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Let $\triangle ABC$ be any triangle with known sides $a$, $b$ and known angle $C$. Determine the remaining side and angles. (The naming convention of angle $A$ being opposite side $a$ etc. is used.)

Attempt:

By the law of cosines we get

$$ c= \sqrt{a^2+b^2-2ab\cos(C)} $$

Now by the law of sines it must hold that

$$ \sin(A) = \frac{a\sin(C)}{c} $$

This has two solutions but when I draw examples of triangles with the known criteria, I only get one triangle. Why is this and which of the two solutions to the equation should I use?

EDIT:

The book I'm using claims that if $\sin(v)=x$ then there are two possible solutions $v=\sin^{-1}(x)$ or $v=180\deg - \sin^{-1}(x)$.

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    $\begingroup$ What do you mean by two solutions?? Are you counting negative one too?? $\endgroup$ – I am Back Jan 7 '17 at 12:52
  • $\begingroup$ It has only one possible solution, as $c>0$ is an implicit condition, considering that it describes a length. Take the positive value of c $\endgroup$ – Dhanvi Sreenivasan Jan 7 '17 at 12:53
  • $\begingroup$ Please see edit. $\endgroup$ – Lundborg Jan 7 '17 at 12:55
  • $\begingroup$ With two sides and the included angle, there is only one possible triangle. You can draw it without ambiguity, so your calculation should also lead you to only one answer. $\endgroup$ – Joffan Jan 7 '17 at 13:14
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For the two values you get from the law of sines, see which ones make sense geometrically. It's possible that one angle will be too large to satisfy the "$180$ total degrees in a triangle" requirement.

To verify this, simply use the fact that $A+B+C = 180^\circ$ to solve for the third angle (which would be $B$ for the example you wrote). If $B < 0$ then you can toss out that corresponding solution for $A$. Note that you'll need to go through this process anyway so it's really not any extra work.

EDIT: I forgot that sometimes there actually is another very small extra step required for the law of sines. See comments on this answer for details.

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  • $\begingroup$ Could we impose a criteria on the known sides and angle wherein we would use a given solution? I would prefer if I could avoid "what makes sense geometrically" arguments. $\endgroup$ – Lundborg Jan 7 '17 at 13:00
  • $\begingroup$ @Neutronic No, but imposing a criterion on the sides and angle sounds like another "what makes sense geometrically" argument anyway. The thing I'm suggesting you do is very simple, and now that I've thought through the whole thing, it's something you'll have to do anyway. See my edit. $\endgroup$ – tilper Jan 7 '17 at 13:05
  • $\begingroup$ I'm not quite sure I understand. Assume $a=10$, $b=6$ and $C=30 \deg$. Then we get $c \approx 5.66$ and $\sin(A) \approx 0.883$. This results in $A \approx 62\deg$ or $A \approx 118 \deg$. How would I determine which one of those is the correct solution without drawing the triangle? $\endgroup$ – Lundborg Jan 7 '17 at 13:08
  • $\begingroup$ @Neutronic do exactly what I said in my edit. Use $A+B+C = 180^\circ$ to find $B$. You'll need to do it twice. Once for each value of $A$ that you found. When you do that, you'll get two positive values of $B$ in this example. Sometimes you'll only get one positive value. But in this case you have two positive values, so there are two solutions to the problem. This is called the ambiguous case. But note that since you have all three sides you can (and should, to avoid the ambiguous case) use the law of cosines again to find the correct triangle. $\endgroup$ – tilper Jan 7 '17 at 13:13
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    $\begingroup$ @Neutronic let me clarify that. Two solutions usually means the ambiguous case. In this case the smaller value of $B$ (corresponding to $A\approx 118^\circ$) must be tossed out because in a triangle, the smallest side must be across from the smallest angle. But for that case, $B$ is the smallest angle while $c$ is the smallest side. Sorry, forgot that extra step. But that just highlights why the law of cosines is preferable. $\endgroup$ – tilper Jan 7 '17 at 13:22
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Hint

Now that you have all sides you can use cosine rule for the other angles and you don't have to worry about signals.

