Week 12 The Law of Cosines and Using Vectors Overview This week we look at a law that handles some of the situations that the Law of Sines could not. The Law of Cosines can be used as well, in part, to develop another formula for the area of a triangle. We will continue with some applications, including bearings. We will also begin a short course on vectors, which give an alternate approach to solving some of the same problems we have been doing. Why another approach? It is easier to work some problems with vectors instead of the Law of Cosines. We will: Solve triangles using the Law of Cosines. Solve applications using the Law of Cosines. Use Heron s Formula to find the area of a triangle. Study the basics of vectors that would help in our problems. Solve applications using vectors. Algebra skills needed: The same ones you have been using. Geometry knowledge needed: Continued use of all angles. Parallelograms. The Law of Cosines The Law of Cosines is derived by placing a triangle, with standard labeling, in the Cartesian coordinate system with vertex A at the origin, B on the x-axis, and C in a quadrant.
2 The focus begins with angle A. The distance formula is used to find a, the side opposite A, with x and y being replaced by bcosa and bsina, and then the whole thing simplified. It takes more steps then I want to do here. You can find it in textbooks. It is good practice for those who plan on taking Calculus to go through the distance formula, substitution for x and y, and simplifying. The same approach could be done with any angle and its opposite side. The Law of Cosines is: a b c 2bc cosa b a c 2ac cosb c a b 2bc cosc It is important to notice that the side being solved for is opposite the angle for which you find the cosine. The other two sides are then used in the calculation on the right side. It is definitely more work than the Law of Sines, but we need it to handle those situations that the Law of Sines does not. There are two basic situations side-angle-side or SAS, and three sides or SSS. We will look at an example for each situation. Example 1: Solve the following triangle. This is a SAS situation. We solve for side a first. a b c b a 2 c cos A 20 28 2(20)(28)cos56 2 a 557.7039* * * a 557.7039 a 23.6 or 24 ***Calculator use: Enter each item as you read it. Although it can all be done in one step, you may want to do as above find a 2 first and then find a as the square root. When in doubt as to how far to carry out, it is best to go more than you might think. Although we use 24 for the final answer, we will use 23.6 in any calculation to keep some accuracy. Let s find B next. Now we can use the Law of Sines. 20 23.6 20sin56 sinb sinb sin56 23.6 1 sin B.70257 B sin.70257 B 44.6 in QI. In QII, B 180 44.6 135. 4
3 The only solution for B is 44.6 or 45, since 135.4 will not work with A = 56. We findc 180 (56 44.6 ) 79.4. We round 79.4 to 79 like A. To summarize: A 56, B 45, C 79 a 24, b 20, c 28 By summarizing, we can be sure we have everything as well as writing the numbers with the same accuracy. We can also check to see that the largest angle is opposite the longest side, etc. The next example shows the SSS situation. It can also be a lot of work to solve such a triangle. Because we will be solving for angles, it is easier to look at what the formula will look like when we solve for an angle. a b c 2bc cos A We solve for cos A. a b c 2bc cos A a b c cos A We can use this or eliminate some negative signs. 2bc a b c 1 c o s A 2bc 1 b c a cos A 2bc Do you need to memorize this? No. You can look at it here while you work problems (and I will take advantage of that), but you can always solve from the original form of the law. Example 2: Solve the following triangle. We have to start somewhere, so it might as well be to find A. Use the above form of the law. b c a cos A 2bc 7 14 10 cos A 2(7)(14).739796 *** 2
4 ***Calculator use: To do the calculating in one step you must enter parentheses around both the numerator and denominator because of the operations. An alternate way is to enter the entire numerator and the enter key, followed by ( 2 x 7 x 14 ) enter. The parentheses are still necessary to make sure you are dividing by the complete denominator and not just the first number. Now we finish finding A: 1 cos A.739796 A cos.739796 A 42.49 or 42.3 for now Why is there not a second possibility for A? The cosine of an obtuse angle would be negative (in QII) and that did not happen. This law will automatically tell you whether it is an acute angle or an obtuse angle according to the sign of the cosine. Because quadrant II angles are in the range of the inverse cosine, the calculator will give you the obtuse angle. To find the next angle B, I prefer the Law of Sines. If you use the Law of Cosines, be sure to substitute the side values in the correct positions. 10 7 7sin 42.3 sinb sin 42.3 sinb 10 1 sin B.47111 B sin.47111 B 28.1 in QI, or B 180 28.1 152.9 in QII. The 152.9 will not work with A = 42.3. Therefore, B = 28.1. We find the last angle C 180 (42.3 28.1 ) 109.6. Rounding all numbers the same, we summarize: A 42, B 28, C 110 a 10, b 7, c 14. Although the sides were given, it helps to see them listed with their opposite angles for checking the answers as reasonable. á Now work the problem set for The Law of Cosines.
