For more information, please see full course syllabus of Trigonometry

For more information, please see full course syllabus of Trigonometry

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### Law of Cosines

**Main formulas**:

- The Law of Cosines:

This holds in*c*^{2}=*a*^{2}+*b*^{2}− 2*ab*cos*C**any*triangle - not just right triangles! This is a generalization of the Pythagorean Theorem. (If*C*is a right angle, then cos*C*= 0, so we get*c*^{2}=*a*^{2}+*b*^{2}.) - Use the Law of Cosines for triangles described in the following ways:
- SAS always has a unique solution (assuming the angle is less than 180
^{° }). - SSS always has a unique solution (assuming each side is less than the sum of the other two).
- (Use Law of Sines for ASA, SAA, and SSA.)

- SAS always has a unique solution (assuming the angle is less than 180
- Heron's Formula:

whereArea =

√

*s*(*s*−*a*)(*s*−*b*)(*s*−*b*), *s*= [1/2](*a*+*b*+*c*) is the*semiperimeter*of the triangle.

**Example 1**:

*ABC*, side

*a*has length 3, side

*b*has length 4, and angle

*C*measures 60

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

**Example 2**:

*ABC*are 5, 7, and 10. Determine how many triangles satisfy these conditions and solve the triangle(s) completely.

**Example 3**:

**Example 4**:

*ABC*are 16, 30, and 34. Determine how many triangles satisfy these conditions and solve the triangle(s) completely.

**Example 5**:

^{° }. Find the length of the third side and the area of the triangle.

### Law of Cosines

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

- Start by drawing a triangle with properly labeled angles and side lengths.
- According to the triangle: SAS and the angle between the sides measures 40
^{°}which is less than 180^{°} - One solution
- Solve using the Law of Cosines
- b
^{2}= a^{2}+ c^{2}− 2ac cosB ⇒ b^{2}= 25^{2}+ 15^{2}− 2(25)(15)cos40^{°}⇒ b^{2}= 275.47 ⇒ b = √{275.47} - b ≈ 16.6
- cosA = [(b
^{2}+ c^{2}− a^{2})/2bc] ⇒ cosA = [(16.6^{2}+ 15^{2}− 25^{2})/2(16.6)(15)] ⇒ cosA = [( − 124.44)/498] ⇒ cosA = − 0.24988 ⇒ A = arccos( − 0.24988) - ∠A ≈ 104.5
^{°} - ∠C = 180
^{°}− 104.5^{°}− 40^{°}= 35.5^{°}

^{°}, ∠C = 35.5

^{°}

- Start by drawing a triangle with properly labeled angles and side lengths.
- According to the triangle: SSS, check to see if each side is less than the sum of the other two sides
- 6 < 8 + 11, 8 < 6 + 11, 11 < 6 + 8
- One solution
- Solve using the Law of Cosines
- cosA = [(b
^{2}+ c^{2}− a^{2})/2bc] ⇒ cosA = [(11^{2}+ 6^{2}− 8^{2})/2(11)(6)] ⇒ cosA = [93/132] ⇒ cosA = 0.7045 ⇒ A = arccos(0.7045) - ∠A ≈ 45.2
^{°} - cosB = [(a
^{2}+ c^{2}− b^{2})/2ac] ⇒ cosB = [(8^{2}+ 6^{2}− 11^{2})/2(8)(6)] ⇒ cosB = [( − 21)/96] ⇒ cosB = ( − 0.21875) ⇒ B = arccos( − 0.21875) - ∠B ≈ 102.6
^{°} - ∠C = 180
^{°}− 45.2^{°}− 102.6^{°}= 32.2^{°}

^{°}, ∠B ≈ 102.6

^{°}, ∠C = 32.2

^{°}

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

- Start by drawing a triangle with properly labeled angles and side lengths.
- According to the triangle: SAS and the angle between the sides measures 15
^{°}which is less than 180^{°} - One solution
- Solve using the Law of Cosines
- c
^{2}= a^{2}+ b^{2}− 2ab cosC ⇒ c^{2}= 6^{2}+ 2^{2}− 2(6)(2)cos15^{°}⇒ c^{2}= 16.82 ⇒ c = √{16.82} - c ≈ 4.1
- cosA = [(b
^{2}+ c^{2}− a^{2})/2bc] ⇒ cosA = [(2^{2}+ 4.1^{2}− 6^{2})/2(2)(4.1)] ⇒ cosA = [( − 15.19)/16.4] ⇒ cosA = − 0.92622 ⇒ A = arccos( − 0.92622) - ∠A ≈ 157.9
^{°} - ∠B = 180
^{°}− 157.9^{°}− 15^{°}= 7.1^{°}

