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Ambiguous Cases - Laws of Cosines and Sines


Date: 04/26/2000 at 14:36:43
From: Kim Feinman
Subject: Are there ambiguous cases with the Law of Cosines?

I am working on solving triangles using the law of cosines and have 
run into a problem where I get different solutions, and I don't 
understand why.

Let's say I have triangle ABC, with sides of length a = 3.2, b = 4.3, 
and c = 5.1. I start off using the law of cosines and I find the 
measure of angle C is 84 degrees. Then I use the law of sines and find 
angle A = 39 degrees.

Here is where I run into trouble. If I just subtract the two angles 
from 180, I get angle B = 57 degrees, but if I go ahead and use the 
law of sines with sides b and c, I get angle B = 33 degrees.

Why is this? Is there an ambiguous case? Because if I try to find the 
measure of angle B using the law of sines using sides a and b, 
instead of b and c as I did, I get the 57 degrees I got earlier just 
by subtracting. Shouldn't both methods work, regardless of which 
angles and sides I use? I think this relates to an ambiguous case of 
the law of sines, but I don't understand why.

Thanks so much!
Kim


Date: 04/26/2000 at 23:06:29
From: Doctor Peterson
Subject: Re: Are there ambiguous cases with the Law of Cosines?

Hi, Kim.

It sounds to me like you just took the inverse cosine when you meant 
to take the inverse sine. That would give you the complement of the 
right answer. Let me know if that's not the explanation.

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/   


Date: 04/27/2000 at 07:23:03
From: Kim Feinman
Subject: Re: Are there ambiguous cases with the Law of Cosines?

Hi,

No, when I use the inverse sin I still get the 37. 

     [3.2*sin(84)]/5.1 = sin A
so 
     sin^-1 [3.2*sin(84)/5.1] = 37

Thanks again,
Kim


Date: 04/27/2000 at 08:49:44
From: Doctor Peterson
Subject: Re: Are there ambiguous cases with the Law of Cosines?

Hi, Kim.

As I understand it, your problem was not in calculating A, but B. Your 
calculation gives A = 38 degrees (approximately); then we find B as

     sin(B) = sin(C) * 4.3/5.1 = 0.8385

     B = sin^-1(0.8385) = 57 degrees

You said you got B = 33 degrees using the law of sines, which is the 
complement of this correct answer. Please check your calculations at 
this step and see whether I'm right that you used the inverse cosine 
here.

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/   


Date: 04/28/2000 at 12:49:40
From: Kim Feinman
Subject: Re: are there ambiguous cases with the Law of Cosines?

Hi again,

Okay. I am _really_ struggling.

In the example you emailed me, I promise I am not using the inverse 
cosine. In my calculator I type [2nd] [sin], which is the sin^-1 
function, right?

Here is another case. Triangle ABC has sides a = 126 and b = 80.1. 
Angle C is 39.4 degrees. 

So I use the law of cosines to find that side c is 81.8.

Then I use the law of sines with side/angles a and c and I come up 
with an angle A of 77.88 degrees.

When I subtract 77.88 and 39.4 from 180, I have an angle B that has to 
be 62.8 degrees.

BUT if I use the law of sines with angles/sides b and c, I come up 
with angle of of 38.4 degrees, and so by subtracting angle A has to be 
102.2 degrees. This is the answer in the book.

What is happening? I really just do NOT understand why I am coming up 
with two different answers using it the two different paths.

Thank you!
-Kim


Date: 04/28/2000 at 13:35:01
From: Doctor Peterson
Subject: Re: are there ambiguous cases with the Law of Cosines?

Hi, Kim. Thanks for writing back and showing your work in detail. 
That's what we've been needing!

In your original problem, I get the same result for B whether I apply 
the law of sines to A and B or to B and C. I still feel that you just 
did a calculation wrong. The triangle is acute, so nothing is 
ambiguous. 

Have you really been recalculating that specific problem to verify 
that it comes out wrong even though you use the inverse sine, or have 
you been assuming that you had the same error in several different 
problems, and checking your work on a different one? It's important to 
work on one problem at a time so we don't get confused.

But this new problem is different. It DOES involve an ambiguous sine. 
Angle A could be either 77.88 or 102.12 degrees in order for sin(A) to 
be 0.977. You have to determine which angles are acute by comparing 
the three sides; since b and c are the smallest sides, angles B and C 
must be acute. Therefore, you should first calculate B and choose the 
acute value, then determine A from that. You may want to look at our 
FAQ on solving triangles at:

http://mathforum.org/dr.math/faq/formulas/faq.trig.html#solveoblitri   

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Trigonometry

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