For more information, please see full course syllabus of Trigonometry

For more information, please see full course syllabus of Trigonometry

### Related Articles:

### Law of Sines

**Main formulas**:

- The Law of Sines:

This holds insin *A**a*= sin *B**b*= sin *C**c**any*triangle - not just right triangles! - Use the Law of Sines for triangles described in the following ways:
- ASA always has a unique solution (assuming the angles sum to less than 180
^{° }). - SAA always has a unique solution (assuming the angles sum to less than 180
^{° }). - SSA might have no solutions, one solution, or two solutions.
- AAA always has infinitely many solutions (assuming the angles sum to 180
^{° }), but the triangles are all similar. - (Use the Law of Cosines for SAS and SSS.)

- ASA always has a unique solution (assuming the angles sum to less than 180

**Example 1**:

*ABC*, angle

*A*measures 50

^{° }, angle

*B*measures 60

^{° }, and side

*c*has length 4. Determine how many triangles satisfy these conditions and solve the triangle(s) completely.

**Example 2**:

*ABC*, side

*a*has length 3, side

*b*has length 4, and angle

*A*measures 40

^{° }. Determine how many triangles satisfy these conditions and solve the triangle(s) completely.

**Example 3**:

*ABC*, side

*a*has length 7, side

*b*has length 12, and angle

*A*measures 45

^{° }. Determine how many triangles satisfy these conditions and solve the triangle(s) completely.

**Example 4**:

*ABC*, angle

*A*measures 40

^{° }, angle

*B*measures 110

^{° }, and side

*a*has length 7. Determine how many triangles satisfy these conditions and solve the triangle(s) completely.

**Example 5**:

*ABC*, side

*a*has length 10, side

*b*has length 8, and angle

*B*measures 20

^{° }. Determine how many triangles satisfy these conditions and solve the triangle(s) completely.

### Law of Sines

^{°}, angle C measures 20

^{°}, and side a has length 200. Determine how many triangles satisfy these conditions and solve the triangle(s) completely.

- Start by drawing a triangle with all the sides and angles labeled properly
- According to the triangle: SAA, and the angles sum to 170
^{°}which is less than 180^{°}so there is one unique solution - Use the Law of Sines: [sinA/a] = [sinB/b] = [sinC/c]
- ∠B = 180
^{°}- 150^{°}- 20^{°}= 10^{°} - [sinA/a] = [sinC/c] ⇒ [(sin150
^{°})/200] = [(sin20^{°})/c] ⇒ c sin150^{°}= 200 sin20^{°}⇒ c = [(200sin20^{°})/(sin150^{°})] ⇒ c ≈ 136.8 - [sinA/a] = [sinB/b] ⇒ [(sin150
^{°})/200] = [(sin10^{°})/b] ⇒ b sin150^{°}= 200 sin10^{°}⇒ b = [(200sin10^{°})/(sin150^{°})] ⇒ b ≈ 69.5

^{°}, c ≈ 136.8, b ≈ 69.5

^{°}. Determine how many triangles satisfy these conditions and solve the triangle(s) completely.

- Start by drawing a triangle with all the sides and angles labeled properly
- According to the triangle: SSA, might have no solution, one solution, or two solutions
- Use the Law of Sines: [sinA/a] = [sinB/b] = [sinC/c]
- [sinA/a] = [sinB/b] ⇒ [(sin36
^{o})/8] = [sinB/5] ⇒ 5sin36^{o}= 8sinB ⇒ sinB = [(5sin36^{°})/8] ⇒ sinB ≈ 0.3674 ⇒ B ≈ 21.6^{°} - ∠ B can also be 180
^{°}- 21.6^{°}= 158.4^{°}but C would be 180^{°}- 36^{°}- 158.4^{°}= - 14.4^{°}so this second triangle does not exist - One solution: ∠ B = 21.6
^{°} - ∠ C = 180
^{°}- 36^{°}- 21.6^{°}= 122.4^{°} - [sinA/a] = [sinC/c] ⇒ [(sin36
^{°})/8] = [(sin122.4^{°})/c] ⇒ c sin36^{°}= 8 sin122.4^{°}⇒ c = [(8sin122.4^{°})/(sin36^{°})] ⇒ c ≈ 11.5

^{°}, ∠C = 122.4

^{°}, c ≈ 11.5

^{°}, angle B measures 75

^{°}, and side a has length 6. Determine how many triangles satisfy these conditions and solve the triangle(s) completely.

- Start by drawing a triangle with all the sides and angles labeled properly
- According to the triangle: SAA, and the angles sum to 118
^{°}which is less than 180^{°}so there is one unique solution - Use the Law of Sines: [sinA/a] = [sinB/b] = [sinC/c]
- ∠ C = 180
^{°}- 75^{°}- 43^{°}= 62^{°} - [sinA/a] = [sinB/b] ⇒ [(sin43
^{°})/6] = [(sin75^{°})/b] ⇒ b sin43^{°}= 6sin75^{°}⇒ b = [(6sin75^{°})/(sin43^{°})] ⇒ b ≈ 8.5 - [sinA/a] = [sinC/c] ⇒ [(sin43
^{°})/6] = [(sin62^{°})/c] ⇒ c sin43^{°}= 6sin62^{°}⇒ c = [(6sin62^{°})/(sin43^{°})] ⇒ c ≈ 7.8

^{°}, b ≈ 8.5, c ≈ 7.8

^{°}. Determine how many triangles satisfy these conditions and solve the triangle(s) completely.

