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588 8 Additional Topics in Trigonometry h d sin csc ( ) tan h h d ★★ 50. Surveying. The layout in the figure at right is used to determine an inaccessible height h when a baseline d in a plane perpendicular to h can be established and the angles , , and can be measured. Show that SECTION 8-2 d Law of Cosines • Law of Cosines Derivation • Solving the SAS Case • Solving the SSS Case If in a triangle two sides and the included angle are given (SAS) or three sides are given (SSS), the law of sines cannot be used to solve the triangle—neither case involves an angle and its opposite side (Fig. 1). Both cases can be solved starting with the law of cosines, which is the subject matter for this section. FIGURE 1 b a b c c (a) SAS case • Law of Cosines (b) SSS case Theorem 1 states the law of cosines. Derivation Theorem 1 Law of Cosines a2 b2 c2 2bc cos b a c b2 a2 c2 2ac cos c2 a2 b2 2ab cos All three equations say essentially the same thing. 8-2 Law of Cosines 589 The law of cosines is used to solve triangles, given: 1. Two sides and the included angle (SAS) 2. Three sides (SSS) We will establish the first equation in Theorem 1. The other two equations then can be obtained from this one simply by relabeling the figure. We start by locating a triangle in a rectangular coordinate system. Figure 2 shows three typical triangles. For an arbitrary triangle located as in Figure 2, the distance-between-two-points formula is used to obtain a (hc)2 (k 0)2 a2 (h c)2 k2 Square both sides. h2 2hc c2 k2 FIGURE 2 Three representative triangles. a b k c (c, 0) (h, k) (h, k) (h, k) b (1) k a (a) a b (c, 0) h c h k (b) (h, 0) h0 c (c, 0) (c) From Figure 2, we note that b2 h2 k2 Substituting b2 for h2 k2 in equation (1), we obtain a2 b2 c2 2hc (2) But cos h b h b cos Thus, by replacing h in equation (2) with b cos , we reach our objective: a2 b2 c2 2bc cos [Note: If is acute, then cos is positive; if is obtuse, then cos is negative.] • Solving the SAS Case For the SAS case, start by using the law of cosines to find the side opposite the given angle. Then use either the law of cosines or the law of sines to find a second angle. Because of the simpler computations, the law of sines will generally be used to find the second angle. 590 8 Additional Topics in Trigonometry EXPLORE-DISCUSS 1 After using the law of cosines to find the side opposite the angle for a SAS case, the law of sines is used to find a second angle. Figure 2 (a) shows that there are two choices for a second angle. (A) If the given angle is obtuse, can either of the remaining angles be obtuse? Explain. (B) If the given angle is acute, then one of the remaining angles may or may not be obtuse. Explain why choosing the angle opposite the shorter side guarantees the selection of an acute angle. (C) Starting with (sin )/a (sin )/b, show that a sinb sin1 (1) (D) Explain why equation (1) gives us the correct angle only if is acute. The above discussion leads to the following strategy for solving the SAS case: Strategy for Solving the SAS Case EXAMPLE 1 Step Find Method 1. Side opposite given angle Law of cosines 2. Second angle (Find the angle opposite the shorter of the two given sides—this angle will always be acute.) Law of sines 3. Third angle Subtract the sum of the measures of the given angle and the angle found in step 2 from 180°. Solving the SAS Case Solve the triangle in Figure 3. FIGURE 3 10.3 cm b 32.4 6.45 cm 8-2 Law of Cosines 591 Solution Solve for b Use the law of cosines: b2 a2 c2 2ac cos Solve for b. b a c 2ac cos 2 2 (10.3)2 (6.45)2 2(10.3)(6.45) cos 32.4° 5.96 cm Solve for Since side c is shorter than side a, must be acute, and the law of sines is used to solve for . sin sin c b sin Solve for sin . c sin b sin1 sin1 Solve for . b sin 32.4° 6.45 5.96 c sin Since is acute, the inverse sine function gives us directly. 35.4° Solve for 180° ( ) 180° (32.4° 35.4°) 112.2° Matched Problem 1 • Solving the SSS Case EXPLORE-DISCUSS 2 Solve the triangle with 77.5°, b 10.4 feet, and c 17.7 feet. Starting with three sides of a triangle, the problem is to find the three angles. Subsequent calculations are simplified if we solve for the obtuse angle first, if present. The law of cosines is used for this purpose. A second angle, which must be acute, can be found using either law, although computations are usually simpler with the law of sines. (A) Starting with a2 b2 c2 2bc cos , show that cos1 a2 b2 c2 2bc (2) (B) Does equation (2) give us the correct angle irrespective of whether is acute or obtuse? Explain. The above discussion leads to the following strategy for solving the SSS case. 592 8 Additional Topics in Trigonometry Strategy for Solving the SSS Case EXAMPLE 2 Step Find Method 1. Angle opposite longest side—this will take care of an obtuse angle, if present. Law of cosines 2. Either of the remaining angles, which will be acute (why?) Law of sines 3. Third angle Subtract the sum of the measures of the angles found in steps 1 and 2 from 180°. Solving the SSS Case Solve the triangle with a 27.3 meters, b 17.8 meters, and c 35.2 meters. Solution Three sides of the triangle are given and we are to find the three angles. This is the SSS case. Sketch the triangle (Fig. 4) and use the law of cosines to find the largest angle, then use the law of sines to find one of the two remaining acute angles. FIGURE 4 17.8 m 27.3 m 35.2 m Since is the largest angle, we solve for it first using the law of cosines. Solve for c2 a2 b2 2ab cos cos a2 b2 c2 2ab cos1 cos1 Solve for cos . a2 b2 c2 2ab (27.3)2 (17.8)2 (35.2)2 2(27.3)(17.8) Solve for . 