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November 12th, 2017, 11:03 AM   #1
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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 12:24 PM.
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November 12th, 2017, 12:36 PM   #2
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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.
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November 12th, 2017, 12:50 PM   #3
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So what you're saying that both ways are correct? But how so?

Last edited by skipjack; November 12th, 2017 at 07:59 PM.
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November 12th, 2017, 08:23 PM   #4
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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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