Applications of Trigonometry functions

applications of trigonometry answer key and applications of trigonometry in engineering
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10 Further Applications of Trigonometry Figure 1 general Sherman, the world’s largest living tree. (credit: mike baird, Flickr) ChAPTeR OUTl Ine 10.1 non-right Triangles: l aw of Sines 10.2 non-right Triangles: l aw of Cosines 10.3 Polar Coordinates 10.4 Polar Coordinates: graphs 10.5 Polar Form of Complex numbers 10.6 Parametric equations 10.7 Parametric equations: graphs 10.8 vectors Introduction 27 e w Th orld’s largest tree by volume, named General Sherman, stands 274.9 feet tall and resides in Northern California. Just how do scientists know its true height? A common way to measure the height involves determining the angle of elevation, which is formed by the tree and the ground at a point some distance away from the base of the tree. This method is much more practical than climbing the tree and dropping a very long tape measure. In this chapter, we will explore applications of trigonometry that will enable us to solve many different kinds of problems, including finding the height of a tree. We extend topics we introduced in Trigonometric Functions and investigate applications more deeply and meaningfully. 27 Source: National Park Service. "The General Sherman Tree." http://www.nps.gov/seki/naturescience/sherman.htm. Accessed April 25, 2014. 761762 CHAPTER 10 f ur ther aP Plic tioa Ns of t rig o Nome ytr l eARnIng Obje CTIveS In this section, you will: • Use the Law of Sines to solve oblique triangles. • Find the area of an oblique triangle using the sine function. • Solve applied problems using the Law of Sines. 10.1 nOn-RIgh T TRIAngle S: lAW OF SIne S Suppose two radar stations located 20 miles apart each detect an aircra b ft etween them. The angle of elevation measured by the first station is 35 degrees, whereas the angle of elevation measured by the second station is 15 degrees. How can we determine the altitude of the aircraft? We see in Figure 1 that the triangle formed by the aircraft and the two stations is not a right triangle, so we cannot use what we know about right triangles. In this section, we will find out how to solve problems involving non-right triangles. 15° 35° 20 miles Figure 1 Using the l aw of Sines to Solve Oblique Triangles In any triangle, we can draw an altitude, a perpendicular line from one vertex to the opposite side, forming two right triangles. It would be preferable, however, to have methods that we can apply directly to non-right triangles without first having to create right triangles. Any triangle that is not a right triangle is an oblique triangle. Solving an oblique triangle means finding the measurements of all three angles and all three sides. To do so, we need to start with at least three of these values, including at least one of the sides. We will investigate three possible oblique triangle problem situations: 1. ASA (angle-side-angle) We know the measurements of two angles and the included side. See Figure 2. β γ α Figure 2 2. AAS (angle-angle-side) We know the measurements of two angles and a side that is not between the known angles. See Figure 3. β γ α Figure 3 3. SSA (side-side-angle) We know the measurements of two sides and an angle that is not between the known sides. See Figure 4. β γ α Figure 4SECTION 10.1 No N-ri ght t ria Ngle s: l w oa f s i Ns e 763 Knowing how to approach each of these situations enables us to solve oblique triangles without having to drop a perpendicular to form two right triangles. Instead, we can use the fact that the ratio of the measurement of one of the angles to the length of its opposite side will be equal to the other two ratios of angle measure to opposite side. Let’s see how this statement is derived by considering the triangle shown in Figure 5. γ a b h α β c Figure 5 h h _ _ Using the right triangle relationships, we know that sin α = and sin β = . Solving both equations for h gives two a b differ ent expressions for h. h = bsin α and h = asin β We then set the expressions equal to each other. bsin α = asin β 1 1 1 _ _ _ (bsin α) = (asin β) Multiply both sides by .     