100.5° Solve for We now solve for either or , using the law of sines. We choose . sin sin a c 8-2 Law of Cosines sin a sin 27.3 sin 100.5 c 35.2 sin1 593 Solve for sin . sin 100.5 27.3 35.2 Solve for . 49.7° is acute. 180° Solve for 180° ( ) 180° (49.7° 100.5°) 29.8° Matched Problem 2 EXAMPLE 3 Solve the triangle with a 1.25 yards, b 2.05 yards, and c 1.52 yards. Finding the Side of a Regular Polygon If a seven-sided regular polygon is inscribed in a circle of radius 22.8 centimeters, find the length of one side of the polygon. Solution FIGURE 5 Sketch a figure (Fig. 5) and use the law of cosines: Actually, you only need to sketch the triangle: 22.8 360 7 22.8 d d 2 22.82 22.82 2(22.8)(22.8) cos d 2(22.8)2 2(22.8)2 cos 360° 7 360° 7 19.8 centimeters Matched Problem 3 If an 11-sided regular polygon is inscribed in a circle with radius 4.63 inches, find the length of one side of the polygon. 594 8 Additional Topics in Trigonometry Answers to Matched Problems 1. a 18.5 ft, 33.3°, 69.2° 2. 37.4°, 95.0°, 47.6° EXERCISE 3. 2.61 in. 8-2 The labeling in the figure below is the convention we will follow in this exercise set. Your answers to some problems may differ slightly from those in the book, depending on the order in which you solve for the sides and angles of a given triangle. 12. a 31.5 meters, b 29.4 meters, c 33.7 meters Problems 13–26 represent a variety of problems involving both the law of sines and the law of cosines. Solve each triangle. If a problem does not have a solution, say so. 13. 92.6°, 88.9°, a 15.2 centimeters b a c A 1. Referring to the figure above, if 47.3°, b 11.7 centimeters, and c 6.04 centimeters, which of the two angles, or , can you say for certain is acute and why? 2. Referring to the figure above, if 93.5°, b 5.34 inches, and c 8.77 inches, which of the two angles, or , can you say for certain is acute and why? Solve each triangle in Problems 3–6. 14. 79.4°, 102.3°, a 6.4 millimeters 15. 126.2°, a 13.8 inches, c 12.5 inches 16. 19.1°, a 16.4 yards, b 28.2 yards 17. a 23.4 meters, b 6.9 meters, c 31.3 meters 18. a 86 inches, b 32 inches, c 53 inches 19. 38.4°, a 11.5 inches, b 14.0 inches 20. 66.4°, b 25.5 meters, c 25.5 meters 21. a 32.9 meters, b 42.4 meters, c 20.4 meters 22. a 10.5 centimeters, b 5.23 centimeters, c 8.66 centimeters 23. 58.4°, b 7.23 meters, c 6.54 meters 24. 46.7°, a 18.1 meters, b 22.6 meters 3. 71.2°, b 5.32 yards, c 5.03 yards 25. 39.8°, a 12.5 inches, b 7.31 inches 4. 57.3°, a 6.08 centimeters, c 5.25 centimeters 26. 47.9°, b 35.2 inches, c 25.5 inches 5. 120°20, a 5.73 millimeters, b 10.2 millimeters 6. 135°50, b 8.44 inches, c 20.3 inches B 7. Referring to the figure at the beginning of the exercise, if a 13.5 feet, b 20.8 feet, and c 8.09 feet, then if the triangle has an obtuse angle, which angle must it be and why? 8. Suppose you are told that a triangle has sides a 12.5 centimeters, b 25.3 centimeters, and c 10.7 centimeters. Explain why the triangle has no solution. Solve each triangle in Problems 9–12 if the triangle has a solution. Use decimal degrees for angle measure. 9. a 4.00 meters, b 10.2 meters, c 9.05 meters 10. a 10.5 miles, b 20.7 miles, c 12.2 miles 11. a 6.00 kilometers, b 5.30 kilometers, c 5.52 kilometers C 27. Show, using the law of cosines, that if 90°, then c2 a2 b2 (the Pythagorean theorem). 28. Show, using the law of cosines, that if c2 a2 b2, then 90°. 29. Show that for any triangle, a2 b2 c2 cos cos cos 2abc a b c 30. Show that for any triangle, a b cos c cos 31. Give a solution to Example 3 that does not use the law of cosines by showing that d 360˚ 22.8 sin . 