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Chapter 6 Additional Topics in Trigonometry 1111572836_0600.qxd 9/29/10 1:43 PM Page 403 Additional Topics 6 in Trigonometry 6.1 Law of Sines 6.2 Law of Cosines 6.3 Vectors in the Plane 6.4 Vectors and Dot Products 6.5 Trigonometric Form of a Complex Number Section 6.3, Example 10 Direction of an Airplane Andresr 2010/used under license from Shutterstock.com 403 Copyright 2011 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 1111572836_0601.qxd 10/12/10 4:10 PM Page 404 404 Chapter 6 Additional Topics in Trigonometry 6.1 Law of Sines Introduction What you should learn ● Use the Law of Sines to solve In Chapter 4, you looked at techniques for solving right triangles. In this section and the oblique triangles (AAS or ASA). next, you will solve oblique triangles—triangles that have no right angles. As standard ● Use the Law of Sines to solve notation, the angles of a triangle are labeled oblique triangles (SSA). A, , and CB ● Find areas of oblique triangles and use the Law of Sines to and their opposite sides are labeled model and solve real-life a, , and cb problems. as shown in Figure 6.1. Why you should learn it You can use the Law of Sines to solve C real-life problems involving oblique triangles. For instance, Exercise 46 b a on page 411 shows how the Law of Sines can be used to help determine A c B the distance from a boat to the shoreline. Figure 6.1 To solve an oblique triangle, you need to know the measure of at least one side and the measures of any two other parts of the triangle—two sides, two angles, or one angle and one side. This breaks down into the following four cases. 1. Two angles and any side (AAS or ASA) 2. Two sides and an angle opposite one of them (SSA) 3. Three sides (SSS) 4. Two sides and their included angle (SAS) The first two cases can be solved using the Law of Sines, whereas the last two cases can be solved using the Law of Cosines (see Section 6.2). Law of Sines (See the proof on page 464.) If ABC is a triangle with sides a, b, and c, then a b c ϭ ϭ . sin A sin B sin C Oblique Triangles Study Tip C C Notice in Figure 6.2 a a that the height h of b h h b each triangle can be found using the formula B h A c B A c ϭ sin A b A is acute. A is obtuse. Figure 6.2 or h ϭ b sin A. The Law of Sines can also be written in the reciprocal form sin A sin B sin C ϭ ϭ . a b c lotusa 2010/used under license from Shutterstock.com Gabriela Trojanowska 2010/used under license from Shutterstock.com Copyright 2011 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 1111572836_0601.qxd 10/12/10 4:10 PM Page 405 Section 6.1 Law of Sines 405 Example 1 Given Two Angles and One Side—AAS For the triangle in Figure 6.3, C ϭ 102.3Њ, B ϭ 28.7Њ, andb ϭ 27.4 feet. Find the remaining angle and sides. C b = 27.4 ft Solution 102.3° a The third angle of the triangle is 28.7° ϭ ЊϪ Ϫ A 180 B C A c B ϭ 180ЊϪ28.7Њ Ϫ 102.3Њ Figure 6.3 ϭ 49.0Њ. Study Tip By the Law of Sines, you have When you are solving a b c ϭ ϭ . a triangle, a careful A B C sin sin sin sketch is useful as a Using b ϭ 27.4 produces quick test for the feasibility of an answer. Remember that b 27.4 ϭ ͑ ͒ ϭ ͑ Њ͒ Ϸ the longest side lies opposite a sin A Њ sin 49.0 43.06 feet sin B sin 28.7 the largest angle, and the and shortest side lies opposite the smallest angle. b 27.4 c ϭ ͑sin C͒ ϭ ͑sin 102.3Њ͒ Ϸ 55.75 feet. sin B sin 28.7 Now try Exercise 9. Example 2 Given Two Angles and One Side—ASA A pole tilts toward the sun at an 8Њ angle from the vertical, and it casts a 22-foot shadow. The angle of elevation from the tip of the shadow to the top of the pole is 43Њ. How tall is the pole? Solution In Figure 6.4,A ϭ 43Њ and C B ϭ 90Њϩ8Њϭ98Њ. So, the third angle is C ϭ ЊϪA Ϫ B b 180 a ° ϭ 180ЊϪ43Њ Ϫ 98Њ 8 ϭ 39Њ. ° By the Law of Sines, you have 43 B c = 22 ft A a c ϭ . sin A sin C Figure 6.4 Because c ϭ 22 feet, the length of the pole is c 22 a ϭ ͑sin A͒ ϭ ͑sin 43Њ͒ Ϸ 23.84 feet. sin C sin 39Њ Now try Exercise 41. For practice, try reworking Example 2 for a pole that tilts away from the sun under the same conditions. Copyright 2011 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 1111572836_0601.qxd 10/12/10 4:10 PM Page 406 406 Chapter 6 Additional Topics in Trigonometry The Ambiguous Case (SSA) In Examples 1 and 2, you saw that two angles and one side determine a unique triangle. However, if two sides and one opposite angle are given, then three possible situations can occur: (1) no such triangle exists, (2) one such triangle exists, or (3) two distinct triangles satisfy the conditions. The Ambiguous Case (SSA) Consider a triangle in which you are given a, b, and A. (Notice that h ϭ b sin A.) A is acute.A is acute.A is acute.A is acute.A is obtuse.A is obtuse. Sketch a a b b a b a h h a b h a h a b b A A A A A A Necessary a < h a ϭ h a Ն b h < a < b a Յ b a > b condition Possible None One One Two None One triangles Example 3 Single-Solution Case—SSA For the triangle in Figure 6.5, C a ϭ 22 inches,b ϭ 12 inches, and A ϭ 42Њ. b = 12 in. a = 22 in. 42° Find the remaining side and angles. A c B Solution Figure 6.5 One solution: a Ն b By the Law of Sines, you have sin B sin A ϭ Reciprocal form b a sin A sin B ϭ b΂ ΃ Multiply each side by b. a sin 42Њ sin B ϭ 12΂ ΃ Substitute for A, a, and b. 22 B Ϸ 21.41Њ. B is acute. Now you can determine that C Ϸ 180ЊϪ42Њ Ϫ 21.41Њ ϭ 116.59Њ. Then the remaining side is given by c a ϭ Law of Sines sin C sin A a c ϭ ͑sin C͒ Multiply each side by sin C. sin A 22 c ϭ ͑sin 116.59Њ͒ Substitute for a, A, and C. sin 42Њ c Ϸ 29.40 inches. Simplify. Now try Exercise 19. Copyright 2011 Cengage Learning. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part. Due to electronic rights, some third party content may be suppressed from the eBook and/or eChapter(s). Editorial review has deemed that any suppressed content does not materially affect the overall learning experience. Cengage Learning reserves the right to remove additional content at any time if subsequent rights restrictions require it. 1111572836_0601.qxd 10/12/10 4:11 PM Page 407 Section 6.1 Law of Sines 407 Example 4 No-Solution Case—SSA a = 15 Show that there is no triangle for which a ϭ 15, b ϭ 25, and A ϭ 85Њ. Solution b = 25 h Begin by making the sketch shown in Figure 6.6. From this figure it appears that no triangle is formed. You can verify this by using the Law of Sines. 85° A sin B sin A ϭ Reciprocal form b a Figure 6.6 No solution: a < h sin A sin B ϭ b΂ ΃ Multiply each side by a sin 85Њ sin B ϭ 25΂ ΃ Ϸ 1.6603 > 1 15 This contradicts the fact that Խsin BԽ Յ 1. So, no triangle can be formed having sides a ϭ 15 and b ϭ 25 and an angle of A ϭ 85Њ.
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