5 1. Solve the following triangle. Problems for The Law of Cosines 2. Solve the following triangle.
6 Answers to Problems for The Law of Cosines 1. We find side b opposite B = 63 by using the Law of Cosines. b a c 2ac cosb b 18 23 2(18)(23)cos63 2 b 477.0959 b 477.0959 b 21.84 or 21.8 18 21.8 18sin63 sin A sin A sin63 21.8 1 sin A.73569 A sin.73569 A 47.37 or 47.4 in QI, A 180 47.4 132.6 in QII. However, A = 132.6 will not work with B = 63. So A = 47.4 is the only solution for A. We findc 180 (63 47.4 ) 69.6. We summarize: A 47, B 63, C 70 a 18, b 22, c 23 2. We might as well start with A and use the Law of Cosines. b c a cos A 2bc 41 25 cos A 2(42)(21) 1 cos A.89569 Acos.89569 A 26.40 (The only solution since the cosine is positive.) To find B: (You can use the Law of Cosines, if you prefer.) 25 42 42sin26.4 sinb sin26.4 sinb 25 sin B.746987 B 48.3 in QI, 180 48.3 131.7 in QII. However, B = 48.3 will not work because B is opposite the longest side and should be the largest angle. With B = 48.3, C = 105.3 becomes the largest angle opposite the longest side. This is not true here. We conclude that B = 131.7 andc 180 (26.4 131.7 ) 21.9. To summarize (rounding to the whole numbers as given): A 26, B 132, C 22 a 25, b 42, c 21 Note: The numbers in these problems were given as whole numbers. Usually that is not the case. If not told how far to go in rounding off, use the given information as a guide. While finding all the information you need, however, keep your numbers out at least one more place.
7 Applications Using the Law of Cosines Many of the applications that use the Law of Cosines look just like the ones for the Law of Sines. The information that is given in a problem dictates which law will be used. Example 1: A tower 46 feet high supports high voltage wires. The tower makes an angle of 75 with the side of a hill on which it stands. Cables attached to the top of the tower are anchored to the hill. If the lower cable is attached to the hill 25 feet below the tower, how long is the cable? We need to find the length of LT. Call it x. We find that LBT = 180-75 = 105. We know SAS and can use the Law of Cosines. x 46 25 2(46)(25)cos 105 2 x 3336. 28 x 57.76 The lower cable is 58 feet long. Example 2: A new street light is to be erected, placing a light at the end of an arm and a bracket used for support, as indicated in the drawing below. Find the angle that the light arm makes with the pole. We use the version of the Law of Cosines that solves for the cosine, with θ across from side 3.4. cos 2.5 1.5 3.4 2( 2.5)(1.5) cos.5100 120.66 1 cos (.5100) The arm makes an angle of 120.7 with the pole. For anyone who substitutes in the original law and solves, it look like this: 3.4 2.5 1.5 2(2.5)(1.5) 2(2.5)(1.5)cos 2.5 1.5 3.4 2.5 1.5 3.4 cos 2(2.5)(1. 5), etc. cos Solve for cos. Problems with bearings may also need the Law of Cosines.
8 Example 3: Two towns, 24 miles apart as the crow flies, are separated by a dense forest. To go the shortest distance from town A to town B, a person has to go 19 miles on a bearing of 325, then turn and continue for 8 miles to reach town B. Find the bearing of town B from town A. We will need to find θ and then convert it to a bearing. First we find CAB using the Law of Cosines. 19 24 8 coscab 2(19)(24 ) cos CAB.95724 CAB CAB 16.8 1 cos.95724 We can find that CAN = 360-325 = 35. Then θ = 35-16.8 = 18.2. We round the angle to 18 and convert to a bearing (in the same form as the one given in the problem) of 360-18 = 342. Town B is at a bearing of 342 from town A. Mark your diagrams carefully, make them larger if necessary, and think.. á Now work the problem set for Applications Using the Law of Cosines.
9 Problems for Applications Using the Law of Cosines 1. To measure the distance across a lake, a surveyor measures a distance from one end of the lake to point A as 115 meters. He then measures the distance from the other end of the lake to point A as 220 meters. If the angle between measurements at A is 112.5, how long is the lake? 2. A billiard ball is struck by a cue stick at point A on the table. The ball travels 2.4 feet and strikes another ball at point B. The ball is then deflected through an angle and moves 4.3 feet to point C on the cushion. If distance AC is 5.4 feet, find the angle of deflection to the nearest whole degree. 3. An airplane flies from airport A at a bearing of 115 for 185 miles to another airport B. From B it flies 110 miles on a bearing of 220 to airport C. How far is airport C from airport A?