^{°}, ∠B = 7.1

^{°}

- Start by drawing a triangle with properly labeled angles and side lengths.
- According to the triangle: SSS, check to see if each side is less than the sum of the other two sides
- 9 < 12 + 15; 12 < 9 + 15; 15 < 9 + 12
- One solution
- Solve using the Law of Cosines
- cosA = [(b
^{2}+ c^{2}− a^{2})/2bc] ⇒ cosA = [(15^{2}+ 9^{2}− 12^{2})/2(15)(9)] ⇒ cosA = [162/270] ⇒ cosA = 0.6 ⇒ A = arccos(0.6) - ∠A ≈ 53.1
^{°} - cosB = [(a
^{2}+ c^{2}− b^{2})/2ac] ⇒ cosB = [(12^{2}+ 9^{2}− 15^{2})/2(12)(9)] ⇒ cosB = [0/216] ⇒ cosB = (0) ⇒ B = arccos(0) - ∠B ≈ 90
^{°} - ∠C = 180
^{°}− 90^{°}− 53.1^{°}= 36.9^{°}

^{°}, ∠B ≈ 90

^{°}, ∠C = 36.9

^{°}

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

- Start by drawing a triangle with properly labeled angles and side lengths.
- According to the triangle: SAS and the angle between the sides measures 55
^{°}which is less than 180^{°} - One solution
- Solve using the Law of Cosines
- a
^{2}= b^{2}+ c^{2}− 2bc cosA ⇒ a^{2}= 3^{2}+ 10^{2}− 2(3)(10)cos55^{°}⇒ a^{2}= 74.5852 ⇒ a = √{74.5852} - a ≈ 8.6
- cosB = [(a
^{2}+ c^{2}− b^{2})/2ac] cosB = [(8.6^{2}+ 10^{2}− 3^{2})/2(8.6)(10)] cosB = [164.96/172] cosB = 0.9591 B = arccos(0.9591) - ∠B ≈ 16.4
^{°} - ∠C = 180
^{°}− 55^{°}− 16.4^{°}= 108.6^{°}

^{°}, ∠C = 108.6

^{°}

- According to the triangle: SSS, check to see if each side is less than the sum of the other two sides
- 6 < 12 + 8; 8 < 6 + 12; 12 < 6 + 8
- One solution
- Solve using the Law of Cosines
- cosA = [(b
^{2}+ c^{2}− a^{2})/2bc] cosA = [(8^{2}+ 12^{2}− 6^{2})/2(8)(12)] cosA = [172/192] cosA = 0.89583 A = arccos(0.89583) - ∠A ≈ 26.4
^{°} - cosB = [(a
^{2}+ c^{2}− b^{2})/2ac] cosB = [(6^{2}+ 12^{2}− 8^{2})/2(6)(12)] cosB = [116/144] cosB = (0.80556) B = arccos(0.80556) - ∠B ≈ 36.3
^{°} - ∠C = 180
^{°}− 26.4^{°}− 36.3^{°}= 36.9^{°}

^{°}, ∠B ≈ 36.3

^{°}, ∠C = 36.9

^{°}

- According to the triangle: SAS and the angle between the sides measures 55
^{°}which is less than 180^{°} - One solution
- Solve using the Law of Cosines
- a
^{2}= b^{2}+ c^{2}− 2bc cosA ⇒ a^{2}= 15^{2}+ 30^{2}− 2(15)(30)cos30^{o}⇒ a^{2}= 345.573 ⇒ a = √{345.573} - a ≈ 18.6
- cosB = [(a
^{2}+ c^{2}− b^{2})/2ac] ⇒ cosB = [(18.6^{2}+ 30^{2}− 15^{2})/2(18.6)(30)] ⇒ cosB = [1020.96/116] ⇒ cosB = 0.91484 ⇒ B = arccos(0.91484) - ∠B ≈ 23.8
^{°} - ∠C = 180
^{°}− 30^{°}− 23.8^{°}= 126.2^{°}

^{°}, ∠C = 126.2

^{°}

- 43 < 9 + 8; 8 < 3 + 9; 9 < 3 + 8
- One solution
- Solve using the Law of Cosines
- cosA = [(b
^{2}+ c^{2}− a^{2})/2bc] ⇒ cosA = [(3^{2}+ 9^{2}− 8^{2})/2(3)(9)] ⇒ cosA = [26/54] ⇒ cosA = 0.48148 ⇒ A = arccos(0.48148) - ∠A ≈ 61.2
^{°} - cosB = [(a
^{2}+ c^{2}− b^{2})/2ac] ⇒ cosB = [(8^{2}+ 9^{2}− 3^{2})/2(8)(9)] ⇒ cosB = [136/144] ⇒ cosB = (0.9444) ⇒ B = arccos(0.9444) - ∠B ≈ 19.2
^{°} - ∠C = 180
^{°}− 61.2^{°}− 19.2^{°}= 99.6^{°}

^{°}, ∠B ≈ 19.2

^{°}, ∠C = 99.6

^{°}

- Solve using the Law of Cosines
- c
^{2}= a^{2}+ b^{2}− 2ab cosC ⇒ c^{2}= 10^{2}+ 5^{2}− 2(10)(5)cos105^{°}⇒ c^{2}= 150.882 ⇒ c = √{150.882} - c ≈ 12.3
- cosA = [(b
^{2}+ c^{2}− a^{2})/2bc] ⇒ cosA = [(5^{2}+ 12.3^{2}− 10^{2})/2(5)(12.3)] ⇒ cosA = [76.29/123] ⇒ cosA = 0.62024 ⇒ A = arccos(0.62024) - ∠A ≈ 51.7
^{°} - ∠B = 180
^{°}− 51.7^{°}− 105^{°}= 23.3^{°}

