Law of sines?

courteous

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The following are given: \(\displaystyle a-b=12\), \(\displaystyle \alpha=105\), \(\displaystyle \gamma=55\). What are a,b,c and ß?

Obviously, \(\displaystyle \beta=20\). Now, I plug in law of sines... and get a wrong solution! Why is that?
 
Use the law of sines and the relation \(\displaystyle a-b=12\) to determine \(\displaystyle a,b\).
 
I believe what Royhaas is saying is that one equation is a-b=12. You can form another equation involving a, b, alpha and beta from the law of signs. That gives you two equations in two unknowns which are solvable.
 
Sure enough, I get: \(\displaystyle \frac{a}{\sin\alpha}=\frac{a-12}{\sin\beta}\) ... which does not give me the right solution \(\displaystyle a=12.58\). I get \(\displaystyle a=18.58\) :x .
 
courteous said:
... I plug in law of sines ... and get a wrong solution! Why is that?

Sure enough, [the Law of Sines] ... does not give me the right solution ...

Hello Courteous:

The answer to your question is, "Because you made a mistake."

The equation that you typed from the Law of Sines is correct. Nobody here will be able to tell you the subsequent error(s) you made unless you show your work.

Cheers,

~ Mark :)
 
\(\displaystyle a*\sin20=a*\sin105-12*\sin105 \Rightarrow a(\sin105-\sin20)=12\sin105 \Rightarrow a=\frac{12\sin105}{\sin105-\sin20}\)
 
~18.58 is correct: either typo in question or book answer is wrong.
 
Agreed. :) Last thing though: if it were the ambigious case (where 'law of sines' gives 2 solutions), so that \(\displaystyle a\) (triangle side, not angle) would have two possible solutions, \(\displaystyle \beta\) would have to be obtuse (not \(\displaystyle \alpha\)), right?
 
courteous said:
... if it were the ambigious case (where 'law of sines' gives 2 solutions), so that a (side, not angle) would have two possible solutions ...

Hi Courteous:

Thank you for asking this question. When I see math students considering more than just "the answer", I feel pleased because I spend a lot of time in my life trying to deal with people who don't think at all. You will go far in life.

It's not quite correct to say that the Law of Sines gives two solutions.

When you set up a problem using the Law of Sines, you will always get one result. How you interpret that result (in a case that's ambiguous) determines whether you've got the correct triangle.

Also, the particular exercise that you posted basically gave you the three angle measurements, so the triangle's shape was already determined (in other words, this triangle is not ambiguous because all of its angles are known).

The ambiguous case deals with situations where we do not know all three angles. We know two of the angles, and we recognize the fact that there are two possibilities for the third angle.

If you are using the Law of Sines to find an angle measure, then you've already got the length for each of the two sides involved. You will get one sine ratio from solving the equation that you set up using the Law of Sines.

It would be great if we could simply enter this ratio into a calculator's arcsine function and get the machine to give us the correct angle measurement. Unfortunately, many different angle measurements generate the same trigonometric ratio; the machine is programmed to simply report the smallest of these angles.

It's up to us to properly interpret the particular situation (i.e., triangle) with which we're dealing in any given exercise.

So, again, I'm glad to see that you're keeping the ambiguous case in the back of your mind (perhaps also wondering when it applies and when it does not). With more practical experience after dealing with many different triangles, you will eventually come to recognize when you've got an ambiguous case.

Please feel free to ask more questions about this if you would like to discuss the ambiguous case some more.

Cheers,

~ Mark :)
 
The non-unique case comes in when two sides are given (a & b) but the angle given is not the included angle (which would have been C but it is not).

In this case, there is a possibility of two solutions. However, you use laws of cosines in such problems. Two solutions arise from the fact that you'll get a quadratic equation.
 
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