## Tuesday, October 29, 2013

### The Law of Sines

The other trig teacher at my school came to me with a problem today.  I, too, am stumped as to how to explain *why* the Law of Sines doesn't work to solve this problem correctly.  After getting two answers I can see why, I just wouldn't have anticipated this issue a priori.  I'm quite embarrassed to ask, but if anyone can offer help beyond "start with the Law of Cosines, stick with the Law of Cosines", I'm listening.

Here's the problem:
Start with the triangle shown, given only the 3 sides, and solve for angle A.

Now use the Law of Sines to solve for angle B, then calculate angle C.

Here's where the issue comes.  Go back to the original triangle having solved for angle A, and instead of solving for angle B next, solve for C instead, and then calculate B.

Here's what I get.  Solving for B first is in blue, solving for C first is in red:

There's only 1 right answer, and it's the blue one.   The Law of Sines ratio doesn't work for all three pairs of angles/sides in the red triangle.

So the question is this: doing an arcsin won't give you the correct answer if your angle is in the 2nd quadrant (because the range of the arcsin function is (-90 degrees, 90 degrees).  How would you know, then, to check for an obtuse angle in this case?  Perhaps the better question is, without checking the Law of Sines ratio for all three pairs, would one know that the red answer is incorrect?

Have I done something procedurally incorrect?  Again, any help is appreciated.

Auntie Ann said...

A bit surprising, but I think Wikipedia actually has the answer. They refer to the ambiguous case, which looks like it might apply here, and there's a nice drawing to show what they mean: http://en.wikipedia.org/wiki/Law_of_sines

They say:

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 A and the sides a and c
-- The angle A is acute (i.e., A < 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 B, where h = c sin A (i.e., a > h).

Anonymous said...

I think the issue is that the law of sines can produce ambiguity, because different angles can have the same sines. This might help (find where it discusses the ambiguous case).
http://www.themathpage.com/aTrig/law-of-sines.htm

PeggyU said...

You can solve for C first, as long as you recognize that it must be obtuse. :) Given the fact that 15^2 > 13^2 + 7^2, C must be obtuse. Therefore, B must be less than 30 degrees.

So, I would think of looking at the sides and using the Pythagorean theorem first to see if you have an obtuse triangle or an acute triangle, and then use the Law of Sines. But maybe that's what you did already!

Stephen Su said...

Mr. Miller,

I believe that the error here is that when solving using the Law of Sines, you are only using Angle A, Side a, and Side b, using 3 elements that constitute the ASS ambiguous case of the Law of Sines.

Law of Cosines: When using this Law, you are using 3 sides, so you get one unique case.

Law of Sines: When using this Law, you are using 2 angles and a side, making up the ASS ambiguous case.

Qs and As:

How would you know, then, to check for an obtuse angle in this case?
The reason you would know to do this is because you are using the ASS congruence theorem in order to complete the Law of Sines, so you have to make sure that there is only one answer.

Perhaps the better question is, without checking the Law of Sines ratio for all three pairs, would one know that the red answer is incorrect?
We know the blue one is correct because using the 180-angleB method of calculating if there is an alternate triangle, you realize that there is only one possible triangle. However, if we do the same thing with the red one, then there are 2 possible triangles. Since the original problem dealt with the SSS, then we know that there is only one possible solution, the blue one.

~Stephen Su

maxutils said...

No problem. The law of sines normally gives a definitive answer, save for the ambiguous case -- where you are give two acute angles and a non included side. Then there are two possibilities, due to the fact that you can have the same reference angle with the same sine value in quadrants 1 and 2. Cosine is different, as you get a positive in 1, and a negative in 2. So ... when given 3 sides and no angles, the easy advice is use law of cosines, because it's all you can do. But the second thing is to eliminate the possibility of a sine ambiguity ... so, choose the angle opposite the longest side ... that's the ONLY place that an obtuse angle could exist... if it's obtuse, fine; if it's not then there can be no sine ambiguity because you've already determined that there can be no obtuse option.

Anonymous said...

Hi Darren, That's a tricky one.

Students should always check the supplement of the angle measure they get from the law of sines. In your blue above, the initial answers for angle B should be 28o or 152o, in which case the 152o can be eliminated as it's too large. This would yield 92o for the measure of angle C.

If a student proceeded first with the scenario you have in red, then their answer from the law of sines for the measure of angle C should be 88o or 92o, in which case both are feasible and must be confirmed or eliminated with the law of cosines.

Joshua Sasmor said...

The fact that you got two cases different triangles is a completely accurate result from the law of sines. The trig book (College Algebra and Trigonometry, by Aufmann, Barker and Nation) I use refers to this as the "ambiguous case". Both of these triangles are possible! If you google "law of sines ambiguous" you'll get a lot of explanations, included a few illustrated and app driven ones.

PeggyU said...

I usually "test" the sides first, before solving, to see if the triangle is acute, right, or obtuse. So since 15^2 > 13^2 + 7^2, C > 90. I think I sent a comment to that effect, but I'm kind of a flake sometimes, and it's possible I left the page without submitting it! :)

Anonymous said...

Probably this is already obvious to you but...
If you consider the sine function, it's only single-valued at 90 degrees (over the range 0-180 degrees, ie. possible angles in a triangle.) Since arcsin gives answers only over the [-90, 90], you'll get the correct answer for an acute angle but not for an obtuse one. So, in practice, avoid using the law of sines when the angle has an opposite side that is relatively long - pick the angle that is more likely to be acute and start there.

Anonymous said...

As you realized, the Law of Sines always gives two solutions for the angle -- both x and 180-x solve sin x = a for 0<=a<=1 (unless x=90 degrees, of course!). In your blue case, the second answer (B=152 degrees) is clearly wrong since we know A=60 degrees. In the red case, you can't choose between C=92 degrees and C=88 degrees without checking the third ratio.

Procedurally, if you're solving for both remaining angles, you can avoid the problem by using the ratio involving the shorter remaining side of the triangle (angle B in your example). That angle will never be obtuse, as an obtuse angle can only occur opposite the longest side of a triangle.

This follows from the Law of Cosines -- if C > 90 degrees, then c^2 = a^2 + b^2 - 2ab*cos(C) > a^2 + b^2 since cos(C)<0, and thus c>a and c>b since a,b,c are all positive.

So, if you have a problem where you are asked to solve for (only) the angle opposite the longest side of the triangle using the Law of Sines, I think you have to check the third ratio unless the obtuse angle solution is excluded by A+B+C = 180 degrees.

Best regards,

Troy

William Sears said...

All your arcsine calculations give two possible solutions (actually an infinite number) but your calculator only displays the lowest and it is up to you to choose the correct one. An accurately drawn diagram helps. This can be a good object lesson for your students or students at any level of education.

CyberChalky said...

Sorry, I don't have time to check in detail, but it looks like a standard case of the ambiguous sine.

Darren said...

This definitely involves the ambiguous case of the Law of Sines, as so many have pointed out. I was asking how one might know, a priori, whether one could determine whether or not to check for the "obtuse" case.

It didn't occur to me to use the Pythagorean Theorem as a simple check, great idea! Just as good is always solving for the angle opposite the smallest side, which *must* be acute.

These are great suggestions and I'll be sure to work them into my instruction in the future, thank you!

maxutils said...

My way guarantees no ambiguous case ... And if you think about it, all triangles, given 3 sides, MUST be congruent: This triangle has only one possibility, and there is only one place an obtuse angle can go. .. across from the longest side.