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592 CHAPTER 7 APPLICATIONS OF TRIGONOMETRY SECTION 7.2 Law of Cosines and Area Law of Cosines Area of a Triangle Heron’s Formula PREPARE FOR THIS SECTION Prepare for this section by completing the following exercises. The answers can be found on page A46. PS1. Evaluate 2a2 + b2 - 2ab cos C for a = 10.0, b = 15.0, and C = 110.0°. Round your result to the nearest tenth. [5.3] PS2. Find the area of a triangle with a base of 6 inches and a height of 8.5 inches. [1.2] PS3. Solve c2 = a2 + b2 - 2ab cos C for C. [6.5] PS4. The semiperimeter of a triangle is defined as one-half the perimeter of the triangle. Find the semiperimeter of a triangle with sides of 6 meters, 9 meters, and 10 meters. [P.1] PS5. Evaluate 1s1s - a21s - b21s - c2 for a = 3, b = 4, c = 5, and a + b + c . [P.2] 2 PS6. State a relationship between the lengths a, b, and c in the triangle shown at the right. [1.3] s = c b a Law of Cosines y The Law of Cosines can be used to solve triangles in which two sides and the included angle (SAS) are known or in which three sides (SSS) are known. Consider the triangle in Figure 7.11. The height BD is drawn from B perpendicular to the x-axis. The triangle BDA is a right triangle, and the coordinates of B are 1a cos C, a sin C2. The coordinates of A are 1b, 02. Using the distance formula, we can find the distance c. B(a cos C, a sin C) c a C D (0, 0) c = 21a cos C - b22 + 1a sin C - 022 A(b, 0) b x c2 = a2 cos2 C - 2ab cos C + b2 + a2 sin2 C c2 = a21cos2 C + sin2 C2 + b2 - 2ab cos C Figure 7.11 c2 = a2 + b2 - 2ab cos C Law of Cosines If A, B, and C are the measures of the angles of a triangle and a, b, and c are the lengths of the sides opposite these angles, then c2 = a2 + b2 - 2ab cos C a2 = b2 + c2 - 2bc cos A b2 = a2 + c2 - 2ac cos B 7.2 EXAMPLE 1 LAW OF COSINES AND AREA 593 Use the Law of Cosines (SAS) In triangle ABC, B = 110.0°, a = 10.0 centimeters, and c = 15.0 centimeters. See Figure 7.12. Find b. A Solution The Law of Cosines can be used because two sides and the included angle are known. b2 = a2 + c2 - 2ac cos B b c = 15.0 cm = 10.02 + 15.02 - 2110.02115.02 cos 110.0° b = 210.02 + 15.02 - 2110.02115.02 cos 110.0° b L 20.7 centimeters 110.0° B a = 10.0 cm C Try Exercise 12, page 598 Figure 7.12 In the next example, we know the length of each side, but we do not know the measure of any of the angles. EXAMPLE 2 Use the Law of Cosines (SSS) In triangle ABC, a = 32 feet, b = 20 feet, and c = 40 feet. Find B. This is the SSS case. Solution b2 = a2 + c2 - 2ac cos B a2 + c2 - b2 cos B = 2ac = 322 + 402 - 202 213221402 B = cos-1 a 322 + 402 - 202 b 213221402 B L 30° • Solve for cos B. • Substitute for a, b, and c. • Solve for angle B. • To the nearest degree Try Exercise 18, page 598 N EXAMPLE 3 N 78° 3.0 mi B A boat sailed 3.0 miles at a heading of 78° and then turned to a heading of 138° and sailed another 4.3 miles. Find the distance and the bearing of the boat from the starting point. 138° 78° A α Solve an Application Using the Law of Cosines 4.3 mi Solution Sketch a diagram (see Figure 7.13). First find the measure of angle B in triangle ABC. b C Figure 7.13 B = 78° + 1180° - 138°2 = 120° (continued) 594 CHAPTER 7 APPLICATIONS OF TRIGONOMETRY Use the Law of Cosines first to find b and then to find A. b2 = a2 + c2 - 2ac cos B = 4.32 + 3.02 - 214.3213.02 cos 120° • Substitute for a, c, and B. b = 24.32 + 3.02 - 214.3213.02 cos 120° b L 6.4 miles cos A = b2 + c2 - a2 2bc A = cos-1 a Study tip The measure of angle A in Example 3 can also be determined by using the Law of Sines. • Solve the Law of Cosines for cos A. b2 + c2 - a2 6.42 + 3.02 - 4.32 b L cos-1 a b L 35° 2bc 12216.4213.02 The bearing of the present position of the boat from the starting point A can be determined by calculating the measure of angle a in Figure 7.13, on page 593. a