- Start by drawing a triangle with all the sides and angles labeled properly
- According to the triangle: SSA, might have no solution, one solution, or two solutions
- Use the Law of Sines: [sinA/a] = [sinB/b] = [sinC/c]
- [sinA/a] = [sinC/c] ⇒ [(sin65
^{°})/9] = [sinC/11] ⇒ 9sinC = 11sin65 ⇒ sinC = [(11sin65^{°})/9] ⇒ sinC ≈ 1.1077

^{°}, angle C measures 54

^{°}, and side c has length 2. Determine how many triangles satisfy these conditions and solve the triangle(s) completely.

- Start by drawing a triangle with all the sides and angles labeled properly
- According to the triangle: SAA, and the angles sum to 170
^{°}which is less than 180^{°}so there is one unique solution - Use the Law of Sines: [sinA/a] = [sinB/b] = [sinC/c]
- ∠B = 180
^{°}− 54^{°}− 24^{°}= 102^{°} - [sinA/a] = [sinC/c] ⇒ [(sin24
^{°})/a] = [(sin54^{°})/2] ⇒ a sin54^{°}= 2sin24^{°}⇒ a = [(2sin24^{°})/(sin54^{°})] ⇒ a ≈ 1.01 - [sinA/a] = [sinB/b] ⇒ [(sin102
^{°})/b] = [(sin54^{°})/2] ⇒ 2sin102^{°}= b sin54^{°}⇒ b = [(2sin102^{°})/(sin54^{°})] ⇒ b ≈ 2.4

^{°}, a ≈ 1.01, b ≈ 2.4

- Use the Law of Sines: [sinA/a] = [sinB/b] = [sinC/c]
- ∠C = 180
^{°}− 30^{°}− 45^{°}= 105^{°} - [sinA/a] = [sinB/b] [(sin30
^{°})/20] = [(sin45^{°})/b] 20sin45^{°}= b sin30^{°}b = [(20sin45^{°})/(sin30^{°})] b ≈ 28.3 - [sinA/a] = [sinC/c] ⇒ [(sin30
^{°})/20] = [(sin105^{°})/c] ⇒ c sin30^{°}= 20sin105^{°}⇒ c = [(20sin105^{°})/(sin30^{°})] ⇒ c ≈ 38.6

^{°}, b ≈ 28.3, c ≈ 38.6

- Use the Law of Sines: [sinA/a] = [sinB/b] = [sinC/c]
- ∠C = 180
^{°}− 10^{°}− 60^{°}= 110^{°} - [sinA/a] = [sinB/b] ⇒ [(sin10
^{°})/7.5] = [(sin60^{°})/b] ⇒ b sin10^{°}= 7.5sin60^{°}⇒ b = [(7.5sin60^{°})/(sin10^{°})] ⇒ b ≈ 37.4 - [sinA/a] = [sinC/c] ⇒ [(sin10
^{°})/7.5] = [(sin110^{°})/c] ⇒ c sin10^{°}= 7.5sin110^{°}⇒ c = [(7.5sin110^{°})/(sin10^{°})] ⇒ c ≈ 40.6

^{°}, b ≈ 37.4, c ≈ 40.6

- Use the Law of Sines: [sinA/a] = [sinB/b] = [sinC/c]
- ∠A = 180
^{°}− 120^{°}− 45^{°}= 15^{°} - [sinA/a] = [sinB/b] ⇒ [(sin15
^{°})/a] = [(sin120^{°})/15] ⇒ a sin120^{°}= 15sin15^{°}⇒ a = [(15sin15^{°})/(sin120^{°})] ⇒ a ≈ 4.5 - [sinB/b] = [sinC/c] ⇒ [(sin120
^{°})/15] = [(sin45^{°})/c] ⇒ c sin120^{°}= 15sin45^{°}⇒ c = [(15sin45^{°})/(sin120^{°})] ⇒ c ≈ 12.2

^{°}, a ≈ 4.5, c ≈ 12.2

- Use the Law of Sines: [sinA/a] = [sinB/b] = [sinC/c]
- ∠A = 180
^{°}− 135^{°}− 10^{°}= 35^{°} - [sinA/a] = [sinB/b] ⇒ [(sin35
^{°})/a] = [(sin135^{°})/45] ⇒ 45sin35^{°}= a sin135^{°}⇒ a = [(45sin35^{°})/(sin135^{°})] ⇒ a ≈ 36.5 - [sinB/b] = [sinC/c] ⇒ [(sin135
^{°})/45] = [(sin10^{°})/c] ⇒ c sin135^{°}= 45sin10^{°}⇒ c = [(45sin10^{°})/(sin135^{°})] ⇒ c ≈ 11.1

^{°}, a ≈ 36.5, c ≈ 11.1

- Use the Law of Sines: [sinA/a] = [sinB/b] = [sinC/c]
- ∠C = 180
^{°}− 108^{°}− 45^{°}= 47^{°} - [sinA/a] = [sinB/b] ⇒ [(sin25
^{°})/12] = [(sin108^{°})/b] ⇒ b sin25^{°}= 12sin108^{°}⇒ b = [(12sin108^{°})/(sin25^{°})] ⇒ b ≈ 27.0 - [sinA/a] = [sinC/c] ⇒ [(sin25
^{°})/12] = [(sin47^{°})/c] ⇒ c sin25^{°}= 12sin47^{°}⇒ c = [(12sin47^{°})/(sin25^{°})] ⇒ c ≈ 20.8

^{°}, b ≈ 27.0, c ≈ 20.8

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

Answer

### Law of Sines

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro
- Law of Sines Formula
- When to Use Law of Sines
- Example 1: How Many Triangles Satisfy Conditions, Solve Completely
- Example 2: How Many Triangles Satisfy Conditions, Solve Completely
- Example 3: How Many Triangles Satisfy Conditions, Solve Completely
- Extra Example 1: How Many Triangles Satisfy Conditions, Solve Completely
- Extra Example 2: How Many Triangles Satisfy Conditions, Solve Completely

