not a solution. Which you will not generally know until you have verified it properly.

 

[Charles Link]

Please see my post 64. @erobz does make an interesting point here that the law of sines, which you would expect to supply you a simple and straightforward answer, does not do that for this case. It is a similar problem with the cross product of two vectors. The cross product only tells you the sine of the angle between the vectors, and there are two solutions. The dot product gives a single solution for the angle between the two vectors.

 

[Mark44]

From post #55:

The "Law of Sines" gives quite simply:
$$\frac{\sin(25^{\circ})}{3} = \frac{\sin \beta}{5} $$
$$\implies \sin \beta = \frac{5}{3}\sin(25^{\circ}) \approx 45^{\circ} $$

Your 2nd equation above has an error: it says that $\sin \beta \dots \approx 45^\circ$. This equation confuses the sine of an angle with the angle itself.

The Law of Sines does not give you the angles -- it is defined in terms of the lengths of two sides and the sines of two angles. It therefore gives you the sines of the angles, an important distinction. It is then up to you to determine what the angle is, keeping in mind that the sine function is not one-to-one. This means that both $\beta$ and $180 -\beta$ have the same sine value.


From post #66:

Using the Law of Cosines (as I have clearly demonstrated), it is not important to work with realistic drawings of triangles.

In your drawing in post #66, did you use the Law of Cosines to find the angle opposite the 56 side (shown as $76.9699^\circ$)? If so, that is not the correct angle for the problem.
Having a realistic drawing would have led you to the correct value for that angle, which is about $103^\circ$, which is the supplement of the angle you found. It can't be emphasized enough that in so many areas of mathematics, making a realistic drawing is a crucial step in the solution process.

The actual solution(s) show up no matter what I did on the paper...other than the mathematics.

Doesn't seem to be the case with regard to the problem stated in the OP.

You are, of course, free to choose which tool to use to solve a problem, but the important thing is to get the right answer.

Since the original problem of this very long thread has been asked and answered, I'm closing this thread. If someone has something to add that hasn't already been said, please DM me and I will consider reopening the thread.