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Additional Topics in Trigonometry Outline Application 7-1 Law of Sines 7-2 Law of Cosines 7-3 Geometric Vectors 7-4 Algebraic Vectors 7-5 Polar Coordinates and Graphs 7-6 Complex Numbers in Rectangular and Polar Forms 7-7 De Moivre’s Theorem Chapter 7 Group Activity: Conic Sections and Planetary Orbits Chapter 7 Review Two fire towers are located 11.5 miles apart on a straight road. Observers at each tower spot a fire and report its location in terms of the angles in the figure. A command post is to be set up on the road as close to the fire as possible. Find the distance (to the nearest tenth of a mile) from the command post to tower A and from the command post to the fire. n this chapter a number of additional topics involving trigonometry are considered. First, we return to the problem of solving triangles— not just right triangles, but any triangle. Then some of these ideas are used to develop the important concept of vector. With our knowledge of trigonometry, we introduce the polar coordinate system, probably the most important coordinate system after the rectangular coordinate system. After considering polar equations and their graphs, we represent complex numbers in polar form. Once a complex number is in polar form, it will be possible to find nth powers and nth roots of the number using an ingenious theorem due to De Moivre. I Preparing for This Chapter Before getting started on this chapter, review the following concepts: Rational Exponents (Appendix A, Section 6) Radicals (Appendix A, Section 7) Complex Numbers (Chapter 2, Section 4) Inverse Functions (Chapter 4, Section 2) Solving Right Triangles (Chapter 5, Section 5) Difference Identities (Chapter 6, Section 2) Significant Digits (Appendix C) Section 7-1 Law of Sines Law of Sines Derivation Solving the ASA and AAS Cases Solving the SSA Case—Including the Ambiguous Case The law of sines (developed in this section) and the law of cosines (developed in the next section) play fundamental roles in solving oblique triangles—triangles without a right angle. Every oblique triangle is either acute, all angles between 0° and 90°, or obtuse, one angle between 90° and 180°. Figure 1 illustrates both types of triangles. FIGURE 1 Oblique triangles. 504 7-1 Law of Sines 505 Note how the sides and angles of the oblique triangles in Figure 1 have been labeled: Side a is opposite angle ␣, side b is opposite angle , and side c is opposite angle ␥. Also note that the largest side of a triangle is opposite the largest angle. Given any three of the six quantities indicated in Figure 1, we are interested in finding the remaining three, if they exist. This process is called solving the triangle. Before proceeding with specific examples, it is important to recall the rules in Table 1 regarding accuracy of angle and side measure. Table 1 is also repeated inside the back cover of the text for easy reference. T A B L E 1 Triangles and Significant Digits Angle to Nearest Significant Digits for Side Measure 1° 10⬘ or 0.1° 1⬘ or 0.01° 10⬙ or 0.001° 2 3 4 5 CALCULATOR CALCULATIONS When solving