ab ab ab sin β sin α _ _ = a b Similarly, we can compare the other ratios. sin γ sin β sin γ sin α _ _ _ _ = and = a c c b Collectively, these relationships are called the Law of Sines. sin β sin γ sin α _ _ _ = = a c b Note the standard way of labeling triangles: angle α (alpha) is opposite side a; angle β (beta) is opposite side b; and angle γ (gamma) is opposite side c. See Figure 6. While calculating angles and sides, be sure to carry the exact values through to the final answer. Generally, final answers are rounded to the nearest tenth, unless otherwise specified. β c a α γ b Figure 6 Law of Sines Given a triangle with angles and opposite sides labeled as in Figure 6, the ratio of the measurement of an angle to the length of its opposite side will be equal to the other two ratios of angle measure to opposite side. All proportions will be equal. The Law of Sines is based on proportions and is presented symbolically two ways. sin β sin γ sin α _ _ _ = = a c b a b c _ _ _ = = sin α sin γ sin β To solve an oblique triangle, use any pair of applicable ratios.764 CHAPTER 10 f ur ther aP Plic tioa Ns of t rig o Nome ytr Example 1 Solving for Two Unknown Sides and Angle of an AAS Triangle Solve the triangle shown in Figure 7 to the nearest tenth. β 10 c 50° 30° γ α b Figure 7 Solution e t Th hree angles must add up to 180 degrees. From this, we can determine that β = 180° − 50° − 30° = 100° To fin d an unknown side, we need to know the corresponding angle and a known ratio. We know that angle α = 50° and its corresponding side a = 10. We can use the following proportion from the Law of Sines to find the length of c. sin(50°) sin(30°) _ _ = c 10 sin(50°) _ c = sin(30°) Multiply both sides by c. 10 10 _ c = sin(30°) Multiply by the reciprocal to isolate c. sin(50°) c ≈ 6.5 Similarly, to solve for b, we set up another proportion. sin(50°) sin(100°) _ _ = 10 b bsin(50°) = 10sin(100°) Multiply both sides by b. 10sin(100°) _ b = Multiply by the reciprocal to isolate b. sin(50°) b ≈ 12.9 er Th efore, the complete set of angles and sides is α = 50° a = 10 β = 100° b ≈ 12.9 γ = 30° c ≈ 6.5 Try It 1 β Solve the triangle shown in Figure 8 to the nearest tenth. a c 98° 43° α 22 Figure 8 Using The l aw of Sines to Solve SSA Triangles We can use the Law of Sines to solve any oblique triangle, but some solutions may not be straightforward. In some cases, more than one triangle may satisfy the given criteria, which we describe as an ambiguous case. Triangles classified as SSA, those in which we know the lengths of two sides and the measurement of the angle opposite one of the given sides, may result in one or two solutions, or even no solution. SECTION 10.1 No N-ri ght t ria Ngle s: l w oa f s i Ns e 765 possible outcomes for SSA triangles Oblique triangles in the category SSA may have four die ff rent outcomes. Figure 9 illustrates the solutions with the known sides a and b and known angle α. No triangle, a h Right triangle, a = h Two triangles, a h, a b One triangle, a ≥ b γ γ γ γ b b b b a a a a a h α β α β α β α β (a) (b) (c) (d) Figure 9 Example 2 Solving an Oblique SSA Triangle Solve the triangle in Figure 10 for the missing side and find the missing angle measures to the nearest tenth. γ 8 6 35° β α Figure 10 Solution Use the Law of Sines to find angle β and angle γ, and then side c. Solving for β, we have the proportion sin β sin α _ _ = a b sin β sin(35°) _ _ = 6 8 8sin(35°) _ = sin β 6 0.7648 ≈ sin β −1 sin (0.7648) ≈ 49.9° β ≈ 49.9° However, in the diagram, angle β appears to be an obtuse angle and may be greater than 90°. How did we get an acute angle, and how do we find the measurement of β? Let’s investigate further. Dropping a perpendicular from γ and viewing the triangle from a right angle perspective, we have Figure 11. It appears that there may be a second triangle that will fit the given criteria. γ' 8 66 35° α' β' βφ Figure 11 e a Th ngle supplementary to β is approximately equal to 49.9°, which means that β = 180° − 49.9° = 130.1°. (Remember that the sine function is positive in both the first and second quadrants.) Solving for γ, we have γ = 180° − 35° − 130.1° ≈ 14.9°766 CHAPTER 10 f ur ther aP Plic tioa Ns of t rig o Nome ytr We can then use these measurements to solve the other triangle. Since γ' is supplementary to α' and β', we have γ' = 180° − 35° − 49.9° ≈ 95.1° Now we need to find c and c'. We have c 6 _ _ = sin(14.9°) sin(35°) 6sin(14.9°) _ c = ≈ 2.7 sin(35°) Finally, c' 6 _ _ = sin(95.1°) sin(35°) 6sin(95.1°) _ c' = ≈ 10.4 sin(35°) To summarize, there are two triangles with an angle of 35°, an adjacent side of 8, and an opposite side of 6, as shown in Figure 12. γ γ' 95.1° 14.9° b' = 8 b = 8 a' = 6 a = 6 130.1° 49.9° 35° 35° α β α' β' c ≈ 2.7 c' ≈ 10.4 (a) (b) Figure 12 However, we were looking for the values