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I've seen this type of problem multiple times on homework, and it's confusing me like mad.

The scenario:

We have a triangle. It is a special case triangle, with one angle, one side, and another angle, in a row, known.
The sum of the two known angles is less than 90, leading one to infer that the third, unknown angle is greater than 90°. What troubles me is this:
The Law of Sines will only work correctly with acute angles, and the Law of Cosines requires two or more sides to be known, in this case one one side is known.

I am currently looking at this specific problem: angles A and B are known, respectively equaling 42° and 38°, and side c being 50, making this an ASA case.
Taking the angle measures from 180 yields 100°, the measure of angle C. Going back to the issue with the Laws of Sines and Cosines, this is an issue, considering I cannot use the Law of Sines with angles A and C, and their respective sides.

What would I do in this case to find sides a, and in turn find side b?

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    $\begingroup$ Why dont you try the law of sines? It works. $\endgroup$ – Hagen von Eitzen Apr 17 '14 at 16:17
  • $\begingroup$ From what my Advanced Algebra II/Trig teacher has said, and my knowledge of trig ratios, LOS won't work because sin 100 = sin 80. $\endgroup$ – EnragedTanker Apr 17 '14 at 16:25
  • $\begingroup$ To find an angle from the law of sines, you need to worry about two possibilities, but it works fine if you consider that. To find the sine from an angle, which is what is needed here, is no problem. $\endgroup$ – Ross Millikan Apr 17 '14 at 17:06
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It works fine. $\frac c{\sin 100^\circ} \approx 50.77$.
Then $50.77 \sin 42^\circ \approx 33.97$ and $50.77 \sin 38^\circ \approx 31.26$.
Check $33.97^2+31.26^2-2\cdot 33.97\cdot 31.26 \cos 100^\circ\approx 2500=50^2$

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  • $\begingroup$ Are you sure? Considering c = 50, I'm getting c/sin 100° ≈ -98.74. $\endgroup$ – EnragedTanker Apr 17 '14 at 16:51
  • $\begingroup$ You are not getting the degrees. If I do 50/sin 100 radians I get -98.74. Note that $\sin 100^\circ$ is close to $1$ and definitely positive. $\endgroup$ – Ross Millikan Apr 17 '14 at 16:55
  • $\begingroup$ Okay, now I enter it on my TI-84 in degree mode and I get about 50.77, so there's that. $\endgroup$ – EnragedTanker Apr 17 '14 at 16:58
  • $\begingroup$ I did what you suggested, and it worked out perfectly. Thank you very much for your assistance. $\endgroup$ – EnragedTanker Apr 17 '14 at 17:02

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