^{°}, ∠B = 23.3

^{°}

- First check the following 2 < 7 + 5; 5 < 7 + 2; 7 < 5 + 2

*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 Cosines

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 Cosines Formula
- When to Use Law of Cosines
- Heron's Formula
- Example 1: How Many Triangles Satisfy Conditions, Solve Completely
- Example 2: How Many Triangles Satisfy Conditions, Solve Completely
- Example 3: Find Area of a Triangle Given All Side Lengths
- Extra Example 1: How Many Triangles Satisfy Conditions, Solve Completely
- Extra Example 2: Length of Third Side and Area of Triangle

- Intro 0:00
- Law of Cosines Formula 0:23
- Graphical Representation
- Relates Sides to Angles
- Any Triangle
- Generalization of Pythagorean Theorem
- When to Use Law of Cosines 2:26
- SAS, SSS
- Heron's Formula 4:49
- Semiperimeter S
- Example 1: How Many Triangles Satisfy Conditions, Solve Completely 5:53
- Example 2: How Many Triangles Satisfy Conditions, Solve Completely 15:19
- Example 3: Find Area of a Triangle Given All Side Lengths 26:33
- Extra Example 1: How Many Triangles Satisfy Conditions, Solve Completely
- Extra Example 2: Length of Third Side and Area of Triangle

### 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 Cosines

*Hi! We are doing some more examples on using the law of cos and Heronâ€™s formula.*0000

*We are given one now where the side lengths triangle (a, b and c) are (16, 30, 34).*0005

*We are asked to determine how many triangles satisfy these conditions and to solve the triangles completely.*0013

*What we are given here are three side lengths that is a (side, side, side) situation.*0020

*Because it is (side, side, side), it has a unique solution if it satisfies that check where each side is less than the sum of the other two.*0026

*Let us check that out for these three sides, let us check if (16 < 30+34) that is certainly true.*0059

*Is (30 < 16+34)? Clearly true, Is (34 < 16+30)? That is certainly true.*0068

*There is a unique triangle satisfying these three lengths of sides.*0083

*Let us try to find out what the angles would be in that triangle because we know what the side lengths are, let me draw the triangle.*0099

*(16, 30, 34) I am going to label an angle here, I will label this angle C and then we are going to use the law of cos to find the angles in the triangle.*0112

*That is why I started out by labeling angle C because I know the way I remember cos is with an angle C in it.*0126

*Let me write that down, c ^{2}=(a^{2}) +(( b^{2} - (2ab)cos(C)).*0133

*The way I have labeled angle C, that makes the opposite side (c) and then these sides must be (a) and (b) here.*0145

*I can fill everything into the law of cos and then I can solve for angle (C), I will fill that in 16 ^{2}= (30^{2}+34^{2}) â€“ 2(30)(34) x cos(C).*0154

*(16 ^{2})=256, (30^{2}=900)+(34^{2}=1156), ((2 x 30 x 34=2040 cos(C)).*0178

*The angles are a little bit messy here, Iâ€™m going to move cos(C) over the other and I will get 2040 cos(C)=900+1156-256 that is 1800 because we moved the 256 over on the other side.*0196

*Cos(C) = 1800/2040, (C)=arcos of that horrible fraction (180/204) but I will go ahead and plug that straight into my calculator.*0220

*(180/204) Iâ€™m using degree mode for this, you have to be careful because otherwise it would give you an answer in radians, but that gives me answer of about 28.1 degrees.*0253

*That fills in one of my angles for my triangle there 28.1 degrees.*0271

*We got one of the angles, we will find out the other two exactly the same way, let me go ahead and work them out for you, I will redraw my triangle.*0281

*(16, 30, 34) this was 28.1, we already figured that out.*0293

*In order to find the other two angles, what Iâ€™m going to do is draw angle (C) in a different place now.*0302

*I am relabeling which angle is which, the point is that is I do not have to rewrite my law of cos switching around the (a, b, c) there.*0306

*This time Iâ€™m going to draw my angle (C) there and that makes side (c) the one opposite and (a), (b) other one next to it.*0317

*I will write down my law of cos, (c ^{2})=(a^{2}+b^{2})-((2ab cos (C)).*0328

*I will work through and I will solve for (C), that is (30 ^{2})=(16^{2}+34^{2})-((2(16)(34)cos(C)).*0338

*Now, it is just a little algebra to simplify this, 900=256+1156, now (2×16×34)=1088, and we still have cos(C).*0355

*I want to simplify this, I want to move the cos(C) on the other side, 1088 cos(C)=(256+1156-900), that simplifies down to 512.*0374

*Cos(C) =512/1088, I will take the inverse cos of that in my calculator.*0399

*I get 61.9 degrees approximately, that tells me one more of my angles, I only have one to go.*0418

