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

[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.
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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.


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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.

 

[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.