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Law of Sines 333202_0601.qxd 12/5/05 10:40 AM Page 430 430 Chapter 6 Additional Topics in Trigonometry 6.1 Law of Sines What you should learn •Use the Law of Sines to solve Introduction oblique triangles (AAS or ASA). In Chapter 4, you studied techniques for solving right triangles. In this section •Use the Law of Sines to solve and the next, you will solve oblique triangles—triangles that have no right oblique triangles (SSA). angles. As standard notation, the angles of a triangle are labeled A, B, and C, and •Find the areas of oblique their opposite sides are labeled a, b, and c, as shown in Figure 6.1. triangles. •Use the Law of Sines to model C and solve real-life problems. Why you should learn it b a You can use the Law of Sines to solve real-life problems involving A c B oblique triangles. For instance, in Exercise 44 on page 438, you FIGURE 6.1 can use the Law of Sines to determine the length of the To solve an oblique triangle, you need to know the measure of at least one shadow of the Leaning Tower side and any two other parts of the triangle—either two sides, two angles, or one of Pisa. 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 require the Law of Cosines (see Section 6.2). Law of Sines If ABC is a triangle with sides a, b, and c, then a b c ϭ ϭ . sin A sin B sin C Hideo Kurihara/Getty Images C C b a a h h b A c B ABc A is acute.A is obtuse. The Law of Sines can also be written in the reciprocal form sin A sin B sin C The HM mathSpace® CD-ROM and ϭ ϭ . Eduspace® for this text contain a b c additional resources related to the concepts discussed in this chapter. For a proof of the Law of Sines, see Proofs in Mathematics on page 489. 333202_0601.qxd 12/5/05 10:40 AM Page 431 Section 6.1 Law of Sines 431 C Example 1 Given Two Angles and One Side—AAS b = 27.4 ft 102.3° a For the triangle in Figure 6.2,C ϭ 102.3Њ, B ϭ 28.7Њ, and b ϭ 27.4 feet. Find the remaining angle and sides. 28.7° Solution A c B The third angle of the triangle is FIGURE 6.2 A ϭ 180ЊϪB Ϫ C ϭ 180ЊϪ28.7Њ Ϫ 102.3Њ ϭ 49.0Њ. By the Law of Sines, you have a b c ϭ ϭ . sin A sin B sin C Using b ϭ 27.4 produces b 27.4 a ϭ ͑sin A͒ ϭ ͑sin 49.0Њ͒ Ϸ 43.06 feet When solving triangles, a sin B sin 28.7Њ careful sketch is useful as a and quick test for the feasibility of b 27.4 an answer. Remember that the ϭ ͑ ͒ ϭ ͑ Њ͒ Ϸ c sin C Њ sin 102.3 55.75 feet. longest side lies opposite the sin B sin 28.7 largest angle, and the shortest Now try Exercise 1. side lies opposite the smallest angle. 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 ϭ Њ ϭ Њϩ Њϭ Њ C From Figure 6.3, note that A 43 and B 90 8 98 . So, the third angle is C ϭ 180ЊϪA Ϫ B ϭ 180ЊϪ43Њ Ϫ 98Њ ϭ Њ b 39 . a 8° By the Law of Sines, you have a c ϭ . sin A sin C 43° ϭ B c = 22 ft A Because c 22 feet, the length of the pole is c 22 a ϭ ͑sin A͒ ϭ ͑sin 43Њ͒ Ϸ 23.84 feet. FIGURE 6.3 sin C sin 39Њ Now try Exercise 35. For practice, try reworking Example 2 for a pole that tilts away from the sun under the same conditions. 