for the triangle with an obtuse angle β. We can see them in the first triangle (a) in Figure 12. Try It 2 Given α = 80°, a = 120, and b = 121, find the missing side and angles. If there is more than one possible solution, show both. Example 3 Solving for the Unknown Sides and Angles of a SSA Triangle In the triangle shown in Figure 13, solve for the unknown side and angles. Round your answers to the nearest tenth. β 12 a α 85° 9 Figure 13 Solution In choosing the pair of ratios from the Law of Sines to use, look at the information given. In this case, we know the angle γ = 85°, and its corresponding side c = 12, and we know side b = 9. We will use this proportion to solve for β. sin β sin(85°) _ _ = Isolate the unknown. 12 9 9sin(85°) _ = sin β 12SECTION 10.1 No N-ri ght t ria Ngle s: l w oa f s i Ns e 767 To find β, apply the inverse sine function. The inverse sine will produce a single result, but keep in mind that there may be two values for β. It is important to verify the result, as there may be two viable solutions, only one solution (the usual case), or no solutions. 9sin(85°) −1_ β = sin   12 −1 β ≈ sin (0.7471) β ≈ 48.3° In this case, if we subtract β from 180°, we find that there may be a second possible solution. Thus, β = 180° − 48.3° ≈ 131.7°. To check the solution, subtract both angles, 131.7° and 85°, from 180°. This gives α = 180° − 85° − 131.7° ≈ − 36.7°, which is impossible, and so β ≈ 48.3°. To fi nd the remaining missing values, we calculate α = 180° − 85° − 48.3° ≈ 46.7°. Now, only side a is needed. Use the Law of Sines to solve for a by one of the proportions. sin(85°) sin(46.7°) _ _ = a 12 sin(85°) _ a = sin(46.7°) 12 12sin(46.7°) __ a = ≈ 8.8 sin(85°) e co Th mplete set of solutions for the given triangle is α ≈ 46.7° a ≈ 8.8 β ≈ 48.3° b = 9 γ = 85° c = 12 Try It 3 Given α = 80°, a = 100, b = 10, find th e missing side and angles. If there is more than one possible solution, show both. Round your answers to the nearest tenth. Example 4 Finding the Triangles That Meet the Given Criteria Find all possible triangles if one side has length 4 opposite an angle of 50°, and a second side has length 10. Solution Using the given information, we can solve for the angle opposite the side of length 10. See Figure 14. sin(50°) sin α _ _ = 10 4 10sin(50°) _________ sin α = 4 sin α ≈ 1.915 α 4 50° 10 Figure 14 We can stop here without finding the value of α. Because the range of the sine function is −1, 1, it is impossible for −1 the sine value to be 1.915. In fact, inputting sin (1.915) in a graphing calculator generates an ERROR DOMAIN. er Th efore, no triangles can be drawn with the provided dimensions. Try It 4 Determine the number of triangles possible given a = 31, b = 26, β = 48°.768 CHAPTER 10 f ur ther aP Plic tioa Ns of t rig o Nome ytr Finding the Area of an Oblique Triangle Using the Sine Function Now that we can solve a triangle for missing values, we can use some of those values and the sine function to find the 1 _ area of an oblique triangle. Recall that the area formula for a triangle is given as Area = bh, where b is base and h is 2 height. For oblique triangles, we must find h before we can use the area formula. Observing the two triangles in Figure 15, one acute and one obtuse, we can drop a perpendicular to represent the height and then apply the trigonometric opposite h _ _ property sin α = t o write an equation for area in oblique triangles. In the acute triangle, we have sin α = c hypotenuse or csin α = h. However, in the obtuse triangle, we drop the perpendicular outside the triangle and extend the base b to form a right triangle. The angle used in calculation is α', or 180 − α. ββ cc hh aa α α' γγ α bb Figure 15 u Th s, 1 1 _ _ Area = (base)(height) = b(csin α) 2 2 Similarly, 1 1 _ _ Area = a(bsin γ) = a(csin β) 2 2 area of an oblique triangle e f Th ormula for the area of an oblique triangle is given by 1 _ Area = bcsin α 2 1 _ = acsin β 2 1 _ = absin γ 2 This i s equivalent to one-half of the product of two sides and the sine of their included angle. Example 5 Finding the Area of an Oblique Triangle Find the area of a triangle with sides a = 90, b = 52, and angle γ = 102°. Round the area to the nearest integer. Solution Using the formula, we have 1 _ Area = absin γ 2 1 _ Area = (90)(52)sin(102°) 2 Area ≈ 2289 square units Try It 5 Find the area of the triangle given β = 42°, a = 7.2 ft, c = 3.4 ft. Round the area to the nearest tenth. Solving Applied Problems Using the l aw of Sines e m Th ore