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November 12th, 2017, 10:03 AM  #1 
Newbie Joined: Oct 2017 From: cali Posts: 8 Thanks: 0  Solving triangles with the given formation using law of sines and cosines So when doing these problems, I can simply just use law of cosines and solve for side b, which is 4.5. That is simple enough to me; however, this is where I start randomly having trouble. I now use law of sines, so here I'll do sin B (angle)/side b = to sin alpha (angle a) over 6.8. I get sin alpha by itself and get 6.8sin10.5/4.5 on the other side, and then find the inverse which gets me 16, which isn't the right answer. However, when I do the same thing except with sin y (angle c) / 2.4 which I show below the line, I get the right answer which is 5.6 for angle c. My question is does it matter which angle I use when using law of sines? because I'm getting different answers using both, when I was under the impression that I could've solved for either of them as long as I had both a side and an angle for one part of the triangle. Last edited by skipjack; November 12th, 2017 at 11:24 AM. 
November 12th, 2017, 11:36 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 18,951 Thanks: 1599 
Note that sin(16°) and sin(164°) have the same value. Once you know the shortest side, you know that the angle opposite it must be acute.

November 12th, 2017, 11:50 AM  #3 
Newbie Joined: Oct 2017 From: cali Posts: 8 Thanks: 0 
So what you're saying that both ways are correct? But how so?
Last edited by skipjack; November 12th, 2017 at 06:59 PM. 
November 12th, 2017, 07:23 PM  #4 
Global Moderator Joined: Dec 2006 Posts: 18,951 Thanks: 1599 
You can get b = 4.46 (to 3 significant figures), then arcsin(sin(10.5°)*2.4/4.46) = 5.63° (to 3 s.f.), and so $\alpha$ = 180°  10.5°  5.63° = 163.9° (approximately), rather than 16.1°. The angle opposite the shortest side is acute (for any triangle). The angle opposite the longest side is obtuse in this problem, but isn't obtuse for every triangle. 

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cosines, formation, law, sines, solving, triangles 
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