333202_0601.qxd 12/5/05 10:40 AM Page 432 432 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, three possible situations can occur: (1) no such triangle exists, (2) one such triangle exists, or (3) two distinct triangles may satisfy the conditions. The Ambiguous Case (SSA) Consider a triangle in which you are given a, b, and A. ͑h ϭ b sin A͒ A is acute.A is acute.A is acute.A is acute.A is obtuse.A is obtuse. Sketch b a b a ba h a b h a a a h h b b A A A A A A Necessary a < h a ϭ h a > b h < a < b a ≤ b a > b condition Triangles None One One Two None One possible Example 3 Single-Solution Case—SSA C For the triangle in Figure 6.4,a ϭ 22 inches,b ϭ 12 inches, and A ϭ 42Њ. Find b = 12 in. a = 22 in. the remaining side and angles. 42° Solution A c B By the Law of Sines, you have One solution: a > b sin B sin A FIGURE 6.4 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 c a ϭ sin C sin A Encourage your students to sketch the a 22 triangle, keeping in mind that the c ϭ ͑sin C͒ ϭ ͑sin 116.59Њ͒ Ϸ 29.40 inches. longest side lies opposite the largest sin A sin 42Њ angle of the triangle. For practice, suggest that students also find h. Now try Exercise 19. 333202_0601.qxd 12/5/05 10:40 AM Page 433 Section 6.1 Law of Sines 433 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. b = 25 h Solution Begin by making the sketch shown in Figure 6.5. From this figure it appears that 85° no triangle is formed. You can verify this using the Law of Sines. A sin B sin A No solution: a h ϭ Reciprocal form < b a FIGURE 6.5 sin A sin B ϭ b΂ ΃ Multiply each side by b. a sin 85Њ sin B ϭ 25΂ ΃ Ϸ 1.660 > 1 15 This contradicts the fact that Խsin BԽ ≤ 1. So, no triangle can be formed having sides a ϭ 15 and b ϭ 25 and an angle of A ϭ 85Њ. Now try Exercise 21. Activities Example 5 Two-Solution Case—SSA Have your students determine the number of triangles possible in each of Find two triangles for which a ϭ 12 meters,b ϭ 31 meters, and A ϭ 20.5Њ. the following cases. 1. A ϭ 62Њ, a ϭ 10, b ϭ 12 Solution (0 triangles) By the Law of Sines, you have 2. A ϭ 98Њ, a ϭ 10, b ϭ 3 (1 triangle) sin B sin A ϭ Reciprocal form 3. A ϭ 54Њ, a ϭ 7, b ϭ 10 b a (0 triangles) sin A sin 20.5Њ Discuss several examples of the two- sin B ϭ b΂ ΃ ϭ 31΂ ΃ Ϸ 0.9047. solution case. a 12 Ϸ Њ Ϸ ЊϪ Њϭ Њ Њ There are two angles B1 64.8 and B2 180 64.8 115.2 between 0 Additional Example and 180Њ whose sine is 0.9047. For B Ϸ 64.8Њ, you obtain Find two triangles for which c ϭ 29, 1 b ϭ 46, and C ϭ 31Њ. C Ϸ 180ЊϪ20.5ЊϪ64.8Њϭ94.7Њ Solution ϭ Њ ϭ Њ ϭ a 12 B 54.8 , A 94.2 , a 56.2 c ϭ ͑sin C͒ ϭ ͑sin 94.7Њ͒ Ϸ 34.15 meters. B ϭ 125.2Њ, A ϭ 23.8Њ, a ϭ 22.7 sin A sin 20.5Њ Ϸ Њ For B2 115.2 , you obtain C Ϸ 180ЊϪ20.5ЊϪ115.2Њϭ44.3Њ a 12 c ϭ ͑sin C͒ ϭ ͑sin 44.3Њ͒ Ϸ 23.93 meters. sin A sin 20.5Њ The resulting triangles are shown in Figure 6.6. b = 31 m b = 31 m a = 12 m a = 12 m 115.2° 20.5° 64.8° 20.5° A B A 1 B2 FIGURE 6.6 Now try Exercise 23. 333202_0601.qxd 12/5/05 10:40 AM Page 434 434 Chapter 6 Additional Topics in Trigonometry Area of an Oblique Triangle The procedure used to prove the Law of Sines leads to a simple formula for the To see how to obtain the height of area of an oblique triangle. Referring to Figure 6.7, note that each triangle has a the obtuse triangle in Figure 6.7, height of h ϭ b sin A.
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