we study trigonometric applications, the more we discover that the applications are countless. Some are flat, diagram-type situations, but many applications in calculus, engineering, and physics involve three dimensions and motion. SECTION 10.1 No N-ri ght t ria Ngle s: l w oa f s i Ns e 769 Example 6 Finding an Altitude Find the altitude of the aircraft in the problem introduced at the beginning of this section, shown in Figure 16. Round the altitude to the nearest tenth of a mile. a 15° 35° 20 miles Figure 16 Solution To find the elevation of the aircraft, we first find the distance from one station to the aircraft, such as the side a, and then use right triangle relationships to find the height of the aircraft, h. Because the angles in the triangle add up to 180 degrees, the unknown angle must be 180° − 15° − 35° = 130°. This angle is opposite the side of length 20, allowing us to set up a Law of Sines relationship. sin(130°) sin(35°) _ _ = a 20 asin(130°) = 20sin(35°) 20sin(35°) _ a = sin(130°) a ≈ 14.98 e di Th stance from one station to the aircraft is about 14.98 miles. Now that we know a, we can use right triangle relationships to solve for h. opposite _ sin(15°) = hypotenuse h _ sin(15°) = a h _ sin(15°) = 14.98 h = 14.98sin(15°) h ≈ 3.88 e a Th ircraft is at an altitude of approximately 3.9 miles. Try It 6 e di Th agram shown in Figure 17 represents the height of a blimp flying over C a football stadium. Find the height of the blimp if the angle of elevation at the southern end zone, point A, is 70°, the angle of elevation from the northern end zone, point B, is 62°, and the distance between the viewing points of the two end zones is 145 yards. 70° 62° AB 145 yards Figure 17 Access the following online resources for additional instruction and practice with trigonometric applications. • l aw of Sines: The basics (http://openstaxcollege.org/l/sinesbasic) • l aw of Sines: The Ambiguous Case (http://openstaxcollege.org/l/sinesambiguous)770 CHAPTER 10 f ur ther aP Plic tioa Ns of t rig o Nome ytr 10.1 SeCTIOn exeRCISeS veRbAl 1. Describe the altitude of a triangle. 2. Compare right triangles and oblique triangles. 3. When can you use the Law of Sines to find a missing 4. In the Law of Sines, what is the relationship between the angle? angle in the numerator and the side in the denominator? 5. What type of triangle results in an ambiguous case? Algeb RAIC For the following exercises, assume α is opposite side a, β is opposite side b, and γ is opposite side c. Solve each triangle, if possible. Round each answer to the nearest tenth. 6. α = 43°, γ = 69°, a = 20 7. α = 35°, γ = 73°, c = 20 8. α = 60°, β = 60°, γ = 60° 9. a = 4, α = 60°, β = 100° 10. b = 10, β = 95°, γ = 30° For the following exercises, use the Law of Sines to solve for the missing side for each oblique triangle. Round each answer to the nearest hundredth. Assume that angle A is opposite side a, angle B is opposite side b, and angle C is opposite side c. 11. Find side b when A = 37°, B = 49°, c = 5. 12. Find side a when A = 132°, C = 23°, b = 10. 13. Find side c when B = 37°, C = 21, b = 23. For the following exercises, assume α is opposite side a, β is opposite side b, and γ is opposite side c. Determine whether there is no triangle, one triangle, or two triangles. Then solve each triangle, if possible. Round each answer to the nearest tenth. 14. α = 119°, a = 14, b = 26 15. γ = 113°, b = 10, c = 32 16. b = 3.5, c = 5.3, γ = 80° 17. a = 12, c = 17, α = 35° 18. a = 20.5, b = 35.0, β = 25° 19. a = 7, c = 9, α = 43° 20. a = 7, b = 3, β = 24° 21. b = 13, c = 5, γ = 10° 22. a = 2.3, c = 1.8, γ = 28° 23. β = 119°, b = 8.2, a = 11.3 For the following exercises, use the Law of Sines to solve, if possible, the missing side or angle for each triangle or triangles in the ambiguous case. Round each answer to the nearest tenth. 24. Find angle A when a = 24, b = 5, B = 22°. 25. Find angle A when a = 13, b = 6, B = 20°. 26. Find angle B when A = 12°, a = 2, b = 9. For the following exercises, find the area of the triangle with the given measurements. Round each answer to the nearest tenth. 27. a = 5, c = 6, β = 35° 28. b = 11, c = 8, α = 28° 29. a = 32, b = 24, γ = 75° 30. a = 7.2, b = 4.5, γ = 43° gRAPhICAl For the following exercises, find the length of side x. Round to the nearest tenth. 31. 32. 33. 45° 25° x 10 6 15 x 120° 75° 70° 50° xSECTION 10.1 s ectio N e xercises 771 35. 34. 36. 8.6 111° x 14 22° 18 x x 50° 42° 110° 40° For the following exercises, find the measure of angle x, if possible. Round to the nearest tenth. 