L 180° - 178° + 35°2 = 67° The distance is approximately 6.4 miles, and the bearing (to the nearest degree) is S67°E. Try Exercise 52, page 599 There are five different cases that we may encounter when solving an oblique triangle. Each case is listed below under the law that can be used to solve the triangle. Choosing Between the Law of Sines and the Law of Cosines Apply the Law of Sines to solve an oblique triangle for each of the following cases. ASA The measures of two angles of the triangle and the length of the included side are known. AAS The measures of two angles of the triangle and the length of a side opposite one of these angles are known. SSA The lengths of two sides of the triangle and the measure of an angle opposite one of these sides are known. This case is called the ambiguous case. It may yield one solution, two solutions, or no solution. Apply the Law of Cosines to solve an oblique triangle for each of the following cases. SSS The lengths of all three sides of the triangle are known. After finding the measure of an angle, you can complete your solution by using the Law of Sines. SAS The lengths of two sides of the triangle and the measure of the included angle are known. After finding the measure of the third side, you can complete your solution by using the Law of Sines. Question • In triangle ABC, A = 40°, C = 60°, and b = 114. Should you use the Law of Sines or the Law of Cosines to solve this triangle? Answer • Because the measure of two angles and the length of the included side are given, the triangle can be solved by using the Law of Sines. 7.2 1 bh can be used to find the area of a triangle when the base and height 2 are given. In this section we will find the areas of triangles when the height is not given. We will use K for the area of a triangle because the letter A is often used to represent the measure of an angle. Consider the areas of the acute and obtuse triangles in Figure 7.14. The formula A = c h C b A Acute triangle B Height of each triangle: c h 595 Area of a Triangle B a LAW OF COSINES AND AREA 1 bh 2 1 K = bc sin A 2 Thus we have established the following theorem. Area of each triangle: a C b h = c sin A A Obtuse triangle Figure 7.14 Study tip K = • Substitute for h. Area of a Triangle Because each formula requires two sides and the included angle, it is necessary to learn only one formula. The area K of triangle ABC is one-half the product of the lengths of any two sides and the sine of the included angle. Thus 1 1 1 K = ac sin B K = bc sin A K = ab sin C 2 2 2 EXAMPLE 4 Find the Area of a Triangle Given angle A = 62°, b = 12 meters, and c = 5.0 meters, find the area of triangle ABC. Solution In Figure 7.15, two sides and the included angle of the triangle are given. Using the formula for area, we have B K = c = 5.0 62° A b = 12 Figure 7.15 C 1 1 bc sin A = 112215.021sin 62°2 L 26 square meters 2 2 Try Exercise 30, page 598 When two angles and an included side are given, the Law of Sines is used to derive a formula for the area of a triangle. First, solve for c in the Law of Sines. c b = sin C sin B c = Substitute for c in the formula K = b sin C sin B 1 bc sin A. 2 K = 1 b sin C 1 bc sin A = ba b sin A 2 2 sin B K = b2 sin C sin A 2 sin B 596 CHAPTER 7 APPLICATIONS OF TRIGONOMETRY In like manner, the following two alternative formulas can be derived for the area of a triangle. K = EXAMPLE 5 a2 sin B sin C 2 sin A and K = c2 sin A sin B 2 sin C Find the Area of a Triangle Given A = 32°, C = 77°, and a = 14 inches, find the area of triangle ABC. Solution To use the preceding area formula, we need to know two angles and the included side. Therefore, we need to determine the measure of angle B. B = 180° - 32° - 77° = 71° Thus K = 142 sin 71° sin 77° a2 sin B sin C = L 170 square inches 2 sin A 2 sin 32° Try Exercise 32, page 598 Math Matters Recent findings