- Intro 0:00
- Law of Sines Formula 0:18
- SOHCAHTOA
- Any Triangle
- Graphical Representation
- Solving Triangle Completely
- When to Use Law of Sines 2:55
- ASA, SAA, SSA, AAA
- SAS, SSS for Law of Cosines
- Example 1: How Many Triangles Satisfy Conditions, Solve Completely 8:44
- Example 2: How Many Triangles Satisfy Conditions, Solve Completely 15:30
- Example 3: How Many Triangles Satisfy Conditions, Solve Completely 28:32
- Extra Example 1: How Many Triangles Satisfy Conditions, Solve Completely
- Extra Example 2: How Many Triangles Satisfy Conditions, Solve Completely

### Trigonometry Online Course

I. Trigonometric Functions | ||
---|---|---|

Angles | 39:05 | |

Sine and Cosine Functions | 43:16 | |

Sine and Cosine Values of Special Angles | 33:05 | |

Modified Sine Waves: Asin(Bx+C)+D and Acos(Bx+C)+D | 52:03 | |

Tangent and Cotangent Functions | 36:04 | |

Secant and Cosecant Functions | 27:18 | |

Inverse Trigonometric Functions | 32:58 | |

Computations of Inverse Trigonometric Functions | 31:08 | |

II. Trigonometric Identities | ||

Pythagorean Identity | 19:11 | |

Identity Tan(squared)x+1=Sec(squared)x | 23:16 | |

Addition and Subtraction Formulas | 52:52 | |

Double Angle Formulas | 29:05 | |

Half-Angle Formulas | 43:55 | |

III. Applications of Trigonometry | ||

Trigonometry in Right Angles | 25:43 | |

Law of Sines | 56:40 | |

Law of Cosines | 49:05 | |

Finding the Area of a Triangle | 27:37 | |

Word Problems and Applications of Trigonometry | 34:25 | |

Vectors | 46:42 | |

IV. Complex Numbers and Polar Coordinates | ||

Polar Coordinates | 1:07:35 | |

Complex Numbers | 35:59 | |

Polar Form of Complex Numbers | 40:43 | |

DeMoivre's Theorem | 57:37 |

### Transcription: Law of Sines

*We are learning about the law of science today, we are trying some more examples of solving triangles completely using the law of science.*0000

*Remember, that means you are given some data about the triangles, some information about the lengths and some of the sides of the triangle, and the measure of some of the angles.*0008

*What you have to do, is first of all figure out if there is a triangle that satisfies that data or maybe if there is more than one.*0019

*And then solve for all the other lengths and sides, and all of the other measures of the angles in the triangle.*0027

*In this example, we are given a triangle- it looks like we are given two angles and a side, let me draw the triangle.*0036

*Capital letters for the corners and the sides you will use lower case letters, it has to be opposite of the capital of the same letter.*0052

*That tells you where the orientation of all the information that you are given, given that (A is 40 degree), angle A=40 degrees, angle B=110 degrees.*0063

*Certainly the way I drew it is not to scale because 110 would be end up twoâ€™s angle, it is bigger than 90 degrees.*0074

*We are given that (a) has side length=7, side (a) has length=7, I fill in the information that I have.*0081

*I want to find out first of all, are we going to have a solution here? We are given here a side and then two angles of a triangle.*0089

*This is (side, angle, angle) the side is not between the angles that is why (side, angle, angle) not (angle, side, angle).*0098

*The thing you want to check there is whether the angles you have been given are legitimate, did they add up to less than 180 degrees.*0109

*We talked about that in the beginning of the lecture, when you are given certain pieces of information how do you know if there is one solution or no solutions or two solutions?*0121

*In (side, angle, angle), you just check whether the angles add up to 180 degrees.*0130

*In this case, 110 + 40 = 150, that is less than 180 so we have not already exceeded the angle limit for the triangle.*0137

*In that case, it has a unique solution, it has one solution, that would be exactly one triangle satisfying the data we have been given.*0150

*That answers the first question, the trickier part is use the law of science to find all the missing quantities here.*0166

*Let me write down the law of science, sin(A)/a = sin(B)/b = sin(C)/c.*0173

*What do we know there? We know (A), so we can find its sin very easily, we know (B) and we know (a).*0189

*We can figure out (C) very quickly because we know two of the angles and it is very easy to find the third one.*0200

*So, (C) = 110 â€“ 40, sorry it is 180 â€“ the two angles we are given (40+110), so that is 180-150=30 degrees, we can fill that in very quickly, (C)=30 degrees.*0207

*We got sin(C), we can figure that out quickly, we need to find (b) and (c), we need to use a lot of sin to find them.*0230

*Sin(A)/a=sin(B)/b and trying to solve for (b) here, I will cross multiply and get b sin(A)=a sin(B) or (b)=a sin(B)/(A).*0241

*Actually, Iâ€™m going to fill in before I cross multiply, I will fill in the quantities that I know, sin(A)=40, (a)=7, (b)=?, (B)=110.*0264

*Now, Iâ€™m going to cross multiply, and I get (b) = (7)sin(110)/sin(40).*0281

*Remember to set your calculator to degree mode so you do not get strange answers because your calculator is thinking in terms of radians.*0291

*I will work out (7)sin(110)/sin(40) what I get there is approximately, Iâ€™m going to round it to 10.23, so that is the length of side (b).*0304

*We found all three angles, two of the sides, we just need to find side (c).*0327

