If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

### Course: Integrated math 3>Unit 8

Lesson 3: Solving general triangles

# Laws of sines and cosines review

Review the law of sines and the law of cosines, and use them to solve problems with any triangle.

## Law of sines

$\frac{a}{\mathrm{sin}\left(\alpha \right)}=\frac{b}{\mathrm{sin}\left(\beta \right)}=\frac{c}{\mathrm{sin}\left(\gamma \right)}$

## Law of cosines

${c}^{2}={a}^{2}+{b}^{2}-2ab\mathrm{cos}\left(\gamma \right)$

## Practice set 1: Solving triangles using the law of sines

This law is useful for finding a missing angle when given an angle and two sides, or for finding a missing side when given two angles and one side.

### Example 1: Finding a missing side

Let's find $AC$ in the following triangle:
According to the law of sines, $\frac{AB}{\mathrm{sin}\left(\mathrm{\angle }C\right)}=\frac{AC}{\mathrm{sin}\left(\mathrm{\angle }B\right)}$. Now we can plug the values and solve:
$\begin{array}{rl}\frac{AB}{\mathrm{sin}\left(\mathrm{\angle }C\right)}& =\frac{AC}{\mathrm{sin}\left(\mathrm{\angle }B\right)}\\ \\ \frac{5}{\mathrm{sin}\left({33}^{\circ }\right)}& =\frac{AC}{\mathrm{sin}\left({67}^{\circ }\right)}\\ \\ \frac{5\mathrm{sin}\left({67}^{\circ }\right)}{\mathrm{sin}\left({33}^{\circ }\right)}& =AC\\ \\ 8.45& \approx AC\end{array}$

### Example 2: Finding a missing angle

Let's find $m\mathrm{\angle }A$ in the following triangle:
According to the law of sines, $\frac{BC}{\mathrm{sin}\left(\mathrm{\angle }A\right)}=\frac{AB}{\mathrm{sin}\left(\mathrm{\angle }C\right)}$. Now we can plug the values and solve:
$\begin{array}{rl}\frac{BC}{\mathrm{sin}\left(\mathrm{\angle }A\right)}& =\frac{AB}{\mathrm{sin}\left(\mathrm{\angle }C\right)}\\ \\ \frac{11}{\mathrm{sin}\left(\mathrm{\angle }A\right)}& =\frac{5}{\mathrm{sin}\left({25}^{\circ }\right)}\\ \\ 11\mathrm{sin}\left({25}^{\circ }\right)& =5\mathrm{sin}\left(\mathrm{\angle }A\right)\\ \\ \frac{11\mathrm{sin}\left({25}^{\circ }\right)}{5}& =\mathrm{sin}\left(\mathrm{\angle }A\right)\end{array}$
Evaluating using the calculator and rounding:
$m\mathrm{\angle }A={\mathrm{sin}}^{-1}\left(\frac{11\mathrm{sin}\left({25}^{\circ }\right)}{5}\right)\approx {68.4}^{\circ }$
Remember that if the missing angle is obtuse, we need to take ${180}^{\circ }$ and subtract what we got from the calculator.
Problem 1.1
$BC=$

Round to the nearest tenth.

Want to try more problems like this? Check out this exercise.

## Practice set 2: Solving triangles using the law of cosines

This law is mostly useful for finding an angle measure when given all side lengths. It's also useful for finding a missing side when given the other sides and one angle measure.

### Example 1: Finding an angle

Let's find $m\mathrm{\angle }B$ in the following triangle:
According to the law of cosines:
$\left(AC{\right)}^{2}=\left(AB{\right)}^{2}+\left(BC{\right)}^{2}-2\left(AB\right)\left(BC\right)\mathrm{cos}\left(\mathrm{\angle }B\right)$
Now we can plug the values and solve:
$\begin{array}{rl}\left(5{\right)}^{2}& =\left(10{\right)}^{2}+\left(6{\right)}^{2}-2\left(10\right)\left(6\right)\mathrm{cos}\left(\mathrm{\angle }B\right)\\ \\ 25& =100+36-120\mathrm{cos}\left(\mathrm{\angle }B\right)\\ \\ 120\mathrm{cos}\left(\mathrm{\angle }B\right)& =111\\ \\ \mathrm{cos}\left(\mathrm{\angle }B\right)& =\frac{111}{120}\end{array}$
Evaluating using the calculator and rounding:
$m\mathrm{\angle }B={\mathrm{cos}}^{-1}\left(\frac{111}{120}\right)\approx {22.33}^{\circ }$

### Example 2: Finding a missing side

Let's find $AB$ in the following triangle:
According to the law of cosines:
$\left(AB{\right)}^{2}=\left(AC{\right)}^{2}+\left(BC{\right)}^{2}-2\left(AC\right)\left(BC\right)\mathrm{cos}\left(\mathrm{\angle }C\right)$
Now we can plug the values and solve:
$\begin{array}{rl}\left(AB{\right)}^{2}& =\left(5{\right)}^{2}+\left(16{\right)}^{2}-2\left(5\right)\left(16\right)\mathrm{cos}\left({61}^{\circ }\right)\\ \\ \left(AB{\right)}^{2}& =25+256-160\mathrm{cos}\left({61}^{\circ }\right)\\ \\ AB& =\sqrt{281-160\mathrm{cos}\left({61}^{\circ }\right)}\\ \\ AB& \approx 14.3\end{array}$
Problem 2.1
$m\mathrm{\angle }A=$
${}^{\circ }$
Round to the nearest degree.