10 Answers to Problems for Applications Using the Law of Cosines 1. We have a SAS situation so we can use the Law of Cosines to find side d (or a if you prefer). d 115 220 2(115)(220)co s112.5 2 d 80988.78 d 284.6 The lake is 285 meters long. 2. We have 3 sides and need to find the angle θ or B. We use the Law of Cosines written to solve for angle B. cosb a c b 4.3 2.4 5.4 2ac cosb cos B.23789 cos 2(4.3)(2.4) 1 B (.23789) B 103.8 The angle of deflection is 104. 3. We need to first find ABC between the known sides. The N-S lines are parallel lines cut by transversal AB, so alternate interior angles are equal. Therefore, ABS = 115. CBS = 220-180 = 40, That gives us ABC = 115-40 = 75. We can now use the Law of Cosines to find d (or b) opposite 75. b a c 2ac cosb b 110 185 2(110)(185)cos75 2 b 35791.1 b189.2 Airport C is 189 miles from airport A.
11 Heron s Formula for the Area of a Triangle The Law of Cosines can be used in conjunction with the previous area formula and quite a bit of Algebra to derive another area formula. Why? This formula uses the sides and no angles or altitude. It is named after the Greek mathematician Heron (around 100 B.C.). With a, b, c as the sides of the triangle, the formula is: Area s( s a)( s b)( s c ) a b c where s 2 The s is half of the perimeter. We have just seen problems where we know three sides, so they are not uncommon. The formula is more work than the others, but the only one possible at times. Example 1: Find the area of a triangle with sides 24, 20, and 38. We don t need a drawing, just let a = 24, b = 20, and c = 38. We find a b c 24 20 38 s s 41. Use the area formula: Area s( s a) ( s b )( s c ) 41(41 24)(41 20)(41 38) 41(17)(21) (3) 43911 209.55 or 210 Example 2: A triangular plot borders a fork in the road. The dimensions of the plot are indicated in the drawing. Find the area of the plot. a b c 73 90 114 s s 138.5 Area s( s a)( s b )( s c ) 138.5(138.5 73)(138.5 90)(138.5 114) 138.5(65.5)(48.5)(24.5) 10779506.9 3283. 2 The area of the plot is about 3283 square meters. á Now work the problem set for Heron s Formula for the Area of a Triangle.
12 Problems for Heron s Formula for the Area of a Triangle 1. Find the area of a triangle with sides of 26, 14, and 32. 2. A gardener is marking off a triangular plot of land in his back yard and will need to determine the area to figure the amount of mulch needed. The dimensions are in the figure below. Find the area of the plot.
13 Answers to Problems for Heron s Formula for the Area of a Triangle 1. a b c 26 14 32 s s 36 Area s( s a)( s b)( s c) 36(36 26)(36 14)(36 32) 36(10)(22)(4) 177.99 or 178 2. a bc 68 51 102 s s 110.5 Area s( s a )( s b )( s c ) 110.5(110.5 68)(110.5 51)(110.5 102) 110.5(42.5)(59.5)(8.5) 2375128.4 1541.1 The area of the plot is 1541 sq. ft.
14 Basic Properties of Vectors For students who plan on taking Calculus and upper level math, you should go through the section(s) in your book on vectors. Most students do not find them difficult once they get used to the terminology and new notation. If you have questions let me know. Since I am addressing the main points of Trigonometry, I will focus on only the characteristics and properties of vectors that are immediately useful for our applications. What is a vector? It is a directed line segment. Its beginning point is the initial point and its end point is its terminal point, indicated by an arrow. They can be placed anywhere in a plane. A vector is determined by its length and direction. How do we refer to a vector? There are two main ways to name a vector. We assign letters for the initial and terminal points and use an arrow over the letters: AC. We can also use letters, usually u and v, in bold print: u, v, v 1. We will want to place vectors in a type of coordinate plane similar to the Cartesian coordinate plane. In the figure, it appears that vectors u and v have the same length and same direction. In the vector system, it is true that two vectors that have the same length and direction are equal: u = v. This is good because it means we can use one vector in place of another, and we will choose to use a vector like v, that has initial point at the origin.
