Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Chapter 6 Additional Topics in Trigonometry 6.1 The Law of Sines Copyright © 2014, 2010, 2007 Pearson Education, Inc. 1 Objectives: • Use the Law of Sines to solve oblique triangles. • Use the Law of Sines to solve, if possible, the triangle or triangles in the ambiguous case. • Find the area of an oblique triangle using the sine function. • Solve applied problems using the Law of Sines. Copyright © 2014, 2010, 2007 Pearson Education, Inc. 2 Oblique Triangles An oblique triangle is a triangle that does not contain a right angle. An oblique triangle has either three acute angles or two acute angles and one obtuse angle. The relationships among the sides and angles of right triangles defined by the trigonometric functions are not valid for oblique triangles. Copyright © 2014, 2010, 2007 Pearson Education, Inc. 3 The Law of Sines Copyright © 2014, 2010, 2007 Pearson Education, Inc. 4 Solving Oblique Triangles Solving an oblique triangle means finding the lengths of its sides and the measurements of its angles. The Law of Sines can be used to solve a triangle in which one side and two angles are known. The three known measurements can be abbreviated using SAA (a side and two angles are known) or ASA (two angles and the side between them are known). Copyright © 2014, 2010, 2007 Pearson Education, Inc. 5 Example: Solving an SAA Triangle Using the Law of Sines Solve the triangle with A = 64°, C = 82°, and c = 14 centimeters. Round lengths of sides to the nearest tenth. A B C 180 64 B 82 180 146 B 180 B 34 a c sin A sin C a sin C c sin A c sin A 14sin 64 12.7 cm a sin C sin82 Copyright © 2014, 2010, 2007 Pearson Education, Inc. 6 Example: Solving an SAA Triangle Using the Law of Sines (continued) Solve the triangle with A = 64°, C = 82°, and c = 14 centimeters. Round lengths of sides to the nearest tenth. B = 34° a » 12.7 cm b » 7.9 cm b c sin B sin C c sin B 14sin34 » 7.9 cm b sin C sin82 b sin C c sin B Copyright © 2014, 2010, 2007 Pearson Education, Inc. 7 Example: Solving an ASA Triangle Using the Law of Sines Solve triangle ABC if A = 40°, C = 22.5°, and b = 12. Round measures to the nearest tenth. B A B C 180 40 B 22.5 180 62.5 B 180 C A 12 B 117.5 b a b sin A 12sin 40 sin B sin A 8.7 a sin B sin117.5 b sin A a sin B Copyright © 2014, 2010, 2007 Pearson Education, Inc. 8 Example: Solving an ASA Triangle Using the Law of Sines (continued) Solve triangle ABC if A = 40°, C = 22.5°, and b = 12. Round measures to the nearest tenth. B B 117.5 a 8.7 C A c 5.2 12 b c sin B sin C c sin B b sin C b sin C 12sin 22.5 5.2 c sin B sin117.5 Copyright © 2014, 2010, 2007 Pearson Education, Inc. 9 The Ambiguous Case (SSA) If we are given two sides and an angle opposite one of the two sides (SSA), the given information may result in one triangle, two triangles, or no triangle at all. SSA is known as the ambiguous case when using the Law of Sines because the given information may result in one triangle, two triangles, or no triangle at all. Copyright © 2014, 2010, 2007 Pearson Education, Inc. 10 The Ambiguous Case (SSA) (continued) Copyright © 2014, 2010, 2007 Pearson Education, Inc. 11 Example: Solving an SSA Triangle Using the Law of Sines (No Solution) Solve triangle ABC if A = 50°, a = 10, and b = 20. There is no angle B for which the sine is greater than 1. There is no triangle with the given measurements. a b sin A sin B a sin B b sin A b sin A 20sin50 1.53 sin B a 10 Copyright © 2014, 2010, 2007 Pearson Education, Inc. 12 Example: Solving an SSA Triangle Using the Law of