for a particular side or angle, carry out all operations within the calculator and then round to the appropriate number of significant digits (as specified in Table 1) at the end of the calculation. Your answers may still differ slightly from those in the book, depending on the order in which you solve for the sides and angles. Law of Sines Derivation FIGURE 2 The law of sines is relatively easy to prove using the right triangle properties studied in Section 5-5. We will also use the fact that sin (180° ⫺ x) ⫽ sin x which is readily obtained using a different identity (a good exercise for you). Referring to the triangles in Figure 2, we proceed as follows: For each triangle, sin ␣ ⫽ h b and sin  ⫽ h a Solving each equation for h, we obtain h ⫽ b sin ␣ and h ⫽ a sin  Thus, b sin ␣ ⫽ a sin  sin ␣ sin  ⫽ a b (1) 506 7 ADDITIONAL TOPICS IN TRIGONOMETRY Similarly, for each triangle in Figure 2, sin ␣ ⫽ m c and sin ␥ ⫽ sin (180° ⫺ ␥) ⫽ m a Solving each equation for m, we obtain m ⫽ c sin ␣ m ⫽ a sin ␥ and Thus, c sin ␣ ⫽ a sin ␥ sin ␣ sin ␥ ⫽ a c (2) If we combine equations (1) and (2), we obtain the law of sines. LAW OF SINES 1 sin ␣ sin  sin ␥ ⫽ ⫽ a b c In words, the ratio of the sine of an angle to its opposite side is the same as the ratio of the sine of either of the other angles to its opposite side. The law of sines is used to solve triangles, given 1. Two angles and any side (ASA or AAS), or 2. Two sides and an angle opposite one of them (SSA). We apply the law of sines to the easier ASA and AAS cases first, then we turn to the more challenging SSA case. Solving the ASA and AAS Cases EXAMPLE 1 FIGURE 3 Solving the ASA Case Solve the triangle in Figure 3. 7-1 Law of Sines Solution Solve for ␥ 507 We are given two angles and the included side, which is the ASA case. Find the third angle, then solve for the other two sides using the law of sines. ␣ ⫹  ⫹ ␥ ⫽ 180° ␥ ⫽ 180° ⫺ (␣ ⫹ ) ⫽ 180° ⫺ (28°0⬘ ⫹ 45°20⬘) ⫽ 106°40⬘ Solve for a sin ␣ sin ␥ ⫽ a c a⫽ ⫽ c sin ␣ sin ␥ 120 sin 28°0⬘ sin 106°40⬘ ⫽ 58.8 meters Solve for b sin  sin ␥ ⫽ b c b⫽ ⫽ c sin  sin ␥ 120 sin 45°20⬘ sin 106°40⬘ ⫽ 89.1 meters MATCHED PROBLEM Solve the triangle in Figure 4. 1 FIGURE 4 Note that the AAS case can always be converted to the ASA case by first solving for the third angle. For the ASA or AAS case to determine a unique triangle, the sum of the two angles must be between 0° and 180°, since the sum of all three angles in a triangle is 180° and no angle can be zero or negative. —Including the Ambiguous Case Solving the SSA Case— We now look at the case where we are given two sides and an angle opposite one of the sides—the SSA case. This case has several possible outcomes, depending on the measures of the two sides and the angle. Table 2 illustrates the various possibilities. 508 7 ADDITIONAL TOPICS IN TRIGONOMETRY T A B L E 2 SSA Variations Number of Triangles ␣ a (h ⴝ b sin ␣) Acute 0⬍a⬍h 0 (a) Acute a⫽h 1 (b) Acute h⬍a⬍b 2 (c) Acute aⱖb 1 (d) Obtuse 0⬍aⱕb 0 (e) Obtuse a⬎b 1 (f) Figure Case Table 2 need not be committed to memory. Usually a rough sketch of a particular situation will indicate which of the variations applies. The case where h ⬍ a ⬍ b is referred to as the ambiguous case, because two triangles, one acute and the other obtuse, are always possible. 