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506 7 Additional Topics in Trigonometry In this chapter a number of additional topics involving trigonometry are considered. First, the problem of solving triangles is returned to—not just right triangles but any triangle. Then some of these ideas are used to develop the important concept of a vector. With knowledge of trigonometry, the polar coordinate system is introduced, probably the most important coordinate system after the rectangular coordinate system. After considering polar equations and their graphs, complex numbers are represented 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. 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. b a c a b c Acute triangle Obtuse triangle (a) TABLE 1 Triangles and Significant Digits Angle to nearest 1° 10 or 0.1° Significant digits for side measure 2 3 1 or 0.01° 4 10 or 0.001° 5 (b) 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. The law of sines is developed in this section, and the law of cosines is developed in the next section. These two laws provide the basic tools for solving oblique triangles. If the given quantities include a side and the opposite angle, the law of sines should be used; otherwise, start with the law of cosines. 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. 7-1 Law of Sines 507 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 answer 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 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 difference identity (a good exercise for you). Referring to the triangles in Figure 2, we proceed as follows: For each triangle, b a h sin h b h a and sin and h a sin m Solving each equation for h, we obtain c h b sin (a) Thus, 180 b a h b sin a sin m sin sin a b c (b) FIGURE 2 (1) Similarly, for each triangle in Figure 2, sin m c sin sin (180° ) and m a Solving each equation for m, we obtain m c sin and m a sin Thus, c sin a sin sin sin a c If we combine equations (1) and (2), we obtain the law of sines. (2) 508 7 Additional Topics in Trigonometry Theorem 1 Law of Sines b sin sin sin a b c a 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. Suppose that an angle of a triangle and its opposite side are known. Then the ratio of Theorem 1 can be calculated. So if one additional part of the triangle, either of the other angles or either of the other sides, is known, then the law of sines can be used to solve the triangle. Thus, the law of sines is used to solve triangles, given: 1. Two sides and an angle opposite one of them (SSA) 2. Two angles and any side (ASA or AAS) If the given information for a triangle consists of two sides and the included angle (SAS) or three sides (SSS), then the law of sines cannot be applied. The key to handling these two cases, the law of cosines, is developed in Section 7-2. We will apply the law of sines to the easier ASA and AAS cases first, then we will turn to the more challenging SSA case. • Solving the ASA and AAS Cases EXAMPLE 1 Solving the ASA Case Solve the triangle in Figure 3. FIGURE 3 b 280 a 4520 120 meters Solution 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. 7-1 Law of Sines 509 180° Solve for 180° ( ) 180° (28°0 45°20) 106°40 sin sin a c Solve for a a c sin sin 120 sin 28°0 sin 106°40 58.8 meters sin sin b c Solve for b b c sin sin 120 sin 45°20 sin 106°40 89.1 meters Matched Problem 1 FIGURE 4 Solve the triangle in Figure 4. b 13 0 c 35 6520 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. • Solving the SSA Case—Including the Ambiguous 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. 510 7 Additional Topics in Trigonometry TABLE 2 SSA Variations a (h b sin ) Acute 0ah Acute Acute Number of triangles Figure 0 ah 1 hab 2 Acute ab 1 Obtuse 0ab 0 Case h b h a (b) b a ab h b a Ambiguous case (c) a (d) (e) a b Obtuse (a) a b 1 b a (f ) 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. EXPLORE-DISCUSS 1 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: (1) sin 1 (2) sin 1 (3) 0 sin 1 EXAMPLE 2 Solving the SSA Case Solve the triangle(s) with 123°, b 23 centimeters, and a 47 centimeters. 