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  • $\begingroup$ Ahhhh, thats kinda clever. $\endgroup$ – Lundborg Jan 7 '17 at 13:09
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For a $\triangle\text{ABC}$, we know that (for EVERY triangle):

$$ \begin{cases} \angle\alpha^\circ+\angle\beta^\circ+\angle\gamma^\circ=180^\circ\\ \\ \frac{\left|\text{A}\right|}{\sin\angle\alpha}=\frac{\left|\text{B}\right|}{\sin\angle\beta}=\frac{\left|\text{C}\right|}{\sin\angle\gamma}\\ \\ \left|\text{A}\right|^2=\left|\text{B}\right|^2+\left|\text{C}\right|^2-2\left|\text{B}\right|\left|\text{C}\right|\cos\angle\alpha\\ \\ \left|\text{B}\right|^2=\left|\text{A}\right|^2+\left|\text{C}\right|^2-2\left|\text{A}\right|\left|\text{C}\right|\cos\angle\beta\\ \\ \left|\text{C}\right|^2=\left|\text{A}\right|^2+\left|\text{B}\right|^2-2\left|\text{A}\right|\left|\text{B}\right|\cos\angle\gamma \end{cases} $$

In your question, you say that $\left|\text{A}\right|$, $\left|\text{B}\right|$ and $\angle\gamma$ are known values.

So, in the system of equations we know:

$$ \begin{cases} \angle\alpha^\circ+\angle\beta^\circ+\color{red}{\angle\gamma^\circ}=180^\circ\\ \\ \frac{\color{red}{\left|\text{A}\right|}}{\sin\angle\alpha}=\frac{\color{red}{\left|\text{B}\right|}}{\sin\angle\beta}=\frac{\left|\text{C}\right|}{\color{red}{\sin\angle\gamma}}\\ \\ \color{red}{\left|\text{A}\right|^2}=\color{red}{\left|\text{B}\right|^2}+\left|\text{C}\right|^2-2\color{red}{\left|\text{B}\right|}\left|\text{C}\right|\cos\angle\alpha\\ \\ \color{red}{\left|\text{B}\right|^2}=\color{red}{\left|\text{A}\right|^2}+\left|\text{C}\right|^2-2\color{red}{\left|\text{A}\right|}\left|\text{C}\right|\cos\angle\beta\\ \\ \left|\text{C}\right|^2=\color{red}{\left|\text{A}\right|^2+\left|\text{B}\right|^2-2\left|\text{A}\right|\left|\text{B}\right|\cos\angle\gamma} \end{cases} $$

So, for example we get:

$$\color{red}{\left|\text{A}\right|^2}=\color{red}{\left|\text{B}\right|^2}+\left(\color{red}{\left|\text{A}\right|^2+\left|\text{B}\right|^2-2\left|\text{A}\right|\left|\text{B}\right|\cos\angle\gamma}\right)-2\color{red}{\left|\text{B}\right|}\cdot\frac{\color{red}{\left|\text{A}\right|\sin\angle\gamma}}{\sin\angle\alpha}\cdot\cos\angle\alpha$$

Using:

$$\frac{1}{\sin\angle\alpha}\cdot\cos\angle\alpha=\cot\angle\alpha$$

We get:

$$\color{red}{\left|\text{A}\right|^2}=\color{red}{\left|\text{B}\right|^2}+\left(\color{red}{\left|\text{A}\right|^2+\left|\text{B}\right|^2-2\left|\text{A}\right|\left|\text{B}\right|\cos\angle\gamma}\right)-2\color{red}{\left|\text{B}\right|}\cdot\color{red}{\left|\text{A}\right|\sin\angle\gamma}\cdot\cot\angle\alpha$$

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