*I could find the third angle by adding up the two angles and subtracting it 180, but I think I would like to practice the law of cos again.*0440

*At the end we can use that adding up to 180 as to check if we did the law of cos right.*0448

*We are going to find the third angle by using the law of cos again and in order to use my law of cos with the same (a,b, c) formula, Iâ€™m going to rotate the (a, b, c) on the triangle.*0453

*Iâ€™m going to cross out my old (a, b, c) and I will re-label the angle that I do not know as (C) which means its opposite side is 34 (c).*0468

*(a)(b) are now the two sides next to it, I will plug into the law of cos to solve for (C).*0481

*c ^{2} is (34^{2})=a^{2}is (16^{2})+b^{2} is now (30^{2})- 2(16)(30) cos(C).*0491

*34 ^{2} is 1156, 16^{2} is 256, 30^{2} is 900, 2×16×30 is 960, cos (C).*0510

*Now an interesting thing happens because I am going to subtract 256+900 for both sides, that is equal to exactly 1156.*0527

*What do I get is 0 is -960 cos (C) and if I divide it by -960, I will get cos(C)=0.*0537

*What angle has cos(0)? Well that is exactly a right triangle. (C) is exactly a 90 degree angle, let me fill that one in.*0547

*We could have noticed that this was in fact a right triangle because 16 ^{2} + 30^{2} = 34^{2}.*0558

*It does not matter because the law of cos works in all triangles.*0568

*It works just as well in right triangles as in other triangles but of course the rules for right triangles SOHCAHTOA do not work in other triangles.*0571

*Now we solved the triangle completely, we got all three sides and we got all three angles but I want to check and see if those three angles actually do add up to 180 degrees.*0581

*To check here, I look at (28.1+61.9+90) if you add those together you will indeed get 180 degrees which means we must have done the problem right.*0594

*Let us just recap what we did there, we are given a (side, side, side) presentation of a triangle.*0612

*The first thing that we did was check that each one of those sides was less than the sum of the other two.*0621

*If that check have not work out, we would have stopped right there and said there is no such triangle, but that check did worked out.*0626

*Then we go on to using the law of cos to find the each of the angles of the triangle.*0632

*We take the law of cos here and we fill in the lengths of the three sides, and then we solve down to find the cos of a missing angle.*0641

*Once we know the cos of the missing angle, we can use our cos to find the angle itself.*0652

*We did that three times just applying it to each angle and term, we got each of the three angles and then we check in the end if they added up to 180.*0656

*A triangle has two sides of length 8 with an included angle of 45 degrees, we want to find the length of the third side and the area of the triangle.*0000

*Let me try drawing that, there is 8, there is 16, that is about a 45 degree angle but it is not really intended to be drawn on the scale.*0010

*Notice here that we are given here a (side, angle, side) presentation of a triangle, (side, angle, side) and the angle is less than 180 degrees.*0026

*We know that there is a unique triangle satisfying this data.*0036

*We want to find the length of the third side, now this is tailor made for the law of cos.*0041

*Let me write down the law of cos to get started, the law of cos says that (c ^{2})=(a^{2}+b^{2}) â€“ ((2ab cos(C)).*0046

*Let me call the missing side (c), which means that its opposite angle is (C) and (a) and (b) will be the sides that we know.*0061

*Now, we can put all that information into the law of cos and we can solve for the missing side (c).*0069

*I plug that in, (c ^{2}=(8^{2} is 64), (16^{2} is 256) â€“ ((2(8)(16) cos(C)).*0077

*(64+256=320) â€“ (2x8x16=256), (C) is the given angle, that was given as 45 a degree angle and I know what is the cos of 45 degree is.*0094

*That is one of the common values that we learned earlier on the trigonometry lectures, the cos of that is the same as pi/4, the cos of the is square root 2/2.*0114

*That simplifies down to 320-128 square root of 2, (C) is equal to the square root of 320-128 square root of 2.*0127

*That is probably something worth checking out on the calculator, I work out the square root of 320-128 square root of 2.*0144

*It tells me that it is approximately 11.8, we have that third side is approximately 11.8 units long.*0157

*That is the first problem of the example here, now we are asked to find the area of the triangle.*0173

*I want to do a little more trigonometry to find that area, I am going to drop altitude from this top angle here and I want to try to find the length of that altitude.*0182

*The reason Iâ€™m trying to find is that is I remember the area formula, area=1/2 base x height.*0192

*I know the base is 16, that is even labeled with side length (b), do I have it labeled on my triangle.*0201

*The height is the length of that altitude, I have got to solve for that length right there.*0211

*Iâ€™m going to use SOHCAHTOA here because I have the hypotenuse of the triangle and I have the angle here.*0217

*I know that by SOHCAHTOA, sin(theta)=opposite/hypotenuse, sin(45)=opposite/8 and so the opposite=(8)sin(45) is equal to 8.*0224

*sin(45) is something I know because it is a common value, it is pi/4.*0257

*That is square root 2/2, that is 4 square root 2=length of the opposite there, that is 4 square root 2.*0261