1 EXAMPLE 2 Discuss which cases in Table 2 apply and why if in the solution process of solving an SSA triangle with ␣ acute it is found that (A) sin  ⬎ 1 (B) sin  ⫽ 1 (C) 0 ⬍ sin  ⬍ 1 Solving the SSA Case Solve the triangle(s) with ␣ ⫽ 123°, b ⫽ 23 centimeters, and a ⫽ 47 centimeters. 7-1 Law of Sines Solution 509 From a rough sketch (Fig. 5), we see that there is only one triangle. FIGURE 5 Solve for  sin  sin ␣ ⫽ b a sin  ⫽ b sin ␣ 23 sin 123° ⫽ a 47  ⫽ sin⫺1 Solve for ␥ 冢23 sin47123° 冣 ⫽ 24° ␣ ⫹  ⫹ ␥ ⫽ 180° ␥ ⫽ 180° ⫺ 123° ⫺ 24° ⫽ 33° Solve for c sin ␣ sin ␥ ⫽ a c c⫽ MATCHED PROBLEM a sin ␥ 47 sin 33° ⫽ ⫽ 31 centimeters sin ␣ sin 123° Solve the triangle(s) with  ⫽ 98°, a ⫽ 62 meters, and b ⫽ 88 meters. 2 EXAMPLE Solving the SSA (Ambiguous) Case 3 Solve the triangle(s) with ␣ ⫽ 26°, a ⫽ 1.0 meter, and b ⫽ 1.8 meters. Solution If we try to draw a triangle with the indicated sides and angle, we find that two triangles, I and II, are possible (Fig. 6). This is verified by the fact that h ⬍ a ⬍ b, where h ⫽ b sin ␣ ⫽ 0.79 meters, a ⫽ 1.0 meter, and b ⫽ 1.8 meters. FIGURE 6 510 7 ADDITIONAL TOPICS IN TRIGONOMETRY Solve for  and ⴕ We start by finding  and ⬘ using the law of sines: sin  sin ␣ ⫽ b a sin  ⫽ b sin ␣ 1.8 sin 26° ⫽ ⫽ 0.7891 a 1.0 Angle  can be either obtuse or acute:  ⫽ 180° ⫺ sin⫺1 0.7891 or ⬘ ⫽ sin⫺1 0.7891 ⫽ 180° ⫺ 52° ⫽ 128° Solve for ␥ and ␥ⴕ ⫽ 52° We next find ␥ and ␥⬘: ␥ ⫽ 180° ⫺ (26° ⫹ 128°) ⫽ 26° ␥⬘ ⫽ 180° ⫺ (26° ⫹ 52°) ⫽ 102° Solve for c and c⬘ Finally, we solve for c and c⬘: sin ␣ sin ␥ ⫽ a c c⫽ ⫽ sin ␣ sin ␥⬘ ⫽ a c⬘ a sin ␥ sin ␣ 1.0 sin 26° sin 26° ⫽ 1.0 meter c⬘ ⫽ ⫽ a sin ␥⬘ sin ␣ 1.0 sin 102° sin 26° ⫽ 2.2 meters In summary, MATCHED PROBLEM Triangle I: ⬘ ⫽ 128° ⬘␥ ⫽ 26° ⬘c ⫽ 1.0 meter Triangle II: ⬘ ⫽ 52° ␥⬘ ⫽ 102° c⬘ ⫽ 2.2 meters Solve the triangle(s) with a ⫽ 8 kilometers, b ⫽ 10 kilometers, and ␣ ⫽ 35°. 3 The law of sines is useful in many applications, as can be seen in Example 4 and the applications in Exercise 7-1. EXAMPLE 4 Surveying To measure the length d of a lake (see Fig. 7), a base line AB is established and measured to be 125 meters. Angles A and B are measured to be 41.6° and 124.3°, respectively. How long is the lake? 7-1 Law of Sines 511 FIGURE 7 Solution Find angle C and use the law of sines. Angle C ⫽ 180° ⫺ (124.3° ⫹ 41.6°) ⫽ 14.1° sin 14.1° sin 41.6° ⫽ 125 d 41.6° 冢sin sin 14.1° 冣 d ⫽ 125 ⫽ 341 meters MATCHED PROBLEM In Example 4, find the distance AC. 4 Answers to Matched Problems 1. ␥ ⫽ 101°40⬘, b ⫽ 141, c ⫽ 152 2. ␣ ⫽ 44°, ␥ ⫽ 38°, c ⫽ 55 m 3.  ⫽ 134°, ⬘ ⫽ 46°, ␥ ⫽ 11°, ␥⬘ ⫽ 99°, c ⫽ 2.7 km, c⬘ ⫽ 14 km 4. 424 m EXERCISE 7-1 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. 4.  ⫽ 43°, ␥ ⫽ 36°, a ⫽ 92 millimeters 5.  ⫽ 112°, ␥ ⫽ 19°, c ⫽ 23 yards 6. ␣ ⫽ 52°, ␥ ⫽ 105°, c ⫽ 47 meters 7. ␣ ⫽ 52°, ␥ ⫽ 47°, a ⫽ 13 centimeters 8.  ⫽ 83°, ␥ ⫽ 77°, c ⫽ 25 miles In Problems 9–16, determine whether the information in each problem allows you to construct 0, 1, or 2 triangles. Do not solve the triangle. Explain which case in Table 2 applies. A 9. a ⫽ 2 inches, b ⫽ 4 inches, ␣ ⫽ 30° 10. a ⫽ 3 feet, b ⫽ 6 feet, ␣ ⫽ 30° Solve each triangle in Problems 1–8. 11. a ⫽ 6 inches, b ⫽ 4 inches, ␣ ⫽ 30° 1. ␣ ⫽ 73°,  ⫽ 28°, c ⫽ 42 feet 12. a ⫽ 8 feet, b ⫽ 6 feet, ␣ ⫽ 30° 2. ␣ ⫽ 41°,  ⫽ 33°, c ⫽ 21 centimeters 13. a ⫽ 1 inch, b ⫽ 4 inches, ␣ ⫽ 30° 3. ␣ ⫽ 122°, ␥ ⫽ 18°, b ⫽ 12 kilometers 14. a ⫽ 2 feet, b ⫽ 6 feet, ␣ ⫽ 30° 512 7 ADDITIONAL TOPICS IN TRIGONOMETRY 15. a ⫽ 3 inches, b ⫽ 4 inches, ␣ ⫽ 30° 34. (A) Use the law of sines and suitable identities to show that for any triangle 16. a ⫽ 5 feet, b ⫽ 6 feet, ␣ ⫽ 30° a⫺b ⫽ a⫹b B Solve each triangle in Problems 17–30. If a problem has no solution, say so. ␣⫺ 2 ␣⫹ tan 2 tan (B) Verify the formula with values from Problem 1. 17. ␣ ⫽ 118.3°, ␥ ⫽ 12.2°, b ⫽ 17.3 feet APPLICATIONS 18.  ⫽ 27.5°, ␥ ⫽ 54.5°, a ⫽ 9.27 inches 19. ␣ ⫽ 67.7°,  ⫽ 54.2°, b ⫽ 123 meters 35. Coast Guard. Two lookout posts, A and B (10.0 miles apart), are established along a coast to watch for illegal ships coming within the 3-mile limit. If post A reports a ship S at angle BAS ⫽ 37°30⬘ and post B reports the same ship at angle ABS ⫽ 20°0⬘, how far is the ship from post A? How far is the ship from the shore (assuming the shore is along the line joining the two observation posts)? 20. ␣ ⫽ 122.7°,  ⫽ 34.4°, b ⫽ 18.3 kilometers 21. ␣ ⫽ 46.5°, a ⫽ 7.9 millimeters, b ⫽ 13.1 millimeters 22. ␣ ⫽ 26.3°, a ⫽ 14.7 inches, b ⫽ 35.2 inches 23.  ⫽ 38.9°, a ⫽ 42.7 inches, b ⫽ 30.0 inches, ␣ acute 24.  ⫽ 27.3°, a ⫽ 244 centimeters, b ⫽ 135 centimeters, ␣ acute 36. Fire Lookout. A fire at F is spotted from two fire lookout stations, A and B, which are 10.0 miles apart. If station B reports the fire at angle ABF ⫽ 53°0⬘ and station A reports the fire at angle BAF ⫽ 28°30⬘, how far is the fire from station A? From station B? 25.  ⫽ 38.9°, a ⫽ 42.7 inches, b ⫽ 30.0 inches, ␣ obtuse 26.  ⫽ 27.3°, a ⫽ 244 centimeters, b ⫽ 135 centimeters, ␣ obtuse ★ 37. Natural Science. The tallest trees in the world grow in Redwood National Park in California; they are taller than a football field is long. Find the height of one of these trees, given the information in the figure. (The 100-foot measurement is accurate to three significant digits.) ★ 38. Surveying. To measure the height of Mt. Whitney in California, surveyors used a scheme like the one shown in the figure in Problem 37. They set up a horizontal baseline 2,000 feet long at the foot of the mountain and found the angle nearest the mountain to be 43°5⬘; the angle farthest from the mountain was found to be 38°0⬘. If the baseline was 5,000 feet above sea level, how high is Mt. Whitney above sea level? 27. ␣ ⫽ 123.2°, a ⫽ 101 yards, b ⫽ 152 yards 28. ␣ ⫽ 137.3°, a ⫽ 13.9 meters, b ⫽ 19.1 meters 29.  ⫽ 29°30⬘, a ⫽ 43.2 millimeters, b ⫽ 56.5 millimeters 30.  ⫽ 33°50⬘, a ⫽ 673 meters, b ⫽ 1,240 meters C 31. Let ␣ ⫽ 42.3° and b ⫽ 25.2 centimeters. Determine a value k so that if 0 ⬍ a ⬍ k, there is no solution; if a ⫽ k, there is one solution; and if k ⬍ a ⬍ b, there are two solutions. 