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Trigonometry Law of sines: Unambiguous case

sweatingbear

Member
May 3, 2013
91
Hello again, forum.

Is it not true that a triangle is unambiguously defined given two sides and the intermediate angle? That is at least what I learned from studying congruence in geometry. Here's the problem: We have the triangle below (the picture is given in the problem and I have redrawn it) and are asked to find the measurement of the other two angles and the side BC.



Now, intuitively, there is only one possible length for BC for which a triangle can be constructed. The angle at C is fixed because changing that very angle would change the length of BA (which is given and thus constrained). Similar arguments apply to angle B.

From these arguments, it is reasonable to expect only one possible triangle from the given data. To start off we can use law of cosines to determine |BC| and from thereon make use law of sines.

The problem though is when I apply the law of sines in order to compute the other two angles, I am able to form two different triangles with different measurements of their angles.

Triangle #1: \(\displaystyle \angle A = 23.8^\circ\), \(\displaystyle \angle B = 34.3^\circ\) and \(\displaystyle \angle C = 121.9^\circ\).

triangle 23.8°,34.3°,121.9° - Wolfram|Alpha

Triangle #2: \(\displaystyle \angle A = 23.8^\circ\), \(\displaystyle \angle B = 145.7^\circ\) and \(\displaystyle \angle C = 10.5^\circ\).

triangle 10.5°,23.8°,145.7° - Wolfram|Alpha

Now this is very strange, I have not come across a triangle (with two given sides and the intermediate angle) which is not unambiguously defined until now. How is this even possible? Clearly in triangle #2, it is not possible for the angle at B to equal 145.7 degrees since this would change the length of AC. But the algebra does not "see" that, simply due to \(\displaystyle \sin (180^\circ - v) \equiv \sin (v)\).

So, forum, how do we resolve this issue? Help much appreciated!
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
I would first use the Law of Cosines to determine:

\(\displaystyle \overline{BC}\approx4.264311887306\)

Then using the Law of Sines, I find:

\(\displaystyle \angle C\approx34.6^{\circ},\,145.4^{\circ}\)

\(\displaystyle \angle B\approx58.4^{\circ},\,121.6^{\circ}\)

The only combination which has the sum of the thee angles as \(\displaystyle 180^{\circ}\) is:

\(\displaystyle \angle A=23.8^{\circ},\,\angle B\approx121.6^{\circ},\,\angle C\approx34.6^{\circ}\)
 

sweatingbear

Member
May 3, 2013
91
Thanks for the reply!

The only difference between my and your approach is that you calculated every angle in the triangle using law of sines, whereas I calculated only one angle and the final one assuming the angle sum is 180. Why exactly did my approach fail in this case? I have had no issues with it before until this very problem.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
I was taught, in this case, to look for both possible values of each of the two unknown angles via the sine identity you cited, and only then to look for that combination of angles who sum is $180^{\circ}$.
 

sweatingbear

Member
May 3, 2013
91
Seemingly a much better approach than mine. Thanks a heap!