*My area, which is Â½ base x height which is Â½(16)(4 square root of 2) is what I figured out the height was.*0275

*That is (8 x 4), that is 32 square root of 2, is my area.*0290

*If you want that to be a decimal, we can approximate that in the calculator as about 45.3.*0297

*That gives us the area of the triangle based on the Â½ base x height calculation, we really done with this one.*0311

*I like to check it using another formula we learned from trigonometry which is heronâ€™s formula.*0319

*Let me remind you what heronâ€™s formula is.*0326

*Heronâ€™s formula says that the area of the triangle, if you know all three sides, which we did figure out, that is the square root of s(s-a)(s-b)(s-c), that was heronâ€™s formula.*0331

*(a, b, c) are the lengths of the sides of the triangle but this mysterious quantity (s) is the semi perimeter, that means Â½ of the perimeter which is the sum of the sides.*0350

*Let us work that out first, we know that two of the sides were 8 and 16, and we worked out on the previous side that the third side is approximately equal to 11.8.*0363

*We put that together, 8 + 16 = 24 +11.8 =35.8 and half of that is 17.9, that is the semi perimeter.*0381

*Let me drop that in to heronâ€™s formula now, the area is equal to (17.9 x 17.9 â€“ a(8)) ((17.9-b(16))((17.9-c(11.8)).*0395

*Now it is just a matter of simplifying that, that is 17.9â€“8 = 9.9, 17.9-16=1.9, 17.9-11.8=6.1.*0430

*Iâ€™m going to multiply those together on my calculator. *0452

*I get 2053.9, take the square root of that and I get approximately 45.3, which is what we figured out on the previous side.*0466

*That is a very useful check that we are doing everything right on the previous side.*0483

*Let us recap what we did there, we are given a triangle with two sides and an included angle (8, 16, and the included angle which is 45).*0488

*We use the law of cos to find the length of the third side of the triangle.*0498

*The law of cos is tailor made if you have (side, angle, side), you use the law of cos to find the length of the third side.*0504

*We then have to find the area of the triangle, I did that the first time by dropping an altitude of the triangle.*0512

*Find the length of the altitude and use the old geometry formula Â½ base x height to find the area.*0519

*The other way we could possibly find the area was to use heronâ€™s formula, which is useful when you know all three sides of the triangle.*0526

*You find the semi perimeter which is what we did here and then you take that and you drop it into heronâ€™s formula for the area and you drop all three sides in there.*0534

*It is just a matter of working through some arithmetic to find the area.*0544

*That is the end of our lecture on the law of cosines and solving triangles and using heronâ€™s formula.*0550

*These are the trigonometry lectures on www.educator.com.*0555

*This is Will Murray for educator.com and we're here today to talk about the law of cosines, which is the second of the two big trigonometric rules.*0000

*Remember, last time we talked about the law of sines.*0007

*You kind of put those together, and together those enable you to find the length of any side and the measure of any angle in a triangle, if you're given enough information to start with.*0011

*Let's start with the formula here.*0022

*The law of cosines is c ^{2}=a^{2}+b^{2}-2abcos(C).*0024

*Let me draw you a a triangle so we can see how that applies.*0031

*Remember, the convention is that you use lowercase a, b, and c for the sides of the triangle, and uppercase A, B, and C for the angles.*0037

*You use the same letter for the angle and the side opposite it.*0048

*My uppercase A goes here, and my B goes here because it's opposite of side b, here's angle C.*0053

*The point of the law of cosines is it relates the lengths of the three sides a, b, and c, little a, little b and little c, to the measure of one of the angles which is capital C here.*0061

*The point is that, first of all, you can use this in any triangle.*0077

*It's not just valid in right triangles.*0080

*Remember the big rule we had, SOH CAH TOA, is only valid in right triangles.*0084

*The law of cosines is valid in any triangle.*0088

*It's a generalization of the Pythagorean theorem, in a sense that, remember the old Pythagorean theorem was just c ^{2}=a^{2}+b^{2}, that only works in a right triangle.*0093

*If you look at the law of cosines, if angle C is a right angle, then the cos(π/2) or the cos(90) is zero.*0105

*If angle C is a right angle, then this term 2ab-cos(c), drops out,the law of cosines just reduces down to the Pythagorean theorem, c ^{2}=a^{2}+b^{2}.*0115

*You can kind of think of the Pythagorean theorem as just being a consequence of the law of cosines.*0128

*The law of cosines is the more general one that applies to any triangle.*0134

*The Pythagorean theorem is the more specific one that just applied when angle C happens to be a right angle.*0137

*Let's see how it's used.*0142

*The law of cosines is really used in two situations.*0146

*First of all, it's used in a side angle side situation.*0151

*That means where you know two sides of a triangle and the included angle.*0155

*The reason it's useful ...*0160

*Let me write the law of cosines again c ^{2}=a^{2}+b^{2}-2abcos(C).*0162

*The point here is that if you label this sides as little a and little b here, that makes this angle capital C and little c is down there.*0173

*If you know the side angle side, in other words, if you know, little a, little b, and capital C, then you know all of the right-hand side of the law of cosines.*0185