37. 38. 39. 98° 37° 13 5 5 8 x x 22° 10 x 11 41. Notice that x is an obtuse angle. 42. 40. 65° 5.7 5.3 12 x 24 59° x 21 10 55° x For the following exercises, find the area of each triangle. Round each answer to the nearest tenth. 44. 45. 18 43. A 25° 30° 16 15 10 32.6 24.1 93° BC 46. 48. 47. 51° 58° 11 9 4.5 25 2.9 18 51° 40° 3.5 30 49. 115° 50 30°772 CHAPTER 10 f ur ther aP Plic tioa Ns of t rig o Nome ytr exTenSIOnS 50. Find the radius of the circle in Figure 18. Round to 51. Find the diameter of the circle in Figure 19. Round the nearest tenth. to the nearest tenth. 8.3 145° 3 110° Figure 18 Figure 19 52. Find m ∠ADC in Figure 20. Round to the nearest 53. Find AD in Figure 21. Round to the nearest tenth. tenth. A A 10 12 8 13 9 44° 53° 60° B D C B D C Figure 21 Figure 20 54. Solve both triangles in Figure 22. Round each 55. Find AB in the parallelogram shown in Figure 23. answer to the nearest tenth. AB A 130° 48° 12 4.2 10 C 130° E 46° B C D 2 48° Figure 23 D Figure 22 56. Solve the triangle in Figure 24. (Hint: Draw a 57. Solve the triangle in Figure 25. (Hint: Draw a perpendicular from H to JK). Round each answer to perpendicular from N to LM). Round each answer the nearest tenth. to the nearest tenth. H L 7 5 20° J K 10 Figure 24 74° MN 4.6 Figure 25 SECTION 10.1 s ectio N e xercises 773 58. In Figure 26, ABCD is not a parallelogram. ∠m is x AB m 35° obtuse. Solve both triangles. Round each answer to n the nearest tenth. 29 45 y h k D 40 65° C Figure 26 ReAl-W ORld A PPl ICATIOnS 59. A pole leans away from the sun at an angle of 7° 60. To determine how far a boat is from shore, two radar to the vertical, as shown in Figure 27. When the stations 500 feet apart find the angles out to the elevation of the sun is 55°, the pole casts a shadow boat, as shown in Figure 28. Determine the distance 42 feet long on the level ground. How long is the of the boat from station A and the distance of the pole? Round the answer to the nearest tenth. boat from shore. Round your answers to the nearest whole foot. 7° C 70° 60° 55° AB 42 ft Figure 27 AB Figure 28 61. Figure 29 shows a satellite orbiting Earth. The 62. A communications tower is located at the top of satellite passes directly over two tracking stations A a steep hill, as shown in Figure 30. The angle of and B, which are 69 miles apart. When the satellite inclination of the hill is 67°. A guy wire is to be is on one side of the two stations, the angles of attached to the top of the tower and to the ground, elevation at A and B are measured to be 86.2° and 165 meters downhill from the base of the tower. The 83.9°, respectively. How far is the satellite from angle formed by the guy wire and the hill is 16°. station A and how high is the satellite above the Find the length of the cable required for the guy ground? Round answers to the nearest whole mile. wire to the nearest whole meter. 16° 165m 86.2° 83.9° 67° AB Figure 30 Figure 29774 CHAPTER 10 f ur ther aP Plic tioa Ns of t rig o Nome ytr 63. e r Th oof of a house is at a 20° angle. An 8-foot solar 64. Similar to an angle of elevation, an angle of panel is to be mounted on the roof and should be depression is the acute angle formed by a horizontal angled 38° relative to the horizontal for optimal line and an observer’s line of sight to an object results. (See Figure 31). How long does the vertical below the horizontal. A pilot is flying over a straight support holding up the back of the panel need to be? highway. He determines the angles of depression to Round to the nearest tenth. two mileposts, 6.6 km apart, to be 37° and 44°, as shown in Figure 32. Find the distance of the plane from point A to the nearest tenth of a kilometer. 8  38° 44° 37° 20° Figure 31 A B Figure 32 65. A pilot is flying over a straight highway. He 66. In order to estimate the height of a building, two determines the angles of depression to two students stand at a certain distance from the mileposts, 4.3 km apart, to be 32° and 56°, as shown building at street level. From this point, they find the in Figure 33. Find the distance of the plane from angle of elevation from the street to the top of the point A to the nearest tenth of a kilometer. building to be 39°. They then move 300 feet closer to the building and find the angle of elevation to be 50°. Assuming that the street is level, estimate the height 32° of the building to the nearest foot. 56° A B Figure 33 67. In order to estimate the height of a