1. Slide 1-1 7 Applications of Trigonometry and Vectors

2. Slide 1-2 7.1 Oblique Triangles and the Law of Sines 7.1 Oblique Triangles and the Law of Sines 7.2 The Ambiguous Case of the Law of Sines 7.2 The Ambiguous Case of the Law of Sines 7.3 The Law of Cosines 7.3 The Law of Cosines 7.4 Vectors, Operations, and the Dot Product 7.4 Vectors, Operations, and the Dot Product 7.5Applications of Vectors 7.5Applications of Vectors Applications of Trigonometry and Vectors

3. Slide 1-3 Oblique Triangles and the Law of Sines 7.1 Congruency and Oblique Triangles ▪ Derivation of the Law of Sines ▪ Solving SAA and ASA Triangles (Case 1) ▪ Area of a Triangle A triangles that do not have a right angle are called the oblique triangle.

4. Slide 1-4 The Law of Sines For this law, start with any arbitrary triangle and from one of the vertices draw a line straight down to the base. This will split the triangle up into two smaller right triangles, such as is shown below,

5. Slide 1-5 Law of Sines for Oblique Triangle

6. Slide 1-6 Triangle Congruences in Geometry

7. Slide 1-7 Example 1 Find a.

8. Slide 1-8 Area of An Oblique Triangle Example:

9. Slide 1-9 Practice Problems

10. Slide 1-10 The Ambiguous Case of the Law of Sines 7.2 Description of the Ambiguous Case ▪ Solving SSA Triangles (Case 2) ▪ Analyzing Data for Possible Number of Triangles

11. Slide 1-11 Note In the ambiguous case, we are given two sides and an angle opposite one of the sides (SSA).

12. Slide 1-12 Example 2

13. Slide 1-13 Example 3

14. Slide 1-14 Example 4

15. Slide 1-15 Four possible cases can occur when solving an oblique triangle.

16. Slide 1-16

17. Slide 1-17 The Law of Cosines 7.3 Derivation of the Law of Cosines ▪ Solving SAS and SSS Triangles (Cases 3 and 4) ▪ Heron’s Formula for the Area of a Triangle ▪ Derivation of Heron’s Formula

18. Slide 1-18 The Law of Cosines Use the Pythagorean theorem

19. Slide 1-19 Law of Cosines In any triangle ABC, with sides a, b, and c,

20. Slide 1-20 Note If C = 90° in the third form of the law of cosines, then cos C = 0, and the formula becomes the Pythagorean theorem.

21. Slide 1-21 Example 5 APPLYING THE LAW OF COSINES (SAS) A surveyor wishes to find the distance between two inaccessible points A and B on opposite sides of a lake. While standing at point C, she finds that b = 259 m, a = 423 m, and angle ACB measures 132°40′. Find the distance c.

22. Slide 1-22 APPLYING THE LAW OF COSINES (SAS) (continued) Use the law of cosines because we know the lengths of two sides of the triangle and the measure of the included angle. The distance between the two points is about 628 m.

23. Slide 1-23 APPLYING THE LAW OF COSINES (SAS) Solve triangle ABC if A = 42.3°, b = 12.9 m, and c = 15.4 m. Example 6

24. Slide 1-24 APPLYING THE LAW OF COSINES (SAS) (continued) Use the law of sines to find the measure of another angle. Now find the measure of the third angle.

25. Slide 1-25 Example 7 APPLYING THE LAW OF COSINES (SSS) Solve triangle ABC if a = 9.47 ft, b = 15.9 ft, and c = 21.1 ft. Solve for cos C.

26. Slide 1-26 APPLYING THE LAW OF COSINES (SSS) (continued) Use the law of sines to find the measure of angle B. Now find the measure of angle A.

27. Slide 1-27 Heron’s Area Formula (SSS) If a triangle has sides of lengths a, b, and c, with semiperimeter then the area of the triangle is

28. Slide 1-28 Example 8 USING HERON’S FORMULA TO FIND AN AREA (SSS) The distance “as the crow flies” from Los Angeles to New York is 2451 mi, from New York to Montreal is 331 mi, and from Montreal to Los Angeles is 2427 mi. What is the area of the triangular region having these three cities as vertices? (Ignore the curvature of Earth.)

29. Slide 1-29 USING HERON’S FORMULA TO FIND AN AREA (SSS) (continued) The semiperimeter s is Using Heron’s formula, the area  is

30. Slide 1-30 Derivation of Heron’s Formula Let triangle ABC have sides of length a, b, and c. Apply the law of cosines. The perimeter of the triangle is a + b + c, so half the perimeter (the semiperimeter) is given by the formula in equation (2) shown next.

31. Slide 1-31 Derivation of Heron’s Formula (continued) Subtract 2b and 2c in a similar way in equation (3) to obtain equations (5) and (6).

32. Slide 1-32 Derivation of Heron’s Formula (continued) Now we obtain an expression for 1 – cos A.

33. Slide 1-33 Derivation of Heron’s Formula (continued) Similarly, it can be shown that

34. Slide 1-34 Derivation of Heron’s Formula (continued) Recall the double- angle identities for cos 2θ.

35. Slide 1-35 Derivation of Heron’s Formula (continued) Recall the double- angle identities for cos 2θ.

36. Slide 1-36 Derivation of Heron’s Formula (continued) The area of triangle ABC can be expressed as follows.

37. Slide 1-37 Derivation of Heron’s Formula (continued) Recall the double-angle identity for sin 2θ.

38. Slide 1-38 Derivation of Heron’s Formula (continued) Heron’s formula