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Different Answers with Sine Rule and Cosine Rule

Date: 08/07/2002 at 02:39:12
From: Thomas Saw
Subject: Different answers with Sine Rule and Cosine Rule

This is really strange. I found this problem in a textbook. A triangle 
ABC has measurements AB = 8.2cm, BC = 9.4cm and AC = 12.8cm and 
angle A = 47 degrees. Find angle B.

Method 1 - Using the Sine Rule;
sin B = AC (sin A)/BC gives B as 84.4 degrees.

Method 2 - Using the Cosine Rule;
AC^2 = BC^2 + AB^2 - 2(BC)(AB) cos B gives B as 93.1 degrees, which 
incidentally is the right answer.

Why does using the Sine Rule here fail to give the right answer?

Thanks for any help you can offer me.


Date: 08/07/2002 at 09:19:58
From: Doctor Rick
Subject: Re: Different answers with Sine Rule and Cosine Rule

Hi, Thomas.

It's strange, isn't it? There are several things going on here at the 
same time, making it a bit hard to untangle.

First, the triangle is overspecified to begin with. Given just the 
lengths of the sides, you can use the law of cosines to find angle A. 
It turns out to be, not 47 degrees exactly, but 47.165 degrees. 

Second, since angle B is close to 90 degrees, a small error in the 
sine of the angle can result in a rather large error in the angle 
itself. In particular, when I apply the law of sines with 47 degrees 
for A, I get sin(B) = 0.99588589; but when I use 47.165 degrees for 
A, I get sin(B) = 0.99857047. That's a small error - but when I take 
the inverse sine of each value, I get 84.80 (I don't know why this is 
slightly different from your answer) and 86.936 degrees respectively.

Now look at the corrected answer of 86.936 degrees using the law of 
sines. Compare it with the correct answer found using the law of 
cosines: 93.064 degrees. Each is 3.064 degrees away from a right angle 
- but in opposite directions! Does this give you an idea of 
what's going on?

When you use the law of cosines to find angle B, you are making use 
of the three sides of the triangle. Remember the SSS congruency 
theorem: three sides determine the shape of a triangle uniquely. This 
is why we are able to find a unique measure for angle B.

When you use the law of sines, you are making use of two sides and an 
angle. Notice how these are related: the angle is NOT between the 
sides. In terms of congruency theorems, we need an SSA theorem - but 
there isn't one. The triangle is NOT uniquely determined by these 
quantities.

Remember that we found that the sine of angle B is 0.99857047. We 
can't just take the inverse sine of this number: that is ONE angle 
with this sine, but 180 degrees minus this angle is another solution. 
The latter solution turns out to be correct in this case. 

You see, there are two triangles that have the same angle A and sides 
AC and BC. The law of sines, applied without thinking, gave one of 
these triangles, but it was the wrong one. You must be careful in 
taking the inverse sine: note both solutions in the range 0 to 180 
degrees, and check whether each leads to a valid solution of the 
triangle. If both do, you need more information (such as side AB) to 
tell which is correct.

See the Dr. Math FAQ on solving triangles:

   Trigonometry Formulas (select Solving Oblique Triangles)
   http://mathforum.org/dr.math/faq/formulas/faq.trig.html 

Also see these items in the Dr. Math Archives:

   Ambiguous Cases - Laws of Cosines and Sines
   http://mathforum.org/library/drmath/view/54143.html 

   Angle-Side-Side Does Not Work
   http://mathforum.org/library/drmath/view/54663.html 

- Doctor Rick, The Math Forum
  http://mathforum.org/dr.math/ 
Associated Topics:
High School Triangles and Other Polygons
High School Trigonometry

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