*You can then solve for little c.*0196

*That's why the law of cosines is useful for side angle side situations, it's because you can fill in everything you know on one side of the law of cosines, then you can solve for little c.*0199

*It's also useful for side side side situations.*0211

*Let me draw that.*0218

*Side side side means you know all three sides of a triangle, but you don't necessary know any of the angles yet.*0219

*The point is, if you know little, little b, and little c, then you know all of these parts of the law of cosines, so you can solve for the cosine of capital C, and you can figure out what that angle capital C is.*0227

*Then you can figure out what one of the angles is.*0245

*You can just kind of rotate the triangle, and relabel what a, b, and c are to find the other two angles.*0248

*If you know all three sides of a triangle, the law of cosines is very useful for finding the angles, one at a time.*0254

*Remember, there's a couple other ways that you can be given information for triangles.*0261

*You can be given, angle side angle, or side angle angle, or side side angle.*0266

*Those two don't really lend themselves very well to solution by the law of cosines.*0273

*If you're given one of those situations then you really want to use the law of sines which we learned about in the previous lecture.*0280

*There's one more formula we're going to be using in this lecture which is Heron's formula.*0289

*The point here is that, if you know all three lengths of sides of a triangle, I'll call them a, b, and c, as usual, then you have a nice formula for the area.*0294

*It's got one more variable in it, this s.*0309

*S is 1/2 (a+b+c), that's the semi perimeter.*0312

*Remember, the perimeter is the distance around the edge of a triangle, that's a+b+c.*0317

*The semi perimeter is just 1/2 of a+b+c.*0323

*We worked that out ahead of time and we call it s.*0327

*We plugged that into this formula, that's a fairly simple formula just involving s and then a, b, and c, and it spits out the are of the triangle for us.*0331

*That's very useful if you know the lengths of the sides.*0339

*You never really have to look at any angles, and you don't have to get into any sines or cosines, no messy numbers there, hopefully.*0342

*Let's try out some examples here for law of cosines.*0350

*First example, we're given a triangle ABC.*0353

*Let me go ahead and draw that.*0355

*I'm going to put a here, and b here, and c here, which forces the angles, remember the angles go opposite the sides of the same letter.*0362

*We're given the a=3,b=4, and angle c measures 60 degrees. *0374

*We want to first of all determine how many triangles satisfy these conditions and then we want to solve the triangles completely.*0383

*To answer the first question, we have a side angle side situation.*0392

*What we know is that side angle side always has a unique solution assuming the angle is less than 180.*0400

*In this case, the angle is 60 which is less than 180, so there's a unique solution.*0408

*There's exactly one triangle satisfying these conditions.*0419

*That answers the first question, how many triangles satisfy those conditions, exactly one.*0431

*Now we have to solve the triangle completely, this is where the law of cosines is going to be useful.*0438

*Let me copy down the law of cosines c ^{2}=a^{2}+b^{2}-2abcos(C).*0444

*This is very useful because we know a and b, and capital C.*0453

*I can just plug all those in and solve for little c.*0460

*Let me do that, a ^{2} would be 9, b^{2} would be 16 because 4^{2} is 16, minus 2×3×4=24, cos(60) ...*0464

*This is 25-24, now cos(60), that's one of my common values, that's π/3, I remember that the cos(60) is 1/2.*0483

*This is 25-24×1/2, 25-12, c ^{2} is 13, c is equal to the square root of 13.*0493

*I can get an approximation for that on my calculator, that's about 3.61.*0513

*That's approximately equal to 3.61.*0521

*I'll fill that in on my triangle.*0530

*Now, I've got the third side of the triangle.*0536

*The only thing that's left is it says to solve the triangles completely.*0540

*I need to find the other two angles A and B.*0544

*To do that, I'm going to use the law of sines.*0547

*Let me write down the law of sines to refresh your memory, sin(A)/a=sin(C)/c.*0551

*I know what side a, side c, and capital C are.*0569

*I'm going to cross multiply this and I get csin(A)=asin(C).*0574

*I'll fill in the values that I know.*0583

*I know little a is 3, sin(C)=sin(60), sin(A), I don't know that yet, and little c, I figured out, is 3.61.*0585

*I'm solving for sin(A)=3sin(60)/3.61.*0604

*I'll just work that out on my calculator.*0616

*What I get is approximately 0.72.*0629

*By the way, it's very important that your calculator be in degree mode if you're using degrees here.*0637

*I gave angle C as 60 degrees.*0642

*It's very important that you set your calculator to degree mode.*0646

*If your calculator's in radian mode, then it will interpret that 60 as a radian measure, so your answers will be way off, so set your calculator to be in degrees before you try this calculation.*0650

*A is arcsin(0.72), I'll work that out on my calculator.*0661

*That tells me that a is just about 46.0 degrees.*0673

*Now, I've got a measure for angle A.*0683

*I'm going to use the law of sines to find the measure for angle B, but I need a little more space.*0688

*Let me redraw my triangle.*0694

*We've got a, b, and c.*0704

*I figured out the c was 3.61, a was given as 3, b was given as 4, C was given as 60.*0711