indicate that Heron’s formula for finding the area of a triangle was first discovered by Archimedes. However, the formula is called Heron’s formula in honor of the geometer Heron of Alexandria (A.D. 50), who gave an ingenious proof of the theorem in his work Metrica. Because Heron of Alexandria was also known as Hero, some texts refer to Heron’s formula as Hero’s formula. Heron’s Formula The Law of Cosines can be used to derive Heron’s formula for the area of a triangle in which three sides of the triangle are given. Heron’s Formula for Finding the Area of a Triangle If a, b, and c are the lengths of the sides of a triangle, then the area K of the triangle is K = 2s1s - a21s - b21s - c2, where s = 1 1a + b + c2 2 Because s is one-half the perimeter of the triangle, it is called the semiperimeter. EXAMPLE 6 Find an Area by Heron’s Formula Find, to two significant digits, the area of the triangle with a = 7.0 meters, b = 15 meters, and c = 12 meters. Solution Calculate the semiperimeter s. s = 7.0 + 15 + 12 a + b + c = = 17 2 2 7.2 LAW OF COSINES AND AREA 597 Use Heron’s formula. K = 1s1s - a21s - b21s - c2 = 117117 - 7.02117 - 152117 - 122 = 11700 L 41 square meters Try Exercise 40, page 598 EXAMPLE 7 Use Heron’s Formula to Solve an Application The original portion of the Luxor Hotel in Las Vegas has the shape of a square pyramid. Each face of the pyramid is an isosceles triangle with a base of 646 feet and sides of length 576 feet. Assuming that the glass on the exterior of the Luxor Hotel costs $35 per square foot, determine the cost of the glass, to the nearest $10,000, for one of the triangular faces of the hotel. Solution The lengths (in feet) of the sides of a triangular face are a = 646, b = 576, and c = 576. s = a + b + c 646 + 576 + 576 = = 899 feet 2 2 K = 1s1s - a21s - b21s - c2 Macduff Everton/CORBIS = 18991899 - 64621899 - 57621899 - 5762 The pyramid portion of the Luxor Hotel in Las Vegas, Nevada = 123,729,318,063 L 154,043 square feet The cost C of the glass is the product of the cost per square foot and the area. C L 35 # 154,043 = 5,391,505 The approximate cost of the glass for one face of the Luxor Hotel is $5,390,000. Try Exercise 60, page 600 EXERCISE SET 7.2 In Exercises 1 to 52, round answers according to the rounding conventions on page 448. In Exercises 1 to 14, find the third side of the triangle. 5. b = 60, c = 84, A = 13° 6. a = 122, c = 144, B = 48° 1. a = 12, b = 18, C = 44° 7. a = 9.0, b = 7.0, C = 72° 2. b = 30, c = 24, A = 120° 8. b = 12, c = 22, A = 55° 3. a = 120, c = 180, B = 56° 9. a = 4.6, b = 7.2, C = 124° 4. a = 400, b = 620, C = 116° 10. b = 12.3, c = 14.5, A = 6.5° 598 CHAPTER 7 APPLICATIONS OF TRIGONOMETRY 11. a = 25.9, c = 33.4, B = 84.0° 34. a = 32, b = 24, c = 36 12. a = 14.2, b = 9.30, C = 9.20° 35. B = 54.3°, a = 22.4, b = 26.9 13. a = 122, c = 55.9, B = 44.2° 36. C = 18.2°, b = 13.4, a = 9.84 14. b = 444.8, c = 389.6, A = 78.44° 37. A = 116°, B = 34°, c = 8.5 38. B = 42.8°, C = 76.3°, c = 17.9 In Exercises 15 to 24, given three sides of a triangle, find the specified angle. 15. a = 25, b = 32, c = 40; find A. 39. a = 3.6, b = 4.2, c = 4.8 40. a = 10.2, b = 13.3, c = 15.4 16. a = 60, b = 88, c = 120; find B. 41. Distance Between Airports A plane leaves airport A and 17. a = 8.0, b = 9.0, c = 12; find C. travels 560 miles to airport B at a bearing of N32°E. The plane leaves airport B and travels to airport C 320 miles away at a bearing of S72°E. Find the distance from airport A to airport C. 18. a = 108, b = 132, c = 160; find A. 19. a = 80.0, b = 92.0, c = 124; find B. 42. Length of a Street A developer owns a triangular lot at the intersection of two streets. The streets meet at an angle of 72°, and the lot has 300 feet of frontage along one street and 416 feet of frontage along the other street. Find the length of the third side of the lot. 20. a = 166, b = 124, c = 139; find B. 21. a = 1025, b = 625.0, c = 1420; find C. 22. a = 4.7, b = 3.2, c = 5.9; find A. 43. Baseball In a baseball game, a batter hits a ground ball 26 feet in the direction of the pitcher’s mound. See the figure below. The pitcher runs forward and reaches for the ball. At that moment, how far is the ball from first base? (Note: A baseball infield is a square that measures 90 feet on each side.) 23. a = 32.5, b = 40.1, c = 29.6; find B. 24. a = 112.4, b = 96.80, c = 129.2; find C. In Exercises 25 to 28, solve the triangle. 