building, two 68. Points A and B are on opposite sides of a lake. Point students stand at a certain distance from the C is 97 meters from A. The measure of angle BAC building at street level. From this point, they find is determined to be 101°, and the measure of angle the angle of elevation from the street to the top of ACB is determined to be 53°. What is the distance the building to be 35°. They then move 250 feet from A to B, rounded to the nearest whole meter? closer to the building and find the angle of elevation to be 53°. Assuming that the street is level, estimate the height of the building to the nearest foot. 1 _ 70. Two search teams spot a stranded climber on a 69. A man and a woman standing 3 mi les apart spot 2 mountain. The first search team is 0.5 miles from a hot air balloon at the same time. If the angle of the second search team, and both teams are at an elevation from the man to the balloon is 27°, and altitude of 1 mile. The angle of elevation from the the angle of elevation from the woman to the first search team to the stranded climber is 15°. The balloon is 41°, find the altitude of the balloon to angle of elevation from the second search team the nearest foot. to the climber is 22°. What is the altitude of the climber? Round to the nearest tenth of a mile. SECTION 10.1 s ectio N e xercises 775 71. A street light is mounted on a pole. A 6-foot-tall 72. Three cities, A, B, and C, are located so that city A man is standing on the street a short distance from is due east of city B. If city C is located 35° west of the pole, casting a shadow. The angle of elevation north from city B and is 100 miles from city A and from the tip of the man’s shadow to the top of his 70 miles from city B, how far is city A from city B? head of 28°. A 6-foot-tall woman is standing on the Round the distance to the nearest tenth of a mile. same street on the opposite side of the pole from the man. The angle of elevation from the tip of her shadow to the top of her head is 28°. If the man and woman are 20 feet apart, how far is the street light from the tip of the shadow of each person? Round the distance to the nearest tenth of a foot. 73. Two streets meet at an 80° angle. At the corner, a 74. Brian’s house is on a corner lot. Find the area of the park is being built in the shape of a triangle. Find front yard if the edges measure 40 and 56 feet, as the area of the park if, along one road, the park shown in Figure 34. measures 180 feet, and along the other road, the House park measures 215 feet. 135° 40  56  Figure 34 75. e B Th ermuda triangle is a region of the Atlantic 76. A yield sign measures 30 inches on all three sides. Ocean that connects Bermuda, Florida, and Puerto What is the area of the sign? Rico. Find the area of the Bermuda triangle if the distance from Florida to Bermuda is 1030 miles, the distance from Puerto Rico to Bermuda is 980 miles, and the angle created by the two distances is 62°. 77. Naomi bought a modern dining table whose top is in the shape of a triangle. Find the area of the table top if two of the sides measure 4 feet and 4.5 feet, and the smaller angles measure 32° and 42°, as shown in Figure 35. 4 feet 4.5 feet 32° 42° Figure 35776 CHAPTER 10 f ur ther aP Plic tioa Ns of t rig o Nome ytr l eARnIng Obje CTIveS In this section, you will: • Use the Law of Cosines to solve oblique triangles. • Solve applied problems using the Law of Cosines. • Use Heron’s formula to fi nd the area of a triangle. 10.2 nOn-RIgh T TRIAngle S: lAW OF COSIne S Suppose a boat leaves port, travels 10 miles, turns 20 degrees, and travels another 8 miles as shown in Figure 1. How far from port is the boat? 8 mi 20° 10 mi Port Figure 1 Unfortunately, while the Law of Sines enables us to address many non-right triangle cases, it does not help us with triangles where the known angle is between two known sides, a SAS (side-angle-side) triangle, or when all three sides are known, but no angles are known, a SSS (side-side-side) triangle. In this section, we will investigate another tool for solving oblique triangles described by these last two cases. Using the l aw of Cosines to Solve Oblique Triangles The tool we need to solve the problem of the boat’s distance from the port is the Law of Cosines, which defines the relationship among angle measurements and side lengths in oblique triangles. Three formulas make up the Law of Cosines. At first glance, the formulas may appear complicated because they include many variables. However, once the