A trigonometric problem with the Laws of sine and cosine

Dec 2019
6
0
Colombia
Greetings, I'm having an interesting case where the solution of a certain angle cannot obtain the same result using the laws of sine and cosine for a triangle. I cannot understand why the answer doesn't match since from the perspective I'm getting of the problem, both laws can be used. In this case, I used cosine law first to find angle A and the sine law to get angle B, however, the answer doesn't match. I'll leave right next to this a picture of the process I have performed. The original information says we only know sides a=5, b=11, and c=7.5 of a triangle.

1 eng.png

Thank you for your time!

With regards and happy new year!
 

greg1313

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Oct 2008
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There are two possible values for $B$. Note that the two values you have for$B$ sum to about $180^\circ$.
 
Dec 2019
6
0
Colombia
Thank you for your fast answer, Greg! I wasn't thinking of that possibility. I thought I was doing something wrong over the algebra or something. This is an interesting thing; I'll keep it in mind from now on!!
 

skipjack

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Dec 2006
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When you calculated angle A, you rounded it to two decimal places. When calculating sin(B), you should use the unrounded value you obtained for A.
 

mathman

Forum Staff
May 2007
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In general the law of cosines gives unambiguous results, since arccos is single valued for angles 0 to 180. Arcsin is double valued over this range.
 
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skipjack

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Dec 2006
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In general, one doesn't have the luxury of being able to choose which method to use.
 
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Dec 2019
6
0
Colombia
Regarding this

When you calculated angle A, you rounded it to two decimal places. When calculating sin(B), you should use the unrounded value you obtained for A.
I have done it again with the unrounded values, but there wasn't a significant change in the results.


In general the law of cosines gives unambiguous results, since arccos is single valued for angles 0 to 180. Arcsin is double valued over this range.
So this is one special case where ambiguous results might happen?


In general, one doesn't have the luxury of being able to choose which method to use.
But theorethically once we have an angle like in this conditions knowing the sides, the results should be equal if I mix the cosine and the sine law, right? Well unless we can have two values for one single angle like this exercise.

Also thanks to everyone for the answers !!! I really appreciate it !
 

greg1313

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Oct 2008
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London, Ontario, Canada - The Forest City
\(\displaystyle \sin(x)=\sin(180^\circ-x)\)

Making a reasonably accurate sketch of the triangle in question, if one has sufficient information to do so, may help determine which angle is correct.
 

skipjack

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Dec 2006
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So this is one special case where ambiguous results might happen?
Potential ambiguity arises when the sine of an angle is used to determine the angle, but it's usually easy to determine the correct value. Some textbooks don't cover this in detail.

In the example originally posted, the cosine method needs to be applied first, as no angle is known initially. Having found that angle A, say, is arccos(203/220), which is approximately 22.672 degrees, one can choose to find sin(B), which turns out to be 0.8479976415..., and arcsin of that is 57.994545... degrees. If angle B had that value, angle C would be obtuse, which is impossible, as AB isn't the longest side of the triangle (if a triangle has an obtuse angle, the side opposite that angle must be longer than each of the other two sides, because the obtuse angle must be greater than each of the other two angles). Hence angle B = 180° - 57.994545... degrees = 122.00545... degrees. If that explanation is considered too cumbersome, one can use the cosine method instead to show that cos(B) = -0.53 exactly, etc.