25. A = 39.4°, b = 15.5, c = 17.2 26. C = 98.4°, a = 141, b = 92.3 26 ft 27. a = 83.6, b = 144, c = 98.1 90 ft 28. a = 25.4, b = 36.3, c = 38.2 44. In Exercises 29 to 40, find the area of the given triangle. Round each area to the same number of significant digits given for each of the given sides. B-2 Bomber The leading edge of each wing of the B-2 Stealth Bomber measures 105.6 feet in length. The angle between the wing’s leading edges 1 ∠ABC2 is 109.05°. What is the wing span (the distance from A to C) of the B-2 Bomber? 29. A = 105°, b = 12, c = 24 B 30. B = 127°, a = 32, c = 25 31. A = 42°, B = 76°, c = 12 A C B-2 Stealth Bomber 32. B = 102°, C = 27°, a = 8.5 45. Angle Between the Diagonals of a Box The rectangular 33. a = 16, b = 12, c = 14 box in the following figure measures 6.50 feet by 3.25 feet 7.2 by 4.75 feet. Find the measure of the angle u that is formed by the union of the diagonal shown on the front of the box and the diagonal shown on the right side of the box. 4.75 ft LAW OF COSINES AND AREA 599 51. Distance to a Plane A plane traveling at 180 miles per hour passes 400 feet directly over an observer. The plane is traveling along a straight path with an angle of elevation of 14°. Find the distance of the plane from the observer 10 seconds after the plane has passed directly overhead. 52. Distance Between Ships A ship leaves a port at a speed of 16 miles per hour at a heading of 32°. One hour later another ship leaves the port at a speed of 22 miles per hour at a heading of 254°. Find the distance between the ships 4 hours after the first ship leaves the port. θ 3.25 ft 53. Distance and Bearing from a Starting Point A plane flew 6.50 ft 46. Submarine Rescue Mission Use the distances shown in the following figure to determine the depth of the submarine below the surface of the water. Assume that the line segment between the surface ships is directly above the submarine. 54. Engine Design An engine has a 16-centimeter connecting rod that is attached to a rotating crank with a 4-centimeter radius. See the following figure. 615 ft 499 ft 181 miles at a heading of 108.5° and then turned to a heading of 124.6° and flew another 225 miles. Find the distance and the bearing of the plane from the starting point. Piston 629 ft C 16 cm 4 cm Submarine c A B 47. Distance Between Ships Two ships left a port at the same time. One ship traveled at a speed of 18 miles per hour at a heading of 318°. The other ship traveled at a speed of 22 miles per hour at a heading of 198°. Find the distance between the two ships after 10 hours of travel. 48. Distance Across a Lake Find the distance across the lake shown in the figure. a. Use the Law of Cosines to find an equation that relates c and A. b. Use the quadratic formula to solve the equation in a. for c. c. Use the equation from b. to find c when A = 55°. Round to A the nearest centimeter. d. Use the Law of Sines to find c when A = 55°. Round to the 136 m B nearest centimeter. How does this result compare with the result in c.? 55. Area of a Triangular Lot Find the area of a triangular piece 78.0° of land that is bounded by sides of 236 meters, 620 meters, and 814 meters. Round to the nearest hundred square meters. 