pattern is understood, the Law of Cosines is easier to work with than most formulas at this mathematical level. Understanding how the Law of Cosines is derived will be helpful in using the formulas. The derivation begins with the Generalized Pythagorean Theorem , which is an extension of the Pythagorean Theorem to non-right triangles. Here is how it works: An arbitrary non-right triangle ABC is placed in the coordinate plane with vertex A at the origin, side c drawn along the x-axis, and vertex C located at some point (x, y) in the plane, as illustrated in Figure 2. Generally, triangles exist anywhere in the plane, but for this explanation we will place the triangle as noted. y C (b cosθ, b sinθ) b y a A θ x x 2 c c (0, 0) B x Figure 2SECTION 10.2 No N-ri ght t ria Ngle s: l w oa f co s i Ns e 777 We can drop a perpendicular from C to the x-axis (this is the altitude or height). Recalling the basic trigonometric identities, we know that x(adjacent) y(opposite) __ __ cos θ = a nd sin θ = b(hypotenuse) b(hypotenuse) In terms of θ, x = bcos θ and y = bsin θ. The ( x, y) point located at C has coordinates (bcos θ, bsin θ). Using the side (x − c) as one leg of a right triangle and y as the second leg, we can find the length of hypotenuse a using the Pythagorean Theorem. Thus, 2 2 2 a = (x − c) + y 2 2 = (bcos θ − c) + (bsin θ) Substitute (bcos θ) for x and (bsin θ) for y. 2 2 2 2 2 = (b cos θ − 2bccos θ + c ) + b sin θ Expand the perfect square. 2 2 2 2 2 2 2 = b cos θ + b sin θ + c − 2bccos θ Group terms noting that cos θ + sin θ = 1. 2 2 2 2 2 = b (cos θ + sin θ) + c − 2bccos θ Factor out b . 2 2 2 a = b + c − 2bccos θ e f Th ormula derived is one of the three equations of the Law of Cosines. The other equations are found in a similar fashion. Keep in mind that it is always helpful to sketch the triangle when solving for angles or sides. In a real-world scenario, try to draw a diagram of the situation. As more information emerges, the diagram may have to be altered. Make those alterations to the diagram and, in the end, the problem will be easier to solve. Law of Cosines The Law of Cosines states that the square of any side of a triangle is equal to the sum of the squares of the other two sides minus twice the product of the other two sides and the cosine of the included angle. For triangles labeled as in Figure 3, with angles α, β, and γ, and opposite corresponding sides a, b, and c, respectively, the Law of Cosines is given as three equations. β 2 2 2 a = b + c − 2bc cos α 2 2 2 b = a + c − 2ac cos β 2 2 2 c = a + b − 2ab cos γ a c To solve for a missing side measurement, the corresponding opposite angle measure is needed. When solving for an angle, the corresponding opposite side α γ b measure is needed. We can use another version of the Law of Figure 3 Cosines to solve for an angle. 2 2 2 2 2 2 2 2 2 b + c − a a + c − b a + b − c _ _ _ cos α = cos β = cos γ = 2bc 2ac 2ab How To… Given two sides and the angle between them (SAS), find the measures of the remaining side and angles of a triangle. 1. Sketch the triangle. Identify the measures of the known sides and angles. Use variables to represent the measures of the unknown sides and angles. 2. Apply the Law of Cosines to find the length of the unknown side or angle. 3. Apply the Law of Sines or Cosines to find the measure of a second angle. 4. Compute the measure of the remaining angle. 778 CHAPTER 10 f ur ther aP Plic tioa Ns of t rig o Nome ytr Example 1 Finding the Unknown Side and Angles of a SAS Triangle Find the unknown side and angles of the triangle in Figure 4. γ ba 5 10 30° α β c 5 12 Figure 4 Solution First, make note of what is given: two sides and the angle between them. This arrangement is classified as SAS and supplies the data needed to apply the Law of Cosines. Each one of the three laws of cosines begins with the square of an unknown side opposite a known angle. For this example, the first side to solve for is side b, as we know the measurement of the opposite angle β. 2 2 2 b = a + c −2accos β 2 2 2 b = 10 + 12 − 2(10)(12)cos(30°) Substitute the measurements for the known quantities. — √3 2 _ b = 100 + 144 − 240 Evaluate the cosine and begin to simplify.   2 — 2 b = 244 − 120 √3 — — b = √244 − 120 √ 3 Use the square root property. b ≈ 6.013 Because we are solving for a length, we use only the positive square root. Now that we know the length b, we can use the Law of Sines to fill in the remaining angles of the triangle. Solving for angle α, we have sin β sin α _ _ = a b sin(30°) sin α _ _ = 10 6.013 10sin(30°) _ sin α = Multiply both sides of the equation by 10. 