*I figured out that A was 46 degrees.*0721

*I'm just trying to find the measure of angle B now.*0727

*I'm going to use again the law of sines.*0729

*Sin(B)/b=sin(C)/c, I'll fill in what I know here, I know that little b is 4, sin(B), I don't know, sin(C)=sin(60), little c is 3.61.*0734

*I'll cross multiply that, 3.61sin(B)=4sin(60), sin(B), that's what we're solving for is equal to 4sin(60)/3.61.*0760

*Let me work that out on my calculator, 4sin(60)/3.61=0.96.*0784

*B is arcsin(0.96) and I'll work that out, that's 73, just about 74 degrees, rounds to 74 degrees.*0801

*Now I figured out angle B, 74 degrees.*0828

*Now, I've solved for all three sides of the triangle, and all three angles of the triangle.*0837

*It's nice at this point, even though we're done with the problem to get some kind of check because we've done lots of calculations here, we could have made a mistake.*0841

*What I'm going to do is add up all three angles in the triangle, and make sure that they come up to be 180 degrees.*0848

*To check on my work here, I'll add up 60+46+74, that does indeed come out to be 180 degrees.*0856

*That suggests that we probably didn't make a mistake in solving all those angles.*0877

*Just to recap this problem here, we're given a side angle side situation, that's a definite tip-off that you're going to be using the law of cosines.*0882

*I filled in my side, the included angle, and a side.*0891

*The first thing I did was I used the law of cosines to find the missing side.*0895

*To solve the triangle completely, I still had two angles that I didn't know, I used the law of sines after that to find the two angles that I didn't know based on knowing the other sides and the other side, and angle.*0903

*Let's try another one now.*0918

*In this one, we're given the side lengths of the triangle, a, b and c are 5, 7 and 10.*0922

*Let me draw a possible triangle like that, 5, 7 and 10.*0926

*We want to find out how many triangles satisfy this conditions and solve the triangles completely.*0935

*The first thing to do with this problem is to identify what we're given.*0941

*We're given a side side side configuration.*0945

*That usually gives you a unique triangle.*0950

*What you have to do is check that each side is less than the sum of the other two.*0952

*Let's check that out.*0960

*Unique if each side is less than the sum of the other two.*0963

*That'll be a quick real check.*0980

*We're comparing 5 with 7+10, 5 is certainly less than 17, so that works.*0983

*7 should be less than 5+10, that certainly works, 7 is less than 15.*0989

*10 should be less than 5+7, that certainly works.*0995

*If any of those checks had failed, for example 10 if it had been 13, 5, and 7, instead of 10, 5, and 7, then the last check would have failed because 13 is not less than 5+7.*1001

*At that point, we would have stopped and said, "This is invalid. There is no triangle that satisfies those conditions."*1015

*Since all those three conditions checked, it does mean that there is a unique solution.*1022

*There's exactly one triangle with those three lengths of sides.*1036

*We've found all the side lengths.*1043

*We need to find the angles, this is where the law of cosines is really handy.*1046

*Let me write that down, the law of cosines says c ^{2}=a^{2}+b^{2}-2abcos(C).*1051

*You can label the sides and angles whichever way you want.*1061

*I'm going to label the top one c.*1067

*Let me write that outside of the triangle.*1075

*That makes the bottom side c and then the two other sides a and b.*1077

*What I can do here is I can plug in little a, little b and little c, and then I can solve for the cosine of capital C, then in turn solve for what angle capital C is.*1085

*Let me do that.*1098

*If c is 10, that's 10 ^{2} equals, a and b is 5^{2}+7^{2}-2×5×7×cos(C).*1100

*Now it's just a little bit of arithmetic, 100=25+49-70cos(C).*1115

*25+49=74, if we pull that over to the other side, we get 26, -70cos(C).*1128

*Cos(C)=-26/70.*1145

*I'm going to figure out that C is arccosine or inverse cosine of -26/70, I'll do that part on my calculator.*1153

*Remember, you have to be in degree mode for this.*1180

*What I get is that C is approximately equal to 111.8 degrees.*1186

*That's 111.8 in that corner.*1196

*Now I'm going to go over in the next page.*1201

*I'm going to keep going to find out the other two angles.*1204

*We'll find them exactly the same way.*1206

*Let me go ahead and redraw my triangle, 5, 7, 10.*1207

*We've already figured out that that angle is 111.8.*1216

*What I'm going to do is relabel the sides and the angles because I still want to use that law of cosines c ^{2}=a^{2}+b^{2}-2abcos(C).*1220

*I'm going to relabel everything here so that I can label C as a new angle, and I can solve for a new angle.*1233

*Relabel that angle as capital C, then my a and b will be 7 and 10.*1242

*Sorry, small C will be 7, and a and b will be 5 and 10.*1256

*We'll go through and we'll work that one out.*1263

*Little c ^{2}, that's 49 is equal to a^{2}, that's 100, plus b^{2} is 25, minus 2ab, that's 2×10×5×cos(C).*1267