162 m 56. Geometry Find the exact area of a parallelogram with sides of exactly 8 feet and 12 feet. The shorter diagonal is exactly 10 feet. C 57. Geometry Find the exact area of a square inscribed in a cir49. Geometry A regular hexagon is inscribed in a circle with a radius of exactly 40 centimeters. Find the exact length of one side of the hexagon. 50. Angle Between Boundaries of a Lot A triangular city lot has sides of 224 feet, 182 feet, and 165 feet. Find the angle between the longer two sides of the lot. cle with a radius of exactly 9 inches. 58. Geometry Find the exact area of a regular hexagon inscribed in a circle with a radius of exactly 24 centimeters. 59. Cost of a Lot A commercial piece of real estate is priced at $2.20 per square foot. Find, to the nearest $1000, the cost of a triangular lot measuring 212 feet by 185 feet by 240 feet. 600 CHAPTER 7 APPLICATIONS OF TRIGONOMETRY 60. Cost of a Lot An industrial piece of real estate is priced at Use Mollweide’s formula to determine whether either or DEF has an incorrect dimension. $4.15 per square foot. Find, to the nearest $1000, the cost of a triangular lot measuring 324 feet by 516 feet by 412 feet. 64. 61. Area of a Pasture Find the number of acres in a pasture ABC Check Dimensions of Trusses The following diagram shows some of the steel trusses in a railroad bridge. whose shape is a triangle measuring 800 feet by 1020 feet by 680 feet. Round to the nearest hundredth of an acre. (An acre is 43,560 square feet.) F D C A 62. Area of a Housing Tract Find the number of acres in a housing tract whose shape is a triangle measuring 420 yards by 540 yards by 500 yards. Round to the nearest tenth of an acre. (An acre is 4840 square yards.) The identity at the right is one of Mollweide’s formulas. It applies to any triangle ABC. a sina b A B 2 b ! c C cos a b 2 The formula is intriguing because it contains the dimensions of all angles and all sides of triangle ABC. The formula can be used to check whether a triangle has been solved correctly. Substitute the dimensions of a given triangle into the formula and compare the value of the left side of the formula with the value of the right side. If the two results are not reasonably close, then you know that at least one dimension is incorrect. The results generally will not be identical because each dimension is an approximation. In Exercises 63 and 64, you can assume that a triangle has an incorrect dimension if the value of the left side and the value of the right side of the above formula differ by more than 0.02. 63. Check Dimensions of Trusses The following diagram shows some of the steel trusses in an airplane hangar. C A F D E B A structural engineer has determined the following dimensions for ABC and DEF: ABC: A = 34.1°, B = 66.2°, C = 79.7° a = 9.23 feet, b = 15.1 feet, c = 16.2 feet DEF: D = 45.0°, E = 56.2°, F = 78.8° d = 13.6 feet, e = 16.0 feet, f = 18.9 feet Use Mollweide’s formula to determine whether either or DEF has an incorrect dimension. ABC 65. Find the measure of the angle formed by the sides P1P2 and P1P3 of a triangle with vertices at P11 -2, 42, P212, 12, and P314, - 32. 66. Given a triangle ABC, prove that a2 = b2 + c2 - 2bc cos A 67. Use the Law of Cosines to show that for any triangle ABC, cos A = 1b + c - a21b + c + a2 2bc - 1 68. Prove that K = xy sin A for a parallelogram, where x and y are the lengths of adjacent sides, A is the measure of the angle between side x and side y, and K is the area of the parallelogram. 69. Find the volume of B E the triangular prism shown in the figure. 72° 4 in. 4 in. 18 in. 70. Show that the area of the circumscribed triangle in the fol- lowing figure is K = rs, where s = a + b + c . 2 B An architect has determined the following dimensions for ABC and DEF: ABC: c A = 53.5°, B = 86.5°, C = 40.0° a = 13.0 feet, b = 16.1 feet, c = 10.4 feet DEF: D = 52.1°, E = 59.9°, F = 68.0° d = 17.2 feet, e = 21.3 feet, f = 22.8 feet a r A b C