*Iâ€™m going to use the law of science. Sin(A)/a=sin(C)/c, angle A=40, a=7, C=30, c= we do not know, that is what we are solving for, so I will cross multiply.*0333

*I will get c=(7)sin(30)/sin(40), there is actually a common value, I do not what the sin is but it is not really that useful here.*0359

*Certainly I do not know sin(40), so Iâ€™m plugging these numbers into my calculator anyway.*0376

*(7)sin(30)/sin(40) is approximately equal to 5.45.*0381

*Let us recap what we needed to do there, we were given two angles and one side of a triangle, I filled in everything I knew, that gave me a (side, angle, angle) configuration.*0408

*It is not (angle, side, angle) because of the orientation of where everything was, the side was not in between the two angles, it was outside of the two angles.*0421

*We have (side, angle, angle) and remember from the list at the beginning of the lecture, if you have (side, angle, angle).*0432

*You just have to check that the angles add up to less than 180 degrees, which they did and then you know you has exactly one solution.*0440

*Once you determine that, you know you are going to find one solution.*0448

*It is quick to find the third angle because you know the angles add up to 180 and then you have to find the two missing sides.*0453

*That takes a little more time but they both come from the law of science.*0460

*You just write down the law of science, you fill in the information that you do know and then you solve for the missing side.*0465

*That worked to find both of those missing sides, then we know all three sides of the triangle and all three angles, we completely solved the triangle.*0472

*On our last example here, we are given a side, and other side, and an angle.*0000

*Let me draw out the information that we have here.*0006

*Capital letters on the angles (A, B, C), lower case letters on the sides, opposite development angles.*0020

*Let us fill in what we have, Side a=10, side b=8, angle B=20 degrees, that is what we are given. *0030

*We are given two sides and an angle, the angle is not in between them so this is (side, side, angle) not (side, angle, side) because the angle is not between them.*0050

*That is a little worrisome because (side, side, angle) is the ambiguous case, it could have no solutions, one solution, or two solutions.*0063

*We do not really know yet until we do some more work, we do not know whether there is going to be one triangle that fits this data, or no triangle fits this data, or maybe two triangles that fits this data.*0089

*We are going to work with the laws of science and try narrowing this down, sin(A)/a=sin(B)/b=sin(C)/c.*0104

*Let me see which of these we know, Iâ€™m going to fill in what we know (a), (b), we are told the measure of angle (B) so I can figure out the sin(B) very quickly.*0116

*The reasonable thing is to try and figure out first of all what is the sin(A), let me go ahead and try to calculate that.*0134

*Sin(A)/a=sin(B)/b, I will fill in what I know there, I do not know sin(a) yet, but I do know angle (A), but I do know that (a)=10, (B)=20, (b)=8.*0142

*So sin(A), if I cross multiply there I will get (10) sin(20)/8.*0168

*Let us see what that comes up to be, 10 x sin(20)/8 turns out to be approximately 0.43.*0178

*Ok, Iâ€™m looking for an angle whose sin is 0.43, let me draw my unit circle again.*0196

*The top half of the unit circle, 0.43 is the y value, remember sin is y value, there is 0.43.*0207

*There are two angles here, there is one and there is another one, they both have sin of 0.43.*0219

*One is (theta) but then there is angle that is actually 180-(theta), they both have sin(0.43).*0226

*My calculator does not really know that, if I work out from the calculator, if I just type arcsin(0.43), let me type that in.*0237

*After we calculate (10) sin(20)/8 and there is 0.43.*0254

*I if I just type in arcsin in my calculator it tells me that it is 25.3 degrees.*0275

*But, that is angle (theta) there is 25.3 degrees, we already know that there is another angle that also have sin(0.43) and that is this other angle 180 â€“ (theta)=154.7 degrees.*0294

*Angle (A) could be either one of those two possibilities, 25.3 degrees or 154.7 degrees.*0324

*That means we have two different possibilities for angle (A), each one of them gives us a set of data that we can solve out the rest of the triangle with.*0335

*We get two possible triangles as our solutions, we have to solve for each one separately.*0349

*We got two solutions here and we are going to solve each one of them separately, I will do that on the next slide.*0365

*Here we found that we have two solutions depending on what angle (A) is.*0372

*I will solve for each one separately, let me draw my triangle.*0383

*(A, B, C) and on the first one we figured out that angle (A) = 25.3 degrees.*0393

*Then we also have some given data a=10, b=8, we do not know what c is, angle (B)=20 degrees, we have not figure out the rest of that.*0409

*Let us go ahead and start figuring out the rest of it, angle (C)=(180-20-25.3)=134.7 degrees.*0428

*In this triangle, 134.7 and finally we have to figure out what side c is, that is (c), we are going to use the law of science on that.*0448

*sin(C)/c=sin(B)/b since those is the simplest values that I can see.*0464

*Sin(134.7) that is angle (C)/we are solving for (c), is the sin(20)/8, if we cross multiply and solve for (c) there, *0475

*(c)sin(20)=(8)sin(134.7), (c)=(8)sin(134.7)/sin(20) you definitely want to go to the calculator for that.*0497

*So I type in (8)sin(134.7)/sin(20).*0515

*It tells me that it is approximately 60.63, what I am seeing there is approximately 60.63.*0530

*I have solved that for all three angles and lengths of that triangle but remember we have another completely different triangle which is based on finding a different value for angle (A).*0549

*We have to solve for those as well, let me draw that one, (A, B,C) (a, b, c) and I will fill in the value that were given.*0561

*Angle (B)=20, (a)=10, (b)=8, and we figured out the other possible value for angle (A) was 154.7 degrees, that is the other possible value.*0580

*And then we want to solve out the rest of the triangle which really means finding angle (C) and (c).*0613