2 14 8-2 Law of Cosines 32. Show that the length d of one side of an n-sided regular polygon, inscribed in a circle of radius r, is given by d 2r sin ★ 180˚ . n 41. Analytic Geometry. If point A in the figure has coordinates (3, 4) and point B has coordinates (4, 3), find the radian measure of angle to three decimal places. y APPLICATIONS A B 33. Surveying. To find the length AB of a small lake, a surveyor measured angle ACB to be 96°, AC to be 91 yards, and BC to be 71 yards. What is the approximate length of the lake? x 0 ★ 42. Analytic Geometry. If point A in the figure has coordinates (4, 3) and point B has coordinates (5, 1), find the radian measure of angle to three decimal places. ★ 43. Engineering. Three circles of radius 2.03, 5.00, and 8.20 centimeters are tangent to one another (see figure). Find the three angles formed by the lines joining their centers (to the nearest 10). C A 595 B 35. Geometry. Two adjacent sides of a parallelogram meet at an angle of 35°10 and have lengths of 3 and 8 feet. What is the length of the shorter diagonal of the parallelogram (to 3 significant digits)? ★ 36. Geometry. What is the length of the longer diagonal of the parallelogram in Problem 35 (to 3 significant digits)? C 2.8 cm 38. Search and Rescue. At noon, two search planes set out from San Francisco to find a downed plane in the ocean. Plane A travels due west at 400 miles per hour, and plane B flies northwest at 500 miles per hour. At 2 P.M. plane A spots the survivors of the downed plane and radios plane B to come and assist in the rescue. How far is plane B from plane A at this time (to 3 significant digits)? 40. Geometry. Find the perimeter of a nine-sided regular polygon inscribed in a circle of radius 7.09 centimeters. 44. Engineering. Three circles of radius 2.00, 5.00, and 8.00 inches are tangent to each other (see figure). Find the three angles formed by the lines joining their centers (to the nearest 10). 45. Geometry. A rectangular solid has sides as indicated in the figure. Find CAB to the nearest degree. 37. Navigation. Los Angeles and Las Vegas are approximately 200 miles apart. A pilot 80 miles from Los Angeles finds that she is 6°20 off course relative to her start in Los Angeles. How far is she from Las Vegas at this time? (Compute the answer to 3 significant digits.) 39. Geometry. Find the perimeter of a pentagon inscribed in a circle of radius 12.6 meters. 34. Surveying. Suppose the figure for this problem represents the base of a large rock outcropping on a farmer’s land. If a surveyor finds ACB 110°, AC 85 meters, and BC 73 meters, what is the approximate length (to one decimal place) of the outcropping? B A 4.3 cm 8.1 cm 46. Geometry. Referring to the figure, find ACB to the nearest degree. ★ 47. Space Science. For communications between a space shuttle and the White Sands tracking station in southern New Mexico, two satellites are placed in geostationary orbit, 130° apart relative to the center of the Earth and 22,300 596 8 Additional Topics in Trigonometry miles above the surface of the Earth (see figure). (A satellite in geostationary orbit remains stationary above a fixed point on the surface of the Earth.) Radio signals are sent from the tracking station by way of the satellites to the shuttle, and vice versa. This system allows the tracking station to be in contact with the shuttle over most of the Earth’s surface. How far to the nearest 100 miles is one of the geostationary satellites from the White Sands tracking station W? The radius of the Earth is 3,964 miles. S ★ 48. Space Science. A satellite S, in circular orbit around the Earth, is sighted by a tracking station T (see figure). The distance TS is determined by radar to be 1,034 miles, and the angle of elevation above the horizon is 32.4°. How high is the satellite above the Earth at the time of the sighting? The radius of the Earth is 3,964 miles. S T Horizon S R W Earth C C SECTION 8-3 Geometric Vectors • • • • Geometric Vectors and Vector Addition Velocity Vectors Force Vectors Resolution of Vectors into Vector Components Many physical quantities, such as length, area, or volume, can be completely specified by a single real number. Other quantities, such as directed distances, velocities, and forces, require for their complete specification both a magnitude and a direction. The former are often called scalar quantities, and the latter are called vector quantities. In this section we limit our discussion to the intuitive idea of geometric vectors in a plane. In Section 8-4 we introduce algebraic vectors, a first step in the generalization of a concept that has far-reaching consequences. Vectors are widely used in many areas of science and engineering. • Geometric Vectors and Vector Addition P v O , or v. FIGURE 1 Vector OP A line segment to which a direction has been assigned is called a directed line segment. A geometric vector is a directed line segment and is represented by an arrow (see Fig. 1). A vector with an initial point O and a terminal point P (the end with . Vectors are also denoted by a boldface letter, such the arrowhead) is denoted by OP as v. Since it is difficult to write boldface on paper, we suggest that you use an arrow over a single letter, such as v , when you want the letter to denote a vector. , denoted by OP , v or v, is the length of the The magnitude of the vector OP directed line segment. Two vectors have the same direction if they are parallel and point in the same direction. Two vectors have opposite direction if they are parallel and point in opposite directions. The zero vector, denoted by 0 or 0, has a magni-