15 How do we determine the direction of the vector? There are two ways. When placed in a plane, with initial point at the origin, the end point will have a type of coordinates that we call components. They will be written as x, y. I am choosing to use x and y so as to relate to other systems, but textbook authors may use pq, or a,b or some other choice of letters. The other way to determine direction is to use the angle that the vector makes with the positive x-axis. (Does that idea sound familiar?) y From this we can see that tan. x How do we determine the length of a vector? The length of a vector is called its magnitude. The magnitude is indicated by the notation v. Looking at the figure above and recalling what we have done before, we have magnitude v x y. Example 1: A vector has components 5,12. Sketch the vector and find its magnitude and direction. v x y ( 5) (12) 2 169 13 (Length is positive.) tan y 12 12 x 5 5 12 5 1 tan 67.38 This negative QIV angle from the range of the inverse tangent gives us 67.38 as the reference angle in QII where the tangent is also negative. Therefore, We can round this to 112.6 or 113. 180 67.38 112.62.
16 Sometimes we know θ and magnitude and are asked to find the components of the vector with initial point at the origin. We cannot find x or y directly. We need another form for the components. Look at the vector figure again. The magnitude of the vector is the same as r, the radius of a circle, or c as the hypotenuse of a right triangle. Using θ and ǁvǁ, we recall x cos x v cos v y sin y v sin v This makes the components x, yv cos, v sin. Example 2: Find the components for a vector with magnitude of 70 and direction of 152. v cos, v sin70cos152, 70sin152 61.8, 329. 32.9 You can check by seeing if tan152. It won t be exactly the same because of 61.8 rounding off, but it should be fairly close. The magnitude in applications is often a force (applied for a constant time) or a speed (distance for one hour). The direction angle is the angle at which the force is applied, or the bearing. Before we go into these applications, we need one more property of vectors. How do we add vectors? Geometrically we can do it two ways. One way is to put the initial point of the second vector at the terminal point of the first vector. The resulting sum vector, called the resultant, is the vector from the first vector s initial point to the second vector s terminal point, as indicated below. Algebraically we add components. If u x, y and v x, y then u v x x, y y 1 1 1 2 1 2 Example 3: Given u = 2,5 and v = -8,3, find u + v. u + v = 2 + -8, 5 + 3 = -6, 8 á Now work the problem set for Basic Properties of Vectors.
17 Problems for Basic Properties of Vectors 1. A vector u has components -7,24. Sketch the vector and find the magnitude and direction angle. 2. Sketch the vector and find the components for a vector with magnitude 150 and direction angle of 212. 3. Given vector u = 9,-13 and v = -5,-10, find the components of u + v, the resultant magnitude, and resultant direction angle. Sketch all three vectors.
18 Answers to Problems for Basic Properties of Vectors 1. u x y ( 7) (24) 625 25 y 24 tan x 7 tan 3.42857 1 tan ( 3.42857) 73. 7 This is a QIV angle from the inverse tangent range. Our direction angle is positive so 73.7 is the acute reference angle. In QII, the direction angle is180 73.7 106.3. 2. v 150, 212 x, y 150cos212,150sin212 127.2, 79. 5 (These should look reasonable.) 3. R u v 9 5, 13 10 4, 23 R 2 2 (4) ( 23) 23 tan 5.75 4 1 tan ( 5.75) 80. 545 23.3 1 While -80.1 is in QIV, direction angles are usually positive. Therefore, 360 80.1 279.9.
19 Using Vectors The other way to add vectors geometrically comes from the idea of a parallelogram. From the diagram, we see that vectors u and v can be considered as sides of a parallelogram with the sum vector as the diagonal of the parallelogram. If we want to, we can then choose to work with the vector parallel (and equal) to v that will have its initial point at the origin. In some problems, this idea can make it easier to label vectors in a drawing with all of the information from the problem as well as what we need to find. It is mostly a matter of choice. I like to label the sum vector or resultant R. In our applications, we will work with components in trigonometric form. Example 1: If vector u has a magnitude of 350 with a direction of 48, and vector v has a magnitude of 65 with a direction of 140, find the magnitude and direction of the resultant. First, we write the components of each vector. u 350cos48,350sin 48 234.2,26 0.1 v 65cos140, 65sin140 49.8, 41.8 Add components to get vector R. R 234.2 49.8, 260.1 41.8 184.4, 301.9 Now we find the magnitude of R by using the components. R 2 x y 1.9 184.4 30 353.8 The direction angle for R is found by the tangent of θ. y 301.9 tan tan.16757 x 184.4 1 tan.16757 58.6 You should see that the magnitude and direction angle for R are reasonable when compared to the other two vectors.