*Angle (C) is pretty easy to find which is (180-20-154.7), which is (180-174.7) = 5.3 degrees.*0618

*Definitely not drawn in the scale here since this angle the way I drawn it is larger than 5.3 but that is ok.*0638

*Now we want to find (c) and we are going to use the law of cos just like we did on the other triangle it is the same arithmetic but with different numbers.*0656

*Sin(C)/(c)=sin(B)/(b), now this time angle (C)=5.3 degrees, (c)=we still do not know, (B)=20 degrees, (b)=8.*0664

*Cross multiply to solve for (c), (c)=(8)sin(5.3)=(c)sin(20), so (c) is (8)sin(5.3)/sin(20).*0687

*I will go to the calculator to figure that out, that tells me that, that is approximately 2.16.*0706

*Now, we have solved out all three angles and sides of that triangle, we are done with that.*0734

*To recap a little bit, let us see what we are given in this triangle.*0741

*We are given a side, another side, and then an angle, this really is a (side, side, angle) triangle.*0746

*Unfortunately, that is the ambiguous case where it could have no solution, one solution, or two solutions.*0756

*We do not really know until you start going to the law of science to figure out what is happening.*0762

*When we apply the law of science, we got a value for sin(A)=0.43, the problem is that there are two angles that have sin(0.43).*0767

*Your calculator will only give you one of them, if you take in there sin(0.43) it will give you this one 25.3 degrees.*0786

*But we know, if you look at the unit circle that sin(180)-(theta) is the same as sin(theta).*0795

*At the same time as looking at 25.3, we have to look at this other possible value 154.7 degrees for (A).*0807

*So we get two different possible angles for (A), that leads us to two different triangles and we have to solve out each one completely.*0816

*In that point, you really can not overwrap the work anymore, you have to split up this two triangles into two different problems.*0824

*In each one you draw it out and you see what pieces of information you are missing.*0833

*In this case, it was the third angle and the third side.*0838

*The third angle, you find out just by subtracting the two angles you have so that was pretty easy and then you use the law of science to solve for the length of the third side.*0845

*In the second triangle we do the same thing, we find the third angle by subtracting from 180 degrees and then we use the law of science to find the third side.*0855

*We got two different triangles, they both satisfy the initial data but using this rule for science, this co function identity sin(180-theta)=sin(theta).*0865

*We figured out the two possible angles and then we can completely solve the two possible triangles separately and we got a whole set of sides and a whole set of angles on each one.*0877

*That was our practice on using the law of science, in the next lecture we will come back and we will look at the law of cos which is a different way to solve out a triangle completely.*0887

*To find out all the angles and all the sides and you use the law of cos when you have a different set of initial data for the triangle.*0897

*We will learn about that in the next lecture, these are the trigonometry series on www.educator.com.*0906

*Hi, these are the trigonometric lectures for educator.com, and today we're going to learn about one of the really big rules of trigonometry which is the law of sines.*0000

*In a later lecture, we're going to learn another rule called the law of cosines.*0010

*They kind of go hand in hand.*0013

*The idea is that these rules help you figure out what the angles and sides are in any triangle.*0015

*I really mean any triangle here.*0024

*Remember a rule we learned before was SOH CAH TOA, that was kind of the old rule.*0028

*The thing about SOH CAH TOA is, it only works in right triangles, in other words, triangles where one angle is a right angle.*0035

*That was our old rule.*0051

*Our new rule is the law of sines, and we're going to be learning the law of cosines in the next lecture.*0053

*The law of sines works in any triangle.*0059

*You can always use it.*0063

*If you happen to have a right triangle, it's probably easier to use SOH CAH TOA.*0064

*If you don't have a right triangle, you want to use the law of sines or the law of cosines, which we'll learn about next.*0069

*Let me draw a picture to get it started here.*0076

*Here's a triangle.*0085

*It doesn't look like a right triangle.*0086

*It's traditional in trigonometry problems to kind of label the sides and corner, the sides and the angles of a triangle, so that you can keep track of what's opposite what.*0089

*Normally, you would label these things as A, B and C, the corners with capital letters A, B and C.*0102

*Then you label the sides with lowercase letters a, b and c.*0111

*You do it in such a way that lowercase, side labeled lowercase a, is opposite angle labeled uppercase A.*0112

*That puts the a over here, the b over here, and c over here.*0122

*What we're going to be doing is looking at a lot of different triangles where you'll be given some information about some of the sides and some of the angles.*0126

*You won't be given all of the information.*0134

*You'll be told the lengths of a couple of sides or maybe one of the angles, or maybe the measure of two of the angles, but not the third angle and the measure of one of the sides, things like that.*0139

*The trick here is you want to use the law of sines and whatever else you can, to find out all the angles and all the sides of the triangle.*0151

*That's called solving the triangle completely, when you figured out what all the angles and what all the sides are.*0162

*There's several different ways that the information could be given to you.*0168

*You'll hopefully remember some of these from your geometry class.*0176

*The first one is Angle Side Angle.*0179

*What that means is that ...*0183

*Let me draw a picture of this.*0185

*Kind of as you walk around the edge of a triangle, you'll know what one angle is and you'll know what one side is, and you'll know what the next angle is.*0187

*You'll be given two angles and a side.*0198

*It's important that they'd be in that order, the angle comes first and then the following side, then the next angle.*0201

*When you're given an angle side angle set up, you know that it always has a solution.*0208

*There's always a solution.*0217

*There's always a triangle that has those properties.*0218

*It's always unique, meaning that there's only one triangle that has those properties.*0221

*You're always looking for just a single triangle.*0226

*There is one thing that you need to check, which is that the two angles that you're given must add up less than 180 degrees.*0230

