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## Precalculus

### Course: Precalculus>Unit 2

Lesson 6: 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

start fraction, a, divided by, sine, left parenthesis, alpha, right parenthesis, end fraction, equals, start fraction, b, divided by, sine, left parenthesis, beta, right parenthesis, end fraction, equals, start fraction, c, divided by, sine, left parenthesis, gamma, right parenthesis, end fraction

## Law of cosines

c, squared, equals, a, squared, plus, b, squared, minus, 2, a, b, cosine, left parenthesis, gamma, right parenthesis

## 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 A, C in the following triangle:
According to the law of sines, start fraction, A, B, divided by, sine, left parenthesis, angle, C, right parenthesis, end fraction, equals, start fraction, A, C, divided by, sine, left parenthesis, angle, B, right parenthesis, end fraction. Now we can plug the values and solve:
\begin{aligned} \dfrac{AB}{\sin(\angle C)}&=\dfrac{AC}{\sin(\angle B)} \\\\ \dfrac{5}{\sin(33^\circ)}&=\dfrac{AC}{\sin(67^\circ)}\\\\ \dfrac{5\sin(67^\circ)}{\sin(33^\circ)}&=AC \\\\ 8.45&\approx AC \end{aligned}

### Example 2: Finding a missing angle

Let's find m, angle, A in the following triangle:
According to the law of sines, start fraction, B, C, divided by, sine, left parenthesis, angle, A, right parenthesis, end fraction, equals, start fraction, A, B, divided by, sine, left parenthesis, angle, C, right parenthesis, end fraction. Now we can plug the values and solve:
\begin{aligned} \dfrac{BC}{\sin(\angle A)}&=\dfrac{AB}{\sin(\angle C)} \\\\ \dfrac{11}{\sin(\angle A)}&=\dfrac{5}{\sin(25^\circ)} \\\\ 11\sin(25^\circ)&=5\sin(\angle A) \\\\ \dfrac{11\sin(25^\circ)}{5}&=\sin(\angle A) \end{aligned}
Evaluating using the calculator and rounding:
m, angle, A, equals, sine, start superscript, minus, 1, end superscript, left parenthesis, start fraction, 11, sine, left parenthesis, 25, degrees, right parenthesis, divided by, 5, end fraction, right parenthesis, approximately equals, 68, point, 4, degrees
Remember that if the missing angle is obtuse, we need to take 180, degrees and subtract what we got from the calculator.
Problem 1.1
• Current
B, C, equals

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, angle, B in the following triangle:
According to the law of cosines:
left parenthesis, A, C, right parenthesis, squared, equals, left parenthesis, A, B, right parenthesis, squared, plus, left parenthesis, B, C, right parenthesis, squared, minus, 2, left parenthesis, A, B, right parenthesis, left parenthesis, B, C, right parenthesis, cosine, left parenthesis, angle, B, right parenthesis
Now we can plug the values and solve:
\begin{aligned} (5)^2&=(10)^2+(6)^2-2(10)(6)\cos(\angle B) \\\\ 25&=100+36-120\cos(\angle B) \\\\ 120\cos(\angle B)&=111 \\\\ \cos(\angle B)&=\dfrac{111}{120} \end{aligned}
Evaluating using the calculator and rounding:
m, angle, B, equals, cosine, start superscript, minus, 1, end superscript, left parenthesis, start fraction, 111, divided by, 120, end fraction, right parenthesis, approximately equals, 22, point, 33, degrees

### Example 2: Finding a missing side

Let's find A, B in the following triangle:
According to the law of cosines:
left parenthesis, A, B, right parenthesis, squared, equals, left parenthesis, A, C, right parenthesis, squared, plus, left parenthesis, B, C, right parenthesis, squared, minus, 2, left parenthesis, A, C, right parenthesis, left parenthesis, B, C, right parenthesis, cosine, left parenthesis, angle, C, right parenthesis
Now we can plug the values and solve:
\begin{aligned} (AB)^2&=(5)^2+(16)^2-2(5)(16)\cos(61^\circ) \\\\ (AB)^2&=25+256-160\cos(61^\circ) \\\\ AB&=\sqrt{281-160\cos(61^\circ)} \\\\ AB&\approx 14.3 \end{aligned}
Problem 2.1
• Current
m, angle, A, equals
degrees
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
• Current
"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?
start text, space, m, end text

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.
• 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.
• How to find out length of the segments of a diagonal of quadrilateral.
• Two triangles can be identified in a quadrilateral with one diagonal drawn. Eight triangles can be identified in a quadrilateral with both diagonals drawn. With the diagonal or diagonals drawn, look for a triangle with enough side and angle measures that you can use the law of sines or law of cosines. Doing so may give you enough information to complete other triangles until you have the measurements you want.
• can someone explain me how Sal gets from this:36=196+144−336cos(∠A)
to this: 336cos(∠A)=304
• 36=196+144-336cos(∠A)
=>36+336cos(∠A)=340 [Adding 336cos(∠A) to both sides]
=>336cos(∠A)=340-36=304 [Subtracting 36 from both sides]
That is how we get to this step. I hope this helps.