Mastering the Law of Sines: Troubleshooting Trig Problems

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The discussion revolves around troubleshooting the Law of Sines in trigonometry, specifically addressing issues with calculating angles. The user initially struggles with finding the correct angle using the formula A/Sin(a) = B/Sin(b) = C/Sin(c) and ends up with an incorrect angle due to the limitations of the arcsine function. It is clarified that arcsine only provides one angle in the range of [-90°, 90°], while the sine function can yield two possible angles, including an obtuse angle. The user learns that the correct approach involves recognizing both potential angles and that additional logic is necessary for solving the problem accurately. Ultimately, the user successfully resolves the issue by applying a different method involving right triangles.
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Homework Statement



So, I am brushing up on my trig...and now I feel a bit slow.
I am attempting to use the law of sines for the attached problem and keep getting the wrong angle. For some reason I never learned the law of sines/cosines before today...

Homework Equations


A/Sin(a) = B/Sin(b) = C/Sin(c)

The Attempt at a Solution


4/sin(27)=5/sin(theta) then...
arcsin(5*sin(27)/4) = theta = 34.6

I know this is the angle beside it, but I don't know what I am doing wrong. Any help would be great!
 

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The downfall of using the law of sines is that you can get the wrong answer when the angle is obtuse as you can see from the picture. Do you know how you can fix the "answer" of 34.58°?
 
The problem is that the arcsine doesn't give the value you want.

Basically, what arcsine does is it takes a value x in [-1,1] and it gives you an angle \theta\in [-90^\circ, 90^\circ] such that \sin(\theta)=x.

But this \theta isn't the only angle such that \sin(\theta)=x! Indeed \pi-\theta is also such an angle. Unfortunately, the arcsine can only give you one answer instead of all the angles that satisfy (otherwise, your calculator would have to give you infinitely many values) and it chooses to ignore \pi-\theta.

The user should know that if he gets \theta as an answer, that \pi-\theta is also a good answer to the question and that \theta might be the wrong answer. This is indeed the case here.
 
Ahhh, ok, so apparently I need to use some logic too...haha. Thank you so much for the help. I was able to solve it by solving for the right triangle and then solving for the attached triangle, but this method tripped me up, unfortunately. Thank you all again for the explanation.
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
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