*That's not so surprising because, of course, if you're given two angles that are bigger than 180 degrees, collectively bigger than 180 degrees, they can't be the two angles of a triangle.*0239

*We all know that three angles of a triangle sum up to 180 degrees.*0248

*The next situation that you might given is Side Angle Angle.*0254

*Let me draw that one.*0260

*This one was angle side angle.*0261

*You might be given side angle angle.*0266

*That means, again, walking around the edge of the triangle.*0272

*You're told what one side is, and then you're told what one angle is, then you're told what the next angle is.*0275

*Assuming that the angles sum up to less than 180 degrees, because if they sum up to more than 180 degrees, you'll never going to get a triangle.*0283

*Assuming that the angles sum up to less than 180 degrees, this again will have always exactly one solution.*0293

*You know you're going to find a triangle and you're just going to have to worry about one triangle.*0301

*The more complicated case, it's called the ambiguous case.*0306

*You might see something in your trigonometry book called the ambiguous case.*0310

*It's Side Side Angle, where you're told the length of one side, then another side, then one angle.*0314

*This one gets a little tricky.*0331

*When you're told two sides in a row, and then an angle that's not the angle in between them, it's one of the other two angles, this does not always have a solution.*0333

*It might have no solution.*0346

*You might try to solve it out and you'll get into some kind of problem.*0348

*We'll see some examples of that so you'll see how it works, or fails to work.*0350

*It might have exactly one triangle that has those properties, or there might be two triangles.*0356

*That gets a little confusing.*0362

*We'll try some examples of that in the examples.*0364

*Angle Angle Angle, you could potentially be given.*0369

*The thing about angle angle angle, if you're just given the three angles of a triangle, well, you could take that angle and blow it up or take that triangle and blow it up or shrink it as much as you want, you'll get similar triangles that have the same three angles.*0376

*That never has unique solution.*0392

*It always has infinitely many solutions, assuming that those three angles sum up to 180 degrees.*0395

*Those triangles, the triangles that you get, are all going to be similar.*0400

*They're all going to be proportionate to each other.*0407

*You could get larger or smaller versions of the same triangle if you're only given angle angle angle.*0410

*Let me fill this in.*0422

*This was the side angle angle, and this was the side side angle case.*0424

*There's two other cases that you might be given, but we're not really going to talk about them in this lecture.*0431

*I'll mention them now and then we'll start solving those triangles in the next lecture.*0436

*The reason is that they really work better for the law of cosines.*0441

*After we learn the law of cosines in the next lecture, then we'll study these two cases.*0444

*Those two cases are side angle side, where you're given two sides and the angle between them, so side angle side.*0450

*That's different from side side angle because of the position of the angles.*0464

*In side angle side, the angle is between the two sides are given.*0469

*In side side angle, it's one of the other angles.*0473

*The other case we'll study in the next lecture is side side side, where you're given all three sides of the triangle.*0477

*Both of those cases don't really lend themselves very well to solutions by the law of sines.*0493

*We'll use the law of sines for these first three cases, angle side angle, side angle angle, and side side angle.*0500

*We'll use the law of sines for those three cases.*0512

*In the next lecture, we'll learn the law of cosines and we'll study side angle side, and side side side.*0515

*We'll get to some examples now.*0523

*First example, we're given an angle, another angle, and a side.*0524

*First thing we need to do is draw a picture of this triangle and then see where we can go from there.*0530

*I've got A, B, C.*0541

*Remember, we'll use capital letters for the angles, and then lowercase letters for the sides.*0545

*You'll always orient them so that lowercase a is opposite angle A.*0550

*That puts B down here, and C over here.*0556

*We're given that c has length 4.*0560

*We're given that angle A measures 50 degrees and angle B measures 60 degrees.*0562

*What we have here is we're given two angles and a side between them, this is angle side angle.*0572

*With angle side angle, that's one has a unique solution if the angles add up to less than 180.*0581

*In this case, the angles sum up to 110, which is less than 180, there is a unique solution.*0592

*We're going to try to solve the triangle, remember, that means finding all the angles.*0617

*Find all the measures of all the angles and the lengths of all the sides.*0625

*There's one that's very easy to start with, which is angle C, that's just 180-a-b, which is 180-110, which is 70 degrees.*0628

*I know that angle C is 70 degrees.*0648

*Looks like my triangle is not really drawn into scale here, because that angle is a little bit smaller than 70 degrees.*0652

*That's okay, we can still do the trigonometry on this.*0658

*I didn't know ahead of time that it was going to be 70 degrees because I haven't solved that part yet.*0662

*That's okay, we'll still work out the rest of the trigonometry.*0667

*Now, we're going to use the law of sines to solve the rest of this.*0672

*Let me write down the law of sines.*0673

*Remember, it says sin(A)/a=sin(B)/b=sin(C)/c.*0675

*Let's see, we've got to figure out what little a and little b are.*0690

*Those are the only things that are missing in our triangle.*0696

*I'm going to use sin(A)/a is equal to, I think I'm going to use sin(C)/c because I already know what little c is.*0697

*If I use B, then I wouldn't know what little b was, I would have too many unknowns.*0714

*Sin(A) is sin(50)/a = sin(70)/4.*0720

*If I cross multiply there, I get 4sin(50)=asin(70), or a=4sin(50)/sin(70).*0735

*Now, it's a matter of plugging those values into my calculator.*0752

*A very important concept that sometimes trigonometry students forget about, I mentioned this in the previous lecture but I want to emphasize it again, is that your calculator needs to be in degree mode if you're working these problems out using degrees.*0757

*If your calculator is in radian mode, then it will know what to do with sin(50) or sin(70) but it won't be what you want.*0772