32. Let ␣ ⫽ 37.3° and b ⫽ 42.8 centimeters. Determine a value k so that if 0 ⬍ a ⬍ k, there is no solution; if a ⫽ k, there is one solution; and if k ⬍ a ⬍ b, there are two solutions. 33. Mollweide’s equation, (a ⫺ b) cos ␥ ␣⫺ ⫽ c sin 2 2 is often used to check the final solution of a triangle, since all six parts of a triangle are involved in the equation. If the left side does not equal the right side after substitution, then an error has been made in solving a triangle. Use this equation to check Problem 1. (Because of rounding errors, both sides may not be exactly the same.) 7-1 Law of Sines 513 39. Engineering. A 4.5-inch piston rod joins a piston to a 1.5inch crankshaft (see the figure). How far is the base of the piston from the center of the crankshaft (distance d) when the rod makes an angle of 9° with the centerline? There are two answers to the problem. 40. Engineering. Repeat Problem 39 if the piston rod is 6.3 inches, the crankshaft is 1.7 inches, and the angle is 11°. 41. Astronomy. The orbits of the Earth and Venus are approximately circular, with the sun at the center. A sighting of Venus is made from Earth, and the angle ␣ is found to be 18°40⬘. If the radius of the orbit of the Earth is 1.495 ⫻ 108 kilometers and the radius of the orbit of Venus is 1.085 ⫻ 108 kilometers, what are the possible distances from the Earth to Venus? (See the figure.) 42. Astronomy. In Problem 41, find the maximum angle ␣. [Hint: The angle is maximum when a straight line joining the Earth and Venus is tangent to Venus’s orbit.] ★ ★ 43. Surveying. A tree growing on a hillside casts a 102-foot shadow straight down the hill (see the figure). Find the vertical height of the tree if, relative to the horizontal, the hill slopes 15.0° and the angle of elevation of the sun is 62.0°. 44. Surveying. Find the height of the tree in Problem 43 if the shadow length is 157 feet and, relative to the horizontal, the hill slopes 11.0° and the angle of elevation of the sun is 42.0°. ★ 45. Life Science. A cross section of the cornea of an eye, a circular arc, is shown in the figure. Find the arc radius R and the arc length s, given the chord length C ⫽ 11.8 millimeters and the central angle ⫽ 98.9°. ★ 46. Life Science. Referring to the figure, find the arc radius R and the arc length s, given the chord length C ⫽ 10.2 millimeters and the central angle ⫽ 63.2°. ★ 47. Surveying. The procedure illustrated in Problems 37 and 38 is used to determine an inaccessible height h when a baseline d on a line perpendicular to h can be established (see the figure) and the angles ␣ and  can be measured. Show that h⫽d 冤 sinsin(␣ sin⫺ ␣) 冥 514 ★★ 7 ADDITIONAL TOPICS IN TRIGONOMETRY 48. Surveying. The layout in the figure 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 h ⫽ d sin ␣ csc (␣ ⫹ ) tan ␥ Section 7-2 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 (a) SAS case (b) SSS case Law of Cosines Derivation Theorem 1 states the law of cosines. LAW OF COSINES 1 a2 ⫽ b2 ⫹ c2 ⫺ 2bc cos ␣ b ⫽ a ⫹ c ⫺ 2ac cos  2 2 2 c2 ⫽ a2 ⫹ b2 ⫺ 2ab cos ␥ All three equations say essentially the same thing. Cases SAS and SSS are most readily solved by starting with the law of cosines.