7-1 Solution 511 Law of Sines From a rough sketch (Fig. 5), we see that there is only one triangle: FIGURE 5 23 cm 47 cm 123 c sin sin b a Solve for b sin 23 sin 123° a 47 23 sin 123° sin1 24° 47 sin 180° 180° 123° 24° 33° Solve for sin sin a c a sin 47 sin 33° c 31 centimeters sin sin 123° Solve for c Matched Problem 2 EXAMPLE 3 Solve the triangle(s) with 98°, a 62 meters, and b 88 meters. Solving the SSA (Ambiguous) Case Solve the triangle(s) with 26°, a 1.0 meter, and b 1.8 meters. Solution FIGURE 6 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. I D 1.8 m 1.0 m A 26 B Solve for and C A 26 c D 1.8 m 1.8 m 1.0 m II D 1.0 m B A 26 We start by finding and using the law of sines: sin sin b a b sin 1.8 sin 26° sin 0.7891 a 1.0 1.0 m c C 512 7 Additional Topics in Trigonometry Angle can be either obtuse or acute: 180° sin1 0.7891 sin1 0.7891 or 180° 52° 128° Solve for and 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 a sin sin 1.0 sin 26° sin 26° 1.0 meter sin sin a c c a sin sin 1.0 sin 102° sin 26° 2.2 meters In summary: Triangle I: Triangle II: Matched Problem 3 128° 52° 26° c 1.0 meter 102° c 2.2 meters Solve the triangle(s) with a 8 kilometers, b 10 kilometers, and 35°. 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 baseline 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 513 FIGURE 7 A B e in rs l e se et 124.3 Ba m 5 2 1 41.6 d C Find angle C and use the law of sines. Solution Angle C 180° (124.3° 41.6°) 14.1° sin 14.1° sin 41.6° 125 d d 125 41.6° sin sin 14.1° 341 meters Matched Problem 4 In Example 4, find the distance AC. 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. b a c A Solve each triangle in Problems 1–8. 1. 73°, 28°, c 42 feet 2. 41°, 33°, c 21 centimeters 3. 122°, 18°, b 12 kilometers 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 9–20, are there zero, one, or two triangles with the given measurements? (Do not solve the triangle.) Which case, if any, of Table 2 applies? Explain. 9. a 5 inches, b 6 inches, 30° 10. a 6 inches, b 5 inches, 30° 11. a 4 feet, b 8 feet, 30° 514 7 Additional Topics in Trigonometry 12. a 2 feet, b 5 feet, 30° 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.) 13. a 5 inches, b 8 inches, 45° 14. a 3 inches, b 4 inches, 45° 36. (A) Use the law of sines and suitable identities to show that for any triangle 15. a 4 feet, b 5 feet, 120° 16. a 6 feet, b 4 feet, 120° ab a b 17. a 8 inches, b 6 inches, 60° 18. a 7 inches, b 8 inches, 60° 19. a 13 feet, b 12 feet, 90° 2 tan 2 tan (B) Verify the formula with values from Problem 1. 20. a 12 feet, b 13 feet, 90° B APPLICATIONS Solve each triangle in Problems 21–32. If a problem has no solution, say so. 37. 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)? 21. 118.3°, 12.2°, b 17.3 feet 22. 27.5°, 54.5°, a 9.27 inches 23. 67.7°, 54.2°, b 123 meters 24. 122.7°, 34.4°, b 18.3 kilometers 38. 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. 46.5°, a 7.9 millimeters, b 13.1 millimeters 26. 26.3°, a 14.7 inches, b 35.2 inches 27. 38.9°, a 30.0 inches, b 42.7 inches 28. 27.3°, a 135 centimeters, b 244 centimeters ★ 29. 123.2°, a 101 yards, b 152 yards 30. 137.3°, a 13.9 meters, b 19.1 meters 31. 29°30, a 43.2 millimeters, b 56.5 millimeters 39. 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 3 significant digits.) 32. 33°50, a 673 meters, b 1,240 meters C 33. 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. 34. 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. 3710 440 100 feet 35. 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 ★ 40. Surveying. To measure the height of Mt. Whitney in California, surveyors used a scheme like the one shown in the figure in Problem 39. They set up a horizontal baseline 2,000 feet long at the foot of the mountain and found the 7-1 Law of Sines 515 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? 41. Engineering. A 4.5-inch piston rod joins a piston to a 1.5inch crankshaft (see 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. 4.5 inches Piston 1.5 inches 9 Crankshaft d ★ 46. Surveying. Find the height of the tree in Problem 45 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°. ★ 47. 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°. 42. Engineering. Repeat Problem 41 if the piston rod is 6.3 inches, the crankshaft is 1.7 inches, and the angle is 11°. 43. 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 figure)? R 44. Astronomy. In Problem 43, find the maximum angle . [Hint: The angle is maximum when a straight line joining the Earth and Venus is tangent to Venus’s orbit.] C R Venus s Cornea Sun Venus Earth ★ 45. Surveying. A tree growing on a hillside casts a 102-foot shadow straight down the hill (see 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°. ★ 48. 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°. ★ 49. Surveying. The procedure illustrated in Problems 39 and 40 is used to determine an inaccessible height h when a baseline d on a line perpendicular to h can be established (see the figure on the next page) and the angles and can be measured. Show that hd sin sin sin ( ) 516 7 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 7-2 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.