*Let me work this out, this is 125, 49 equals 125 minus 2×10×5, that's 100, cos(C).*1285

*If I subtract 125 from both sides, I get -76 is -100cos(C), divide both sides by -100, we get 76/100=cos(C).*1300

*C=arccos(76/100), I'll plug that into my calculator.*1324

*It tells me that that's approximately 40.5 degrees for that angle right there.*1342

*Now, there's one angle left to find.*1362

*Again, I'm going to relabel the sides and the angles so that I can continue to use the law of cosines in its standard form.*1364

*I don't have to change around what a, b and c are in the law of cosines.*1373

*I'll do this in red.*1375

*In red, I'm going to call this angle capital C, and that means I have to relabel my sides, that means little c is equal to 5, a and b are 7 and 10.*1380

*Now I'm going to plug those values into the law of cosines.*1399

*c ^{2} is 25, equals 49, 7^{2}, plus b^{2} is 100, minus 2×a×b, 7×10, times cos(C).*1403

*Now it's a matter of solving that for capital C again.*1422

*I get 25 equals 149, minus 2×7×10, that's 140, cos(C).*1424

*If I subtract 149 from both sides, I get -124, is equal to -140cos(C).*1440

*Cos(C)=124/140, C=arccos(124/140), I'll go to the calculator on that.*1455

*I get 27.7 degrees.*1481

*Let me fill that in, 27.7 degrees.*1490

*Now we've solved the triangle completely.*1494

*We started out with all three side lengths and we found all three angles in the triangle.*1497

*We're really done but it's always good to find a way to check your work.*1503

*Let me check my work in blue here.*1506

*Again, I found all three angles.*1508

*I'm going to add them together and see if I get 180 to check that out.*1511

*I'm going to add up 111.8+40.5+27.7, what I get is exactly 180 degrees.*1514

*That tells me that I must have been right in getting those three angles for the triangle.*1535

*Just to recap here, what we were given was three sides of a three angle.*1540

*We were given three side lengths, that was a side side side situation.*1546

*We had to check that the three lengths, that we didn't have a situation where two sides added up to be less than the third side.*1554

*We had to check that each side was less than the sum of the other two.*1560

*Once we did that, we knew we have exactly one solution.*1564

*Then we filled in the side lengths of the triangle, and we used the law of cosines.*1568

*The law of cosines lets you fill in three side lengths and then solve for the cosine of one of the angles.*1574

*You work it through, solve for the cosine, then you get each one of the angles by taking the arccosine, then you just go through a separate procedure like that for each one of the angles.*1580

*In our third problem, we're trying to find the area of a triangle whose side lengths are 5, 7 and 10.*1595

*Now, this one is really a set up for Heron's formula, because Heron's formula works perfectly when you know the three sides of a triangle.*1603

*We're going to use Heron here.*1612

*Heron says that the area is equal to the square root of s times s minus a, s minus b, and s minus c.*1618

*You've got to figure out what s is.*1632

*S is the semi-perimeter of a triangle which means 1/2 of the perimeter, 1/2 of the sum of the three side lengths.*1635

*That's 1/2 of 5+7+10, which is 1/2 of 22, which is 11.*1646

*I'm going to plug that in to Heron's formula, wherever I see an s, that's 11×(11-5)×(11-7)×(11-10).*1658

*Then I'll just work on simplifying that, that's 11×6×4×1, that's the square root of 264, 11×24.*1677

*That problem was really pretty quick if you remember Heron's formula.*1697

*Heron's formula is very useful.*1700

*If you know three side lengths of a triangle, then what you do is you work out the semi-perimeter, you just drop the side lengths into this formula for the semi-perimeter, then you drop the semi-perimeter and the three side lengths for the Heron's formula for the area.*1703

*It simplifies down pretty quickly to give you the area of the triangle.*1720

*We'll try some more examples later on.*1722

1 answer

Last reply by: Dr. William Murray

Fri Mar 20, 2015 6:29 PM

Post by Glenn O'Neill on March 19, 2015

in Example 2 you use the Law of Cosines to find each angle by relabeling all the angles to fit the formula... wouldn't it be a lot easier to label the triangle once, solve for angle C and then use the law of sines to find other other 2 angles?

2 answers

Last reply by: Dr. William Murray

Thu Jul 18, 2013 8:22 AM

Post by John Hunter on July 6, 2013

I hope i haven't missed the answer to this question somewhere in the lecture. You gave us the number of solutions for ASA, SAA, SSA, and AAA for the law of sines, but is there a shortcut to tell how many solutions a triangle will have when using the law of cosines with SAS and SSS? Many Thanks.

1 answer

Last reply by: Dr. William Murray

Thu May 30, 2013 4:01 PM

Post by Shyann Williams on March 29, 2011

Angle B = (angle A-C)

1 answer

Last reply by: Dr. William Murray

Thu May 30, 2013 3:59 PM

Post by Mark Mccraney on January 18, 2010

There is a typo in Heron's formula as stated in Quicknotes: the fourth term, (s-b), should be (s-c). This will take in account for all three sides of the triangle, not 2 and side b being used twice.