Sines (Two Solutions) Solve triangle ABC if A = 35°, a = 12, and b = 16. Round lengths of sides to the nearest tenth and angle measures to the nearest degree. a b sin A sin B a sin B b sin A b sin A 16sin35 0.7648 sin B a 12 sin 1 0.7648 50 There are two angles between 0° and 180° for which sinB = 0.7648 Copyright © 2014, 2010, 2007 Pearson Education, Inc. 13 Example: Solving an SSA Triangle Using the Law of Sines (Two Solutions) (continued) Solve triangle ABC if A = 35°, a = 12, and b = 16. Round lengths of sides to the nearest tenth and angle measures to the nearest degree. sin 1 0.7648 50 B1 50 A B1 35 50 85 B2 180 50 130 A B2 35 130 165 A B1 180 and A B2 180, there are two possible solutions. Copyright © 2014, 2010, 2007 Pearson Education, Inc. 14 Example: Solving an SSA Triangle Using the Law of Sines (Two Solutions) (continued) Solve triangle ABC if A = 35°, a = 12, and b = 16. Round lengths of sides to the nearest tenth and angle measures to the nearest degree. B1 50 B2 130 A B1 C1 180 35 50 C1 180 85 C1 180 A B2 C2 180 35 130 C2 180 165 C2 180 C2 15 C1 95 Copyright © 2014, 2010, 2007 Pearson Education, Inc. 15 Example: Solving an SSA Triangle Using the Law of Sines (Two Solutions) (continued) Solve triangle ABC if A = 35°, a = 12, and b = 16. Round lengths of sides to the nearest tenth and angle measures to the nearest degree. B2 130 C2 15 B1 50 C1 95 b c2 b c1 sin B2 sin C2 sin B1 sin C1 b sin C2 c2 sin B2 b sin C1 c1 sin B1 b sin C2 16sin15 c2 5.4 b sin C1 16sin95 sin B2 sin130 c1 20.8 sin B1 sin50 Copyright © 2014, 2010, 2007 Pearson Education, Inc. 16 Example: Solving an SSA Triangle Using the Law of Sines (Two Solutions) (continued) Solve triangle ABC if A = 35°, a = 12, and b = 16. Round lengths of sides to the nearest tenth and angle measures to the nearest degree. There are two triangles. In one triangle, the solution is B1 50 C1 95 c1 20.8 In the other triangle, the solution is B2 130 C2 15 c2 5.4 Copyright © 2014, 2010, 2007 Pearson Education, Inc. 17 The Area of an Oblique Triangle The area of a triangle equals one-half the product of the lengths of two sides times the sine of their included angle. In the figure, this wording can be expressed by the formulas 1 1 1 Area bc sin A ab sin C ac sin B 2 2 2 Copyright © 2014, 2010, 2007 Pearson Education, Inc. 18 Example: Finding the Area of an Oblique Triangle Find the area of a triangle having two sides of length 8 meters and 12 meters and an included angle of 135°. Round to the nearest square meter. C 1 Area ab sin C 8m 12 m 2 135 A B 1 (12)(8)(sin135) 2 34 sq m Copyright © 2014, 2010, 2007 Pearson Education, Inc. 19 Example: Application Two fire-lookout stations are 13 miles apart, with station B directly east of station A. Both stations spot a fire. The bearing of the fire from station A is N35°E and the bearing of the fire from station B is N49°W. How far, to the nearest tenth of a mile, is the fire from station B? A 90 35 55 C B 90 49 41 35 49 A B C 180 55 41 C 180 A B 96 C 180 13 C 84 Copyright © 2014, 2010, 2007 Pearson Education, Inc. 20 Example: Application (continued) Two fire-lookout stations are 13 miles apart, with station B directly east of station A. Both stations spot a fire. The bearing of the fire from station A is N35°E and the bearing of the fire from station B is N49°W. How far, to the nearest tenth of a mile, is the fire from station B? c a sin C C 35 A sin A c sin A a sin C c sin A 13sin 55 » 10.7 a sin C sin84 49 B The fire is approximately 10.7 miles from station B. Copyright © 2014, 2010, 2007 Pearson Education, Inc. 21