2. Copyright © 2009 Pearson Education, Inc. CHAPTER 8: Applications of Trigonometry 8.1The Law of Sines 8.2The Law of Cosines 8.3Complex Numbers: Trigonometric Form 8.4Polar Coordinates and Graphs 8.5Vectors and Applications 8.6Vector Operations

3. Copyright © 2009 Pearson Education, Inc. 8.1 The Law of Sines  Use the law of sines to solve triangles.  Find the area of any triangle given the lengths of two sides and the measure of the included angle.

4. Slide 8.1 - 4 Copyright © 2009 Pearson Education, Inc. Solving Oblique Triangles 1.AAS: Two angles of a triangle and a side opposite one of them are known. 2.ASA: Two angles of a triangle and the included side are known.

5. Slide 8.1 - 5 Copyright © 2009 Pearson Education, Inc. Solving Oblique Triangles 3.SSA: Two sides of a triangle and an angle opposite one of them are known. (In this case, there may be no solution,one solution,or two solutions. The latter is known as the ambiguous case.)

6. Slide 8.1 - 6 Copyright © 2009 Pearson Education, Inc. Solving Oblique Triangles 4.SAS: Two sides of a triangle and the included are known.

7. Slide 8.1 - 7 Copyright © 2009 Pearson Education, Inc. Solving Oblique Triangles 5.SSS: All three sides of the triangle are known.

8. Slide 8.1 - 8 Copyright © 2009 Pearson Education, Inc. Law of Sines The Law of Sines applies to the first three situations. The Law of Sines In any triangle ABC, A B C a b c

9. Slide 8.1 - 9 Copyright © 2009 Pearson Education, Inc. Example In, e = 4.56, E = 43º, and G = 57º. Solve the triangle. Solution: Draw the triangle. We have AAS.

10. Slide 8.1 - 10 Copyright © 2009 Pearson Education, Inc. Example Solution continued Find F: F = 180º – (43º + 57º) = 80º Use law of sines to find the other two sides.

11. Slide 8.1 - 11 Copyright © 2009 Pearson Education, Inc. Example Solution continued We have solved the triangle.

12. Slide 8.1 - 12 Copyright © 2009 Pearson Education, Inc. Solving Triangles SSA When two sides of a triangle and an angle opposite one of them are known, the law of sines can be used to solve the triangle. There are various possibilities as show in the following 8 cases. Angle B is acute Case 1: No solution b < c; side b is too short to reach the base. No triangle formed.

13. Slide 8.1 - 13 Copyright © 2009 Pearson Education, Inc. Solving Triangles SSA Angle B is acute Case 2: One solution b < c; side b just reaches the base and is perpendicular to it. Angle B is acute Case 3: Two solutions b < c; an arc of radius b meets the base at two points. This is called the ambiguous case.

14. Slide 8.1 - 14 Copyright © 2009 Pearson Education, Inc. Solving Triangles SSA Angle B is acute Case 4: One solution b = c; an arc of radius b meets the base at just one point other than B. Angle B is acute Case 5: One solution b > c; an arc of radius b meets the base at just one point.

15. Slide 8.1 - 15 Copyright © 2009 Pearson Education, Inc. Solving Triangles SSA Angle B is obtuse Case 6: No solution b < c; side b is too short to reach the base. No triangle formed. Angle B is obtuse Case 7: No solution b = c; an arc of radius b meets the base only point B. No triangle is formed.

16. Slide 8.1 - 16 Copyright © 2009 Pearson Education, Inc. Solving Triangles SSA Angle B is obtuse Case 8: One solution b > c; an arc of radius b meets the base at just one point. The eight cases lead us the three possibilities in the SSA situation: no solution, one solution, or two solutions.

17. Slide 8.1 - 17 Copyright © 2009 Pearson Education, Inc. Example In, b = 15, c = 20, and B = 29º. Solve the triangle. Solution Draw a triangle. Find C.

18. Slide 8.1 - 18 Copyright © 2009 Pearson Education, Inc. Example Solution continued There are two angles less than 180º with a sine of 0.6464: 40º and 140º. So there are two possible solutions. Possible Solution I If C = 40º, then A = 180º – (29º + 40º) = 111º

19. Slide 8.1 - 19 Copyright © 2009 Pearson Education, Inc. These measures make a triangle as shown. Thus we have a solution. Example Solution continued Then we find a:

20. Slide 8.1 - 20 Copyright © 2009 Pearson Education, Inc. Example Solution continued Then we find a: Possible Solution II If C = 140º, then A = 180º – (29º + 140º) = 11º These measures make a triangle as shown. Thus we have a solution.

21. Slide 8.1 - 21 Copyright © 2009 Pearson Education, Inc. The Area of a Triangle The area of any is one half the product of he lengths of two sides and the sine of the included angle:

22. Slide 8.1 - 22 Copyright © 2009 Pearson Education, Inc. Example A university landscaping architecture department is designing a garden for a triangular area in a dormitory complex. Two sides of the garden, formed by the sidewalks in front of buildings A and B, measure 172 ft and 186 ft, respectively, and together form a 53º angle. The third side of the garden,formed by the sidewalk along Crossroads Avenue, measures 160 ft. What is the area of the garden to the nearest square foot?

23. Slide 8.1 - 23 Copyright © 2009 Pearson Education, Inc. Example Solution: Use the area formula. The area of the garden is approximately 12,775 ft 2.