This is a rather basic problem that I am lacking a good explanation for. Suppose you have scalene triangle $ABC$ that you want to solve, with $A=61^\circ$, $b=4$ and $c=6$.

First, you must use Law of Cosines because you know two sides and one angle. Doing this, you can solve for the third side $c$, and find that $c\approx 5.3599569$.

At this point, you know all three sides of the triangle, and one angle, so we are able to use both Law of Sines and Law of Cosines.

If we use Law of Sines to find angle $C$, we will have:

$\sin(C)\approx .9790597793$, so $C\approx78.254$. However, since we do not yet know the third angle of the triangle, it is also possible that $C$ equal the supplement of the angle we just found, so $C_2\approx 101.746$. We check to see that this is a viable measure, and it is because $C_2+A<180^\circ$, so there are two triangles that correspond with this value of sine.

However, if we instead used the Law of Cosines, we would find that $C\approx 78.254$. So my question is, does Law of Cosines "miss" an answer that Law of Sines finds, or does this result from Law of Cosines show that $C_2$ is not a valid answer, even though it otherwise checks out?

Thanks in advance.

  • $\begingroup$ An update: I have been thinking more, and finding using Law of Sines to fin angle $B$ instead of $C$ can prevent this problem, because it is impossible to have ambiguity, since $B$ must be acute because it corresponds to the shortest side of the triangle. However, I would still appreciate any other explanations, especially if a contradiction can be derived within working with angles $C$ and $C_2. $\endgroup$ – MathStudent1324 Dec 10 '18 at 3:32
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    $\begingroup$ $C_2$ is invalid. If you also find $B$ you'll see that $C_2$ does not "check out" since it will make $A+B+C>\pi$. $\endgroup$ – Lord Shark the Unknown Dec 10 '18 at 4:11
  • $\begingroup$ I'd use the law of tangents to find the angles, bypassing the side $a$ that you found using cosine rule. (it is side $a$ right? not side $c$) $\endgroup$ – user10354138 Dec 10 '18 at 4:24
  • $\begingroup$ You meant you used the Law of Cosines to find the third side $a$. You already know sides $b$ and $c$. $\endgroup$ – N. F. Taussig Dec 10 '18 at 11:23

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