*You need to convert your calculator into degree mode before you solve any of these problems if you're using degrees.*0783

*I've converted my calculator, I've used the Mode button to convert it from radians into degrees.*0788

*Now, I'm going to work out 4×sin(50)/sin(70), what I get is approximately, I'm rounding this a little bit, 3.26.*0793

*That tells me that a is approximately 3.26.*0816

*So, I've sold for a, now I'm going to solve for little b the same way.*0825

*Sin(B)/b = sin(C)/c, I'll fill in what I know here.*0831

*I know that capital B is 60 degrees, don't know little b yet.*0841

*Capital C is 70 degrees, and little c is 4.*0849

*Again, I'll cross multiply, bsin(70)=4sin(60), b=4sin(60)/sin(70).*0856

*I'll plug that into my calculator, 4sin(60)/sin(70).*0870

*That gives me approximately 3.69.*0880

*Now, I've solved the triangle.*0895

*I've found the lengths of all three sides, and I've found the measures of all three angles.*0897

*The key to solving this problem is to identify which quantities you're given.*0902

*We're given two angles and a side.*0907

*Once we knew that it was angle side angle, we knew that we had a unique solution because the angles didn't add up too big.*0910

*We could find the third angle and then we kind of worked our way around using the law of sines to solve down and find the missing side lengths.*0918

*We'll try some more examples to that.*0927

*Now, we're given triangle ABC.*0931

*Let's see.*0937

*We're given side a, side b, and angle A.*0938

*Let me draw that out.*0942

*My drawing might not be to scale, that's alright.*0944

*So ABC, I'll use capital letters for the angles, and lowercase letters on the opposite sides, a, b and c.*0952

*We're given that angle A is 40 degrees, we're given that side a has length 3, side b has length 4, and we want to solve for the other quantities here.*0961

*The first thing is to identify what kind of information we've been given.*0975

*Two sides and an angle that is not in the middle, this is a side side angle configuration because the angle is not in the middle, that would be side angle side.*0984

*Side side angle, if you remember back to beginning of the lecture, it might have ...*0993

*Side side angle is the ambiguous case.*1006

*It might have no solutions, one solution, or two solutions.*1008

*That's a little disturbing, that this thing could have more than one solution or it might not have solution.*1030

*We have to solve it out and see what we can find.*1035

*We're going to use the law of sines, that says sin(A)/a=sin(B)/b=sin(C)/c.*1038

*We'll solve that out and we'll see what we can find.*1053

*Let's start with ...*1056

*I see that I know angle A and side a, I know side b but not angle B, and I don't know anything about the C.*1059

*I'm going to start out with the a's and b's.*1067

*I'll start out with sin(A)/a=sin(B)/b.*1068

*Now, sin(40)/3=sin(B)/4.*1080

*I'm solving this thing for sin(B), if I multiply the 4 over, I get sin(B)=4sin(40)/3.*1098

*If I work out what 4sin(40)/3 is, it tells me that it's about 0.86.*1113

*I'm looking for an angle whose sine is around 0.86.*1139

*By the way, I've got my calculator in degree mode since the angles were given in degrees.*1144

*I'm looking for an angle whose sine is about 0.86.*1149

*Here's the thing.*1153

*If I just type arcsin(0.86) on my calculator, it will tell me that B is approximately equal to 59.0 degrees.*1154

*This is where it gets really tricky because, remember, if you're looking for an angle whose sine is 0.86, remember that means its y-value, sine is the y-value, is 0.86 ...*1175

*There are two angles that have that value as its sine, there's a θ and then there's this big angle which is 180-θ.*1204

*So, sin(180-θ) is the same as sin(θ).*1222

*Since we know that sin(B) is 0.86, B could be 59 degrees, angle B could also be 180-59 degrees, in other words, 121 degrees.*1233

*We've got two different possibilities for angle B.*1259

*That means we're going to have two different triangles, we have two solutions here.*1265

*We have two solutions depending on which angle B is.*1276

*For each one, we're going to have to solve around and find the other information in the triangle using that value for angle B. That's a little disconcerting.*1282

*Let me go to a new slide and we will draw out each one of those triangles with each one of those solutions.*1290

*We already figured out with this problem that there are two solutions.*1297

*Let me solve each one of those separately.*1310

*Here's A, B and C.*1317

*We're given that A was 40 degrees, and side a has length 3, side b has length 4.*1320

*In the first solution that we've figured out on the previous page, b had measured 59 degrees.*1329

*Then we want to figure out the values of the other angles and the lengths in the triangle.*1342

*First of all, angle C is 180-40-59, that works out to 81.0 degrees.*1348

*Now, let's figure out what the length of side c is, we'll use the law of sines for that.*1376

*Sin(C)/c=sin(A)/a, sin(81.0)/c=sin(40)/3, if we cross multiply that, we'd get c=3sin(81.0)/sin(40).*1384

*Now, we'll work that part out on my calculator, remember, put your calculator in degree mode for this.*1427

*3sin(81.0)/sin(40), I get an approximate answer of 4.61 for side c there.*1437

*Now, I've solved that triangle completely.*1457

*I've found all three sides and all three angles, but there's another solution, remember, with a different angle for B.*1460

*I have to start all over with that possible triangle.*1468

*Let me redraw the triangle from scratch because I don't want to get confused with any of my earlier work.*1471

*Even though I'm drawing it in a similar fashion, remember this is not drawn to scale, this triangle will actually look quite different if we drew it to scale.*1477

*On the other solution, angle B measured 121 degrees.*1490

*A was still 40 because that was given in the original problem.*1502

*Now we have to find the values for this new triangle.*1505

*It's the same solution techniques, we'll be going through the same kinds of calculations but we're using completely different numbers now.*1509

