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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 $v=180\deg - \sin^{-1}(x)$.

Lundborg
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    What do you mean by two solutions?? Are you counting negative one too?? – Vidyanshu Mishra Jan 07 '17 at 12:52
  • 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 – Dhanvi Sreenivasan Jan 07 '17 at 12:53
  • Please see edit. – Lundborg Jan 07 '17 at 12:55
  • 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. – Joffan Jan 07 '17 at 13:14
  • Although the law of sines permits two solutions, for an SAS triangle we must select only one of them. If you use the law of sines to calculate the angle opposite the smaller side the first (smaller) principal solution will always be right since it is always acute. If you calculate the angle opposite the bigger side with law of sines you will have to check whether it is acute or not. For $a \le b$ and given angle $C$, $\angle B$ if any of the following is true: 1) $C\ge 90^o$, 2) a=b, 3) $cos(C)<a/b$. If $a=6,b=10,C=30^o$, the last test fails since $0.866>9/16$, so the angle will be obtuse. – smichr Apr 27 '24 at 15:44

4 Answers4

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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.

vcastro
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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.

  • 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. – Lundborg Jan 07 '17 at 13:00
  • @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. –  Jan 07 '17 at 13:05
  • 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? – Lundborg Jan 07 '17 at 13:08
  • @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. –  Jan 07 '17 at 13:13
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    @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. –  Jan 07 '17 at 13:22
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For a $\triangle\text{ABC}$, we know that (for EVERY triangle):

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

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:

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

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:

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

Jan Eerland
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For SAS we know that there is only one triangle possible, so the key to avoiding the "thorns" of multiple solutions for the angle computed from the law of sines, is to use the law of sines for the case where you know that the primary solution will be correct.

Fortunately we know that there can only be, at most, only one obtuse angle in a triangle. We also know that all numbers used in the law of sines will be positive so the angle computed with $sin^{-1}$ will be positive and in the first quadrant (i.e. acute). So the key is to use the law of sines to calculate an angle that we know must be acute.

Given $SAs$ (where $S$ refers to the larger number and $s$ refers to the smaller number), the smaller of the two unknown angles $\angle s$ is opposite side $s$ and must be acute. So

  1. determine the missing side with the law of cosines
  2. calculate the smaller unknown angle using the law of sines
  3. compute the larger of the unknown angles by subtracting the other two from $180^o$

Consider the triangle with $SAs = 4,11.87^o,3$:

  1. the missing side is $\sqrt{(4^2 + 3^2 - 2(4)(3)cos(11.87^o)}=1.23$
  2. the smaller of the two remaining angles is given by $sin(\angle s)/3=sin(11.87^o)/1.23\rightarrow\angle s\approx 30.1^o$
  3. the other angle is $\angle S = 180^o-11.87^o-30.1^o \approx 138^o$

If we had used the law of sines in step 2 to calculate the larger of the two angles we would have obtained approximately $42^o$.

One cannot apply the law of sines without giving some thought to the context at hand. But if one is willing to put some thought into which angle is calculated using this law, it is a shorter path to a solution than using the law of cosines to determine the 2nd angle in the triangle.

We can also by-pass computing the missing side of the triangle by using the formula given here (along with caveats about numerical instability for small angles and sides that are nearly the same):

$$tan(A) = \frac{sin(C)}{\frac{b}{a}-cos(C)}$$

where $a,b,C$ are the given sides and angle and $A$ is the angle opposite side $a$. For example, using $SAs = 4,11.87^o,3$

$$tan(A) = sin(11.87^o)/(4/3-cos(11.87^o)) \approx 0.56\rightarrow A \approx 30.1^o$$

As with law of sines, this angle will always be right if we calculate the angle opposite the smaller side. If used to calculate the larger side it might be negative (if the angle is obtuse) and $\pi$ would have to be added to make it positive.

smichr
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