6.013 10sin(30°) 10sin(30°) −1_ _ α = sin Find the inverse sine of .   6.013 6.013) α ≈ 56.3° e o Th ther possibility for α would be α = 180° − 56.3° ≈ 123.7°. In the original diagram, α is adjacent to the longest side, so α is an acute angle and, therefore, 123.7° does not make sense. Notice that if we choose to apply the Law of Cosines, we arrive at a unique answer. We do not have to consider the other possibilities, as cosine is unique for angles between 0° and 180°. Proceeding with α ≈ 56.3°, we can then find the third angle of the triangle. γ = 180° − 30° − 56.3° ≈ 93.7° e co Th mplete set of angles and sides is α ≈ 56.3° a = 10 β = 30° b ≈ 6.013 γ ≈ 93.7° c = 12 Try It 1 Find the missing side and angles of the given triangle: α = 30°, b = 12, c = 24. SECTION 10.2 No N-ri ght t ria Ngle s: l w oa f co s i Ns e 779 Example 2 Solving for an Angle of a SSS Triangle Find the angle α for the given triangle if side a = 20, side b = 25, and side c = 18. Solution For this example, we have no angles. We can solve for any angle using the Law of Cosines. To solve for angle α, we have 2 2 2 a = b + c −2bccos α 2 2 2 20 = 25 + 18 −2(25)(18)cos α Substitute the appropriate measurements. 400 = 625 + 324 − 900cos α Simplify in each step. 400 = 949 − 900cos α −549 = −900cos α Isolate cos α. −549 _ = cos α −900 0.61 ≈ cos α −1 cos (0.61) ≈ α Find the inverse cosine. α ≈ 52.4° See Figure 5. γ b = 25 α 52.4° a = 20 c = 18 β Figure 5 Analysis Because the inverse cosine can return any angle between 0 and 180 degrees, there will not be any ambiguous cases using this method. Try It 2 Given a = 5, b = 7, and c = 10, find t he missing angles. Solving Applied Problems Using the l aw of Cosines Just as the Law of Sines provided the appropriate equations to solve a number of applications, the Law of Cosines is applicable to situations in which the given data fits the cosine models. We may see these in the fields of navigation, surveying, astronomy, and geometry, just to name a few. Example 3 Using the Law of Cosines to Solve a Communication Problem On many cell phones with GPS, an approximate location can be given before the GPS signal is received. This is accomplished through a process called triangulation, which works by using the distances from two known points. Suppose there are two cell phone towers within range of a cell phone. The two towers are located 6,000 feet apart along a straight highway, running east to west, and the cell phone is north of the highway. Based on the signal delay, it can be determined that the signal is 5,050 feet from the first tower and 2,420 feet from the second tower. Determine the position of the cell phone north and east of the first tower, and determine how far it is from the highway. 780 CHAPTER 10 f ur ther aP Plic tioa Ns of t rig o Nome ytr Solution For simplicity, we start by drawing a diagram similar to Figure 6 and labeling our given information. 2,420 . 5,050 . θ 6,000 . Figure 6 Using the Law of Cosines, we can solve for the angle θ. Remember that the Law of Cosines uses the square of one side to find the cosine of the opposite angle. For this example, let a = 2420, b = 5050, and c = 6000. Th us, θ corresponds to the opposite side a = 2420. 2 2 2 a = b + c − 2bccos θ 2 2 2 (2420) = (5050) + (6000) − 2(5050)(6000)cos θ 2 2 2 (2420) − (5050) − (6000) = − 2(5050)(6000)cos θ 2 2 2 (2420) − (5050) − (6000) ___ = cos θ −2(5050)(6000) cos θ ≈ 0.9183 −1 θ ≈ cos (0.9183) θ ≈ 23.3° To answer the questions about the phone’s position north and east of the tower, and the distance to the highway, drop a perpendicular from the position of the cell phone, as in Figure 7. This forms two right triangles, although we only need the right triangle that includes the first tower for this problem. 5,050 . y 23.3° x Figure 7 Using the angle θ = 23.3° and the basic trigonometric identities, we can find the solutions. Thus x _ cos(23.3°) = 5050 x = 5050 cos(23.3°) x ≈ 4638.15 feet y _ sin(23.3°) = 5050 y = 5050sin(23.3°) y ≈ 1997.5 feet e ce Th ll phone is approximately 4,638 feet east and 1,998 feet north of the first tower, and 1,998 feet from the highway. Example 4 Calculating Distance Traveled Using a SAS Triangle Returning to our problem at the beginning of this section, suppose a boat leaves port, travels 10 miles, turns 20 degrees, and travels another 8 miles. How far from port is the boat? The diagram is repeated here in Figure 8.