*C=180-40-121.0, that comes out to be 140-121=19.0 degrees.*1517

*Now, we've found C is 19.0, but we still have to find side little c, we'll use the law of sines for that.*1542

*Sin(C)/c=sin(A)/a, if we fill in what we know, Sin(19.0)/c=sin(40)/3.*1553

*If we cross multiply, we'd get c=3sin(19.0s)/sin(40).*1573

*Got my calculator set on degree mode, so 3sin(19)/sin(40), my calculator tells me that that's approximately 1.52.*1584

*Now, I've solved that triangle completely.*1612

*I've got three angles and three side lengths, totally different from the angles and side lengths that we've solved for in the first triangle, even though they both satisfy the initial data given in the problem.*1615

*That's kind of the curse of the side side angle case.*1627

*It is ambiguous and you can get more than one solution.*1634

*The way we figured out there were two solutions was when we initially tried to solve for angle B.*1638

*We've figured out sin(B)/b=sin(A)/a, then we solved for sin(B) was equal to a particular number there but I think it was around 0.8.*1645

*The problem is there was more than one angle that has that sine, and we have to investigate the possibility of both of them.*1664

*That's what led us to the two solutions.*1670

*We went with each one of those angles, we find the one by the arcsine button on my calculator or the inverse sine, the other one we found subtracting that angle from 180 because those angles have the same sine.*1673

*With each of those angles, those set us on two completely different roads to solve down the triangles using the law of sines and find the other values of the triangles, and get us two completely different answers.*1693

*Both of which are valid, both of which satisfy the initial data given in the problem.*1705

*For our next example, we're given a triangle ABC.*1714

*We have, again, two side lengths and angle not in between them.*1719

*Let me draw this out.*1724

*Angles are always capital letters A, B, C.*1731

*The sides are always lowercase letters opposite the angle with the same letter, there's little a, b, and c.*1735

*This time, we're told that side a has length 7, side b has length 12, and angle A measures 45 degrees.*1742

*It says, determine how many triangles there are with these conditions and solve them completely.*1753

*The first thing is to figure out which of those situations were in, side side angle, side angle side, angle side angle.*1757

*What we want to do is look at this and we've got two sides and an angle, but the angle is not in between them, so this is side side angle.*1770

*Side side angle, unfortunately, that's the ambiguous case.*1779

*It could have, no solutions, no triangles at all, it could have one solution or it could have two solutions.*1791

*We don't really know yet without doing a little extra examination here.*1811

*We're going to start out using our law of sines and see what happens.*1818

*Let me remind you what the law of sines is sin(A)/a=sin(B)/b=sin(C)/c.*1821

*Which of those pieces of information have we been given?*1834

*We know what capital A is, so I can find its sine.*1836

*I know what little a is.*1840

*I know what little b is.*1842

*That's it.*1844

*It makes sense to solve for sin(B).*1846

*Let me rewrite that, sin(A)/a=sin(B)/b, so sin(40)/7=sin(B)/12.*1850

*If you cross multiply that and then solve for sin(B), we get sin(B)=12sin(45)/7.*1872

*I'll plug that into my calculator, 12sin(45)/7 comes out to 1.21 approximately.*1886

*This is very significant because, let's look at this, 1.21=sin(B).*1906

*Remember, the sine of any angle is always between -1 and 1.*1917

*Here we've got the sine of an angle equal to 1.21, that's bigger than 1.*1925

*That's a real problem.*1933

*Sin(B) here is greater than 1, that's a contradiction of what we know about sines and cosines, which means that there can not be any triangle satisfying these conditions.*1937

*As soon as we get to sin(B) being bigger than 1, we know that no such triangle exists.*1956

*If you try to find the arcsine on your calculator of something bigger than 1, my calculator gives me an error, because it knows that sine should always be between -1 and 1.*1968

*Immediately, I know that something's gone wrong, and what's gone wrong is that we must have been given bogus initial data.*1982

*We know that no such triangle exists.*1989

*We know that there are no solutions here, which in a sense is fortunate because it means we don't have to do a lot of work to go ahead and try to find the other sides and angles, because we know there isn't any such triangle in the first place.*1992

*We'll try some more examples like this later.*2007

3 answers

Last reply by: Dr. William Murray

Fri Jul 22, 2016 1:08 PM

Post by Peter Ke on June 22 at 06:29:20 PM

This is just a curious question. For SSA, you could have no solution, 1 solution, and 2 solution. If the value of sin(theta) is in between -1 and 1 it will have 2 solution. If the value of sin(theta) is less than -1 or greater than 1 there is no solution. Then what value of sin(theta) will produce 1 solution because you did not show that in the video?

1 answer

Last reply by: Dr. William Murray

Thu Jun 5, 2014 11:46 AM

Post by Charlie Jiang on June 1, 2014

Isn't the law of sines this:

a/SinA = b/SinB = c/SinC ?

1 answer

Last reply by: Dr. William Murray

Tue Mar 11, 2014 3:55 PM

Post by Heather Magnuson on March 6, 2014

There are not any examples of a situation with only one solution. Would this happen when the triangle is a perfect right triangle?

1 answer

Last reply by: Dr. William Murray

Tue Aug 13, 2013 5:12 PM

Post by Taylor Wright on July 19, 2013

How do you determine which angle for SSA is going to be the one which could have 2 values?

1 answer

Last reply by: Dr. William Murray

Sun Apr 28, 2013 11:18 AM

Post by JANE CONNER on February 19, 2012

If angle A is obtuse, how can you tell that sine B will be less than 1. Is it because sin b is the supplement of sin A?