1. Copyright © 2011 Pearson Education, Inc. Slide 10.2-1

2. Copyright © 2011 Pearson Education, Inc. Slide 10.2-2 Chapter 10: Applications of Trigonometry and Vectors 10.1The Law of Sines 10.2The Law of Cosines and Area Formulas 10.3Vectors and Their Applications 10.4Trigonometric (Polar) Form of Complex Numbers 10.5Powers and Roots of Complex Numbers 10.6Polar Equations and Graphs 10.7More Parametric Equations

3. Copyright © 2011 Pearson Education, Inc. Slide 10.2-3 SAS or SSS cases form a unique triangle Triangle Side Restriction In any triangle, the sum of the lengths of any two sides must be greater than the length of the remaining side. 10.2The Law of Cosines and Area Formulas

4. Copyright © 2011 Pearson Education, Inc. Slide 10.2-4 10.2Derivation of the Law of Cosines Let ABC be any oblique triangle drawn with its vertices labeled as in the figure below. The coordinates of point A become (c cos B, c sin B). Figure 10 pg 10-28

5. Copyright © 2011 Pearson Education, Inc. Slide 10.2-5 10.2Derivation of the Law of Cosines Point C has coordinates (a, 0) and AC has length b. This result is one form of the law of cosines. Placing A or C at the origin would have given the same result, but with the variables rearranged.

6. Copyright © 2011 Pearson Education, Inc. Slide 10.2-6 10.2The Law of Cosines The Law of Cosines In any triangle ABC, with sides a, b, and c, That is, the square of a side in any triangle is equal to the sum of the squares of the other two sides, minus twice the product of those two sides and the cosine of the angle included between them.

7. Copyright © 2011 Pearson Education, Inc. Slide 10.2-7 10.2Using the Law of Cosines to Solve a Triangle (SAS) ExampleSolve triangle ABC if A = 42.3°, b = 12.9 meters, and c = 15.4 meters. Solution Start by finding a using the law of cosines.

8. Copyright © 2011 Pearson Education, Inc. Slide 10.2-8 10.2Using the Law of Cosines to Solve a Triangle (SAS) B must be the smaller of the two remaining angles since it is opposite the shorter of the two sides b and c. Therefore, it cannot be obtuse. Caution If we had chosen to find C rather than B, we would not have known whether C equals 81.7 ° or its supplement, 98.3 °.

9. Copyright © 2011 Pearson Education, Inc. Slide 10.2-9 10.2Using the Law of Cosines to Solve a Triangle (SSS) ExampleSolve triangle ABC if a = 9.47 feet, b =15.9 feet, and c = 21.1 feet. Solution We solve for C, the largest angle, first. If cos C < 0, then C will be obtuse.

10. Copyright © 2011 Pearson Education, Inc. Slide 10.2-10 10.2Using the Law of Cosines to Solve a Triangle (SSS) Verify with either the law of sines or the law of cosines that B  45.1°. Then,

11. Copyright © 2011 Pearson Education, Inc. Slide 10.2-11 10.2Summary of Cases with Suggested Procedures Oblique TriangleSuggested Procedure for Solving Case 1: SAA or ASA1.Find the remaining angle using the angle sum formula (A + B + C = 180°). 2.Find the remaining sides using the law of sines. Case 2: SSAThis is the ambiguous case; 0, 1, or 2 triangles. 1.Find an angle using the law of sines. 2.Find the remaining angle using the angle sum formula. 3.Find the remaining side using the law of sines. If two triangles exist, repeat steps 2 and 3.

12. Copyright © 2011 Pearson Education, Inc. Slide 10.2-12 10.2Summary of Cases with Suggested Procedures Oblique TriangleSuggested Procedure for Solving Case 3: SAS1.Find the third side using the law of cosines. 2.Find the smaller of the two remaining angles using the law of sines. 3.Find the remaining angle using the angle sum formula. Case 4: SSS1.Find the largest angle using the law of cosines. 2.Find either remaining angle using the law of sines. 3.Find the remaining angle using the angle sum formula.

13. Copyright © 2011 Pearson Education, Inc. Slide 10.2-13 10.2Area Formulas. Heron’s Formula If a triangle has sides of lengths a, b, and c, and if the semiperimeter is Then the area of the triangle is That is, the area is the square root of the product of four factors: the semiperimeter, and the three differences of the semiperimeter and the individual side lengths.

14. Copyright © 2011 Pearson Education, Inc. Slide 10.2-14 10.2Using Heron’s Formula to Find an Area ExampleThe distance “as the crow flies” from Los Angeles to New York is 2451 miles, from New York to Montreal is 331 miles, and from Montreal to Los Angeles is 2427 miles. What is the area of the triangular region having these three cities as vertices? (Ignore the curvature of the earth.) Solution

15. Copyright © 2011 Pearson Education, Inc. Slide 10.2-15 10.2Area of a Triangle Given SAS The area of any triangle is given by A = ½bh, where b is its base and h is its height.

16. Copyright © 2011 Pearson Education, Inc. Slide 10.2-16 10.2Area of a Triangle Given SAS Area of a Triangle (SAS) In any triangle ABC, the area A is given by the following formulas: That is, the area is half the product of the lengths of two sides and the sine of the angle included between them.

17. Copyright © 2011 Pearson Education, Inc. Slide 10.2-17 10.2Finding the Area of a Triangle (SAS) ExampleFind the area of triangle ABC in the figure. SolutionWe are given B = 55°, a = 34 feet, and c = 42 feet.