20 The above problem may seem like a lot of work, but it is fairly quick. The calculations are short when compared with the Law of Cosines. The diagram above should look similar to some of the applications with bearings. That is because the use of vectors is an alternate approach to that type of problem. Let s look at two basic applications. Example 2: A force of 26.9 pounds acts together with a force of 14.3 pounds at an angle of 48.3. Find the magnitude of the resultant and the direction angle θ with the smaller force. We have two vectors that we can place where we want as long as the angle between them is 48.3. We will choose to put the shorter vector along the positive x-axis (because it gives us an easy direction angle of 0 ). The other vector will have a direction angle of 48.3 The components for the shorter vector are 14.3cos0, 14.3sin 14.3, 0. The components for the other given vector are 26.9cos48.3,26.9sin 48. 3 17. 89, 20.01. The components of the resultant are 14.3+17.89, 0+20.01 = 32.19, 20.01. The magnitude of the resultant is R 32.19 20.01 37.9. To find the direction angle θ: y 20.01 tan.62162 x 32.19 1 tan.62162 31.9 The magnitude of the resulting force is 37.9 pounds at an angle of 31.9. (The magnitude and angle direction for the resultant are reasonable compared to the other vectors.) After doing a few problems, you should be fairly quick with these calculations. I strongly urge you to do a quick check of some kind, even if it is just whether your answers are reasonable or not. (You could also use the Law of Cosines to find the magnitude of the resultant.)
21 Example 3: An airplane s velocity with respect to the air is 580 miles per hour and it is headed N58 W. The wind, at the altitude of the plane, is from the southwest and has a velocity of 60 miles per hour. What is the true direction of the plane, and what is its true speed? We draw the vectors with their bearings, with a wind from the southwest and going northeast at a direction of N45 E. The speeds (for one hour) act as magnitudes. Before writing components, we need to convert bearings to direction angles for the vectors. The plane s direction angle is 90 + 58 = 148. The wind s direction angle is 45. We can now write the components. The plane: 580cos148, 580sin148. (We don t need the calculation yet.) The wind: 60cos45,60sin45. The resultant: 580cos148 60cos45,580sin148 60sin45 449.4,349.8. The magnitude: R ( 449.4) (349.8 ) 569.49 To find the actual bearing β: y 349.8 tan.77837 is the resultant direction angle. x 449. 4 1 tan (.77837) 37.9 in QIV. This will give us the acute reference angle of 37.9. The direction angle for the resultant is 180-37.9 = 142.1. For our bearing β, 142.1 90 52.1 or 52. The plane s actual speed is 569 mph, and its actual bearing is N52 W. á Now work the problem set for Using Vectors.
22 Problems for Using Vectors 1. If u has a magnitude of 345 with direction angle of 105 and v has a magnitude of 250 with direction angle of 29, find the magnitude and direction angle of the resultant. Make a sketch showing all of the vectors and labeling with the information given. 2. Two forces, one of 36 pounds and the other of 50 pounds, act on the same object. The angle between the forces is 24. Find the magnitude of the resultant and the angle it makes with the smaller force. 3. An airplane flies with airspeed of 260 miles per hour and a compass heading of 110. If a 36 mile-per- hour wind is blowing out of the north (bearing 180 ), what is the plane s actual speed and heading?
23 Answers to Problems for Using Vectors 1. Write components for each vector and then add. u : 345cos105, 345sin105 v : 250cos29, 250sin29 R : 345cos105 250cos29, 345sin105 250sin29 R : 129.4, 454.4 R : 129.4 454. 4 223223.7 472.5 454.4 74.1 1294. 1 1 tan tan 3.51159 2. 36 : 36cos0,36sin0 50 : 50cos24, 50sin24 R : 36cos0 50cos24, 36sin0 50sin24 81.7, 20.3 R 81.7 20.3 7086.98 20.3 81.7 1 1 tan tan.24847 84.2 13.95 The resulting force is 84 pounds at an angle of 14. 3. We need to convert bearings to direction angles (the angle made with the positive x-axis). The airplane s direction angle is -20 (or 340 if you prefer). The wind s direction angle is -90 (or 270 if you prefer). Finding the components: The plane: 260cos( 20 ), 260sin( 20) The wind: 36cos( 90 ), 36sin( 90 ) R : 260cos( 20 ) 36cos( 90 ), 260sin( 20 ) 36sin( 90) R : 244.3, 124.9 R 244.3 ( 124.9) 124.9 244. 3 274.4 1 1 tan tan (.5 11257) 27.1 The direction angle is the acute angle of 27.1 (just in a negative direction), so 90 27.1 117.1. The plane s actual speed is 274 mph at a bearing of 117.