Want to try more problems like this? Check out this exercise.

## Practice set 3: General triangle word problems

Problem 3.1
"Only one remains." Ryan signals to his brother from his hiding place.
Matt nods in acknowledgement, spotting the last evil robot.
"$34$ degrees." Matt signals back, informing Ryan of the angle he observed between Ryan and the robot.
Ryan records this value on his diagram (shown below) and performs a calculation. Calibrating his laser cannon to the correct distance, he stands, aims, and fires.
To what distance did Ryan calibrate his laser cannon?

Want to try more problems like this? Check out this exercise.

## Want to join the conversation?

• I want to know why this article says "Remember that if the missing angle is obtuse, we need to take 180 degrees and subtract what we got from the calculator" when using the law of sines to find a missing angle. Are there videos that explain why this is? I don't understand why during calculations it won't just give the obtuse angle, and instead gives an acute angle.

I'm only taking this geometry course and there is nothing along the way that explains this, not even the videos that introduce the law of sines.

Thanks.
• There can be two since sin(theta) = sin(180-theta) for all values of theta that are real numbers e.g. -1000.98, sqrt(2) etc.

Since you are using the sin^-1 function you will only ever get 1 angle as the range is defined from -90 to 90 degrees(which is -pi/2 to pi/2 in radians). You can it sketch on desmos to see what it look likes.

So if you need to find an obtuse angle then you need to use 180-theta to find the obtuse angle.

It might be helpful to take look at videos related function as well.
• in problem 3.3 I had to open the explanation and do not understand why the law of sines in this solution is switched to sin(c)/length of side. isnt it usually the other way around??
• I can't find anything here about ambiguous triangles. What if a question asks you to solve from a description where two triangles exist? Like "Determine the unknown side and angles in each triangle, if two solutions are possible, give both: In triangle ABC, <C = 31, a = 5.6, and c = 3.9." I solved for height and see that two solutions exist, and the answer key in my textbook agrees, but I can't figure out how to get either. From a set of questions that's only supposed to be on sine law.
• Use the Law of Sines to get one possible angle A:
sin(A)/a=sin(C)/c
sin(A)/5.6=sin(31)/3.9
sin(A)=5.6sin(31)/3.9
A=arcsin(5.6sin(31)/3.9)=47.6924

Subtract 31 (C) and this angle (A) from 180 to find the third angle (B=101.3076) and use the Law of Sines again to find the third side. If you use the given angle-side pair (C and c) you will be less likely to incur error from your own rounding of angle A:
b/sin(B)=c/sin(C)
b/sin(101.3076)=3.9/sin(31)
b=3.9sin(101.3076)/sin(31)=7.4253

But if you know that supplementary angles share a sine value, you know that A can also be an obtuse angle with the same sine as 47.6924:
A=180-47.6924=132.3076

And again, subtract 31 (C) and the obtuse angle A from 180 to find the other possible third angle (B=16.6924) and use the Law of Sines to find the other possible third side, again using angle C and side c to avoid errors from rounding:
b/sin(B)=c/sin(C)
b/sin(16.6924)=3.9/sin(31)
b=3.9sin(16.6924)/sin(31)=2.1750

It all comes from knowing that there are two angles, one obtuse and one acute, for every sine value. And you find the obtuse one by subtracting the acute one from 180.

If you try to do this with a unique triangle (one without two possible sets of angles and sides) your given angle and the obtuse angle you find will add up to more than 180, and so if you try to find a third angle to go with the obtuse one, your subtraction will tell you the third angle is negative, at which point you know you're going down a nonsensical mathematical road, and there weren't two possible triangles to begin with.

Good luck!
• So, obviously, there is the law of sines and the law of cosines. That is what this entire section has been about. However, I'm curious about if there is such a thing as the law of tangents. Since there is both sine and cosine, wouldn't it make sense if there was something like the law of tangents?
• Yes there is. Though I will admit that the only way I know that is by looking it up. I assumed that was but wasn't certain. You can probably find the exact statement of the law on Wikipedia or some math site.
• What is the law of tangents used for??
(1 vote)
• The law of tangents is used to keep people from getting to the point. Old people tend to use it a lot :)
• In the Video 'Solving An Angle With The Law Of Sines' at , Sal said that Law of Sines is 'sin(a)/A = sin(b)/B = sin(c)/C (lower case letters are the angles and the upper case letters are the side opposite to the angle), in this article it says 'A/sin(a) = B/sin(b) = C/sin(c)'. So which one should I choose?
• The Law of Sines can be written either way! You can put the angles in the numerators and the sides in the denominators, or the other way around.

1/2 = 2/4 = 3/6

It's still true if we reverse the numerators and denominators:

2/1 = 4/2 = 6/3
• If law of sines is a/sin(a)=b/sin(b)=c/sin(c), does sin(a)/a=sin(b)/b=sin(c)/c work?
• Yes. Just raise one of those equations to the -1 power, and you get the other equation. They're equivalent.
• Is this a correct way to check whether you use the law of sines or law of cosines? If you have two angles and a side or two sides and an angle, use the law of sines, if you have 3 sides use the law of cosines. Is there anything you need to add to this list?
• The law of sines works only if you know an angle, a side opposite it, and some other piece of information. If you know two sides and the angle between them, the law of sines won't help you.

In any other case, you need the law of cosines.