1. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-1 Oblique Triangles and the Law of Sines 7.1 The Law of Sines ▪ Solving SAA and ASA Triangles (Case 1) ▪ Area of a Triangle

2. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-2 Solve triangle ABC if A = 28.8°, C = 102.6°, and c = 25.3 in. 7.1 Example 1 Using the Law of Sines to Solve a Triangle (SAA) (page 304) a b

3. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-3 Jerry wishes to measure the distance across the Big Muddy River. He determines that C = 117.2°, A = 28.8°, and b = 75.6 ft. Find the distance a across the river. 7.1 Example 2 Using the Law of Sines in an Application (ASA) (page 305) The distance across the river is about 65.1 ft. B = 180˚ - 28.8˚ - 117.2˚ = 34˚

4. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-4 The bearing of a lighthouse from a ship was found to be N 52° W. After the ship sailed 5.8 km due south, the new bearing was N 23° W. Find the distance between the ship and the lighthouse at each location. 7.1 Example 3 Using the Law of Sines in an Application (ASA) (page 305) Let x = the distance to the lighthouse at bearing N 52° W and y = the distance to the lighthouse at bearing N 23° W. The distance between the ship and the second location is about 9.4 km. The distance between the ship and the first location is about 4.7 km. 52˚ 5.8km 23˚ 128˚ 29˚ x y

5. Need to Memorize: Area of Triangle Formulas Area of a triangle SASK = ½ bc sin A = ½ ab sin C = ½ ac sin B AAS,ASA K = ½ c 2 sin A sin B / sin C = ½ a 2 sin B sin C / sin A = ½ b 2 sin A sin C / sin B Hero’s Formula SSS where s = ½ (a + b + c) Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-5

6. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-6 7.1 Example 4 Finding the Area of a Triangle (SAS) (page 306) Find the area of triangle DEF in the figure. K = ½ a 2 sin B sin C / sin A Find the area of triangle ABC if B = 58°10′, a = 32.5 cm, and C = 73°30′. A ≈ 43.13 ft 2

7. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-7 Oblique Triangles and the Law of Sines 7.1 The Law of Sines Solving SAA and ASA Triangles Area of a Triangle SASK = ½ bc sin A = ½ ab sin C = ½ ac sin B AAS,ASA K = ½ c 2 sin A sin B / sin C = ½ a 2 sin B sin C / sin A = ½ b 2 sin A sin C / sin B

8. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-8 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

9. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-9 Solve triangle ABC if a = 17.9 cm, c = 13.2 cm, and C = 75°30′. 7.2 Example 1 Solving the Ambiguous Case (No Such Triangle) (page 314) Use the Law of Sines to find A. Since sin A > 1 is impossible, no such triangle exists. sin A

10. 7-10 Solve triangle ABC if A = 61.4°, a = 35.5 cm, and b = 39.2 cm. 7.2 Example 2 Solving the Ambiguous Case (Two Triangles) (page 314) sin B There are two angles between 0° and 180° such that sin B ≈.9695: B 1 ≈ 75.8˚ and B 2 ≈ 104.2˚ 61.4° A 35.5 B C 39.2 B 1 ≈ B 2 ≈ C 1 ≈ C 2 ≈ c 1 ≈ c 2 ≈ c1c1 B 1 ≈ 75.8˚ B 2 ≈ 104.2˚ C 1 ≈ C 2 ≈ c 1 ≈ c 2 ≈ B 1 ≈ 75.8˚ B 2 ≈ 104.2˚ C 1 ≈ 42.8˚ C 2 ≈ 14.4˚ c 1 ≈ 27.5 c 2 ≈ B 1 ≈ 75.8˚ B 2 ≈ 104.2˚ C 1 ≈ 42.8˚ C 2 ≈ 14.4˚ c 1 ≈ 27.5 c 2 ≈ 10.1 B 1 ≈ 75.8˚ B 2 ≈ 104.2˚ C 1 ≈ 42.8˚ C 2 ≈ 14.4˚ c 1 ≈ c 2 ≈ c2c2 ≈10.1

11. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-11 7.2 Example 3 Solving the Ambiguous Case (One Triangle) (page 316) Solve triangle ABC if B = 68.7°, b = 25.4 in., and a = 19.6 in. 68.7° A 19.6 B C 25.4 sin A There are two angles between 0° and 180° such that sin A ≈.7189: A 1 ≈ A 2 ≈ C 1 ≈ C 2 ≈ c 1 ≈ c 2 ≈ A 1 ≈ 46.0˚ A 2 ≈ 134.0˚ C 1 ≈ C 2 ≈ c 1 ≈ c 2 ≈ A 1 ≈ 46.0˚ A 2 ≈ 134.0˚ C 1 ≈ 65.3˚ C 2 ≈ NP c 1 ≈ c 2 ≈ c1c1 A 1 ≈ 46.0˚ A 2 ≈ 134.0˚ C 1 ≈ 65.3˚ C 2 ≈ NP c 1 ≈ 24.8 c 2 ≈

12. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-12 Without using the law of sines, explain why no triangle exists satisfying B = 93°, b = 42 cm, and c = 48 cm. 7.2 Example 4 Analyzing Data Involving an Obtuse Angle (page 317) Because B is an obtuse angle, it must be the largest angle of the triangle. Thus, b must be the longest side of the triangle. We are given that c > b, so no such triangle exists.

13. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-13 The Ambiguous Case of the Law of Sines 7.2 Description of the Ambiguous Case Solving SSA Triangles Analyzing Data for Possible Number of Triangles

14. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-14 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

15. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-15 Two boats leave a harbor at the same time, traveling on courses that make an angle of 82°20′ between them. When the slower boat has traveled 62.5 km, the faster one has traveled 79.4 km. At that time, what is the distance between the boats? 7.3 Example 1 Using the Law of Cosines in an Application (SAS) (page 320) The harbor is at C. The slower boat is at A, and the faster boat is at B. We are seeking the length of side c. Law of Cosines: c 2 = a 2 + b 2 – 2ab cosC The two boats are about 94.3 km apart. c 2 = 79.4 2 + 62.5 2 – 2(79.4)(62.5) cos82˚20’ c 2 = 8886.5

16. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-16 Solve triangle ABC if B = 73.5°, a = 28.2 ft, and c = 46.7 ft. 7.3 Example 2 Using the Law of Cosines to Solve a Triangle (SAS) (page 321) c 2 = a 2 + b 2 – 2ab cosC 46.7 2 = 28.2 2 + 47.2 2 – 2(28.2)(47.2) cosC Calculator Steps: 46.7 2 – 28.2 2 – 47.2 2 enter then divide by (-228.247.2) enter then cos -1 (ans) b ≈ C ≈ A ≈ b ≈ 47.2 ft C ≈ A ≈ b ≈ 47.2 ft C ≈ 71.6˚ A ≈ 34.9˚ b ≈ 47.2 ft C ≈ 71.6˚ A ≈

17. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-17 Solve triangle ABC if a = 25.4 cm, b = 42.8 cm, and c = 59.3 cm. 7.3 Example 3 Using the Law of Cosines to Solve a Triangle (SSS) (page 321) 42.8 2 = 25.4 2 + 59.3 2 – 2 25.4 59.3cos B B ≈ 39.3˚ A ≈ 22.1˚

18. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-18 7.3 Example 4 Designing a Roof Truss (SSS) (page 322) Find the measure of angle C in the figure. Use the calculator to find C: 9 2 – 11 2 – 6 2 enter Ans / (-2*11*6) enter Cos -1 ans enter

19. 7.3 Example 5 Using Heron’s Formula to Find an Area (SSS) (page 323) The distance “as the crow flies” from Chicago to St. Louis is 262 mi, from St. Louis to New Orleans is 599 mi, and from New Orleans to Chicago is 834 mi. What is the area of the triangular region having these three cities as vertices? The semiperimeter The area of the triangular region is about 40,800 mi 2.

20. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-20 The Law of Cosines 7.3 The Law of Cosines Solving SAS and SSS Triangles Heron’s Formula for the Area of a Triangle c 2 = a 2 + b 2 – 2ab cosC, where s = semi perimeter

21. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-21 Vectors, Operations, and the Dot Product 7.4 Basic Terminology ▪ Algebraic Interpretation of Vectors ▪ Operations with Vectors ▪ Dot Product and the Angle Between Vectors

22. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-22 Find the magnitude and direction angle for 7.4 Example 1 Finding Magnitude and Direction Angle (page 335) Magnitude: Direction angle: Reference angle: tan -1 ( 4 / 5 ) ≈ 38.66˚ θ ≈ 141.34˚ Graphing calculator solution Rectangular to Polar

23. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-23 Vector v has magnitude 14.5 and direction angle 220°. Find the horizontal and vertical components. 7.4 Example 2 Finding Horizontal and Vertical Components (page 336) Horizontal component: –11.1 Vertical component: –9.3 Graphing calculator solution Polar to Rectangular x y

24. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-24 Write each vector in the form  a, b . 7.4 Example 3 Writing Vectors in the Form  a, b  (page 336) u: magnitude 8, direction angle 135° v: magnitude 4, direction angle 270° w: magnitude 10, direction angle 340° Ref angle 45˚, in quad II Ref angle 20˚, in quad IV Bottom half of y-axis

25. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-25 Two forces of 32 and 48 newtons act on a point in the plane. If the angle between the forces is 76°, find the magnitude of the resultant vector. 7.4 Example 4 Finding the Magnitude of a Resultant (page 337) 76˚ 32N 48N 104˚ 32N

26. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-26 Let u =  6, –3  and v =  –14, 8 . Find the following. 7.4 Example 5 Performing Vector Operations (page 338) Graphing calculator solution =  6, –3  +  –14, 8  = - 1 / 2  –14, 8  =  7,–4 

27. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-27 Find each dot product. 7.4 Example 6 Finding Dot Products (page 339) Orthogonal Vectors are two vectors which are perpendicular and their dot product will = 0 if they are orthogonal.

28. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-28 Find the angle θ between the two vectors u =  5, –12  and v =  4, 3 . 7.4 Example 7 Finding the Angle Between Two Vectors (page 340)

29. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-29 Vectors, Operations, and the Dot Product 7.4 Vocabulary: magnitude (length of vector) horizontal and vertical components polar to rectangular (rcosθ,rsinθ) rectangular to polar Operations with Vectors sum/difference scalar multiplication Dot Product The Angle Between Vectors  a,b   c,d  = ac + bd

30. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-30 Applications of Vectors 7.5 The Equilibrant ▪ Incline Applications ▪ Navigation Applications

31. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-31 Find the magnitude of the equilibrant of forces of 54 Newton's and 42 Newton's acting on a point A, if the angle between the forces is 98°. Then find the angle between the equilibrant and the 42-newton force. 7.5 Ex 1 Finding Magnitude and Direction of an Equilibrant (pg 344) The equilibrant is the green vector. Need to find α: α = 180 - θ 54 2 = 42 2 + 63.63 2 – 24263.63cosθ 63.63 newtons θ ≈ 57.18˚ α ≈ 122.82˚ 98˚ 42N 54N 82˚ 42N α θ θ

32. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-32 Find the force required to keep a 2500-lb car parked on a hill that makes a 12° angle with the horizontal. 7.5 Example 2 Finding a Required Force (page 345) The vertical force BA represents the force of gravity. Vector BC represents the force with which the weight pushes against the hill. Vector BF represents the force that would pull the car up the hill. Since vectors BF and AC are equal, gives the magnitude of the required force.

33. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-33 7.5 Example 2 Finding a Required Force (cont.) Vectors BF and AC are parallel, so the measure of angle EBD equals the measure of angle A. Since angle BDE and angle C are right angles, triangles CBA and DEB have two corresponding angles that are equal and, thus, are similar triangles. Therefore, the measure of angle ABC equals the measure of angle E, which is 12°. From right triangle ABC, A force of approximately 520 lb will keep the car parked on the hill.

34. Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-34 A force of 18.0 lb is required to hold a 74.0-lb crate on a ramp. What angle does the ramp make with the horizontal? 7.5 Example 3 Finding an Incline Angle (page 345) Vector BF represents the force required to hold the crate on the incline. In right triangle ABC, the measure of angle B equals θ, the magnitude of vector BA represents the weight of the crate, and vector AC equals vector BF. The ramp makes an angle of about 14.1° with the horizontal.

35. port Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-35 A ship leaves port on a bearing of 25.0° and travels 61.4 km. The ship then turns due east and travels 84.6 km. How far is the ship from port? What is its bearing from port? 7.5 Example 4 Applying Vectors to a Navigation Problem (page 346) 123.76 km 84.6 2 = 61.4 2 +123.76 2 – 261.4123.76cosθ θ ≈ 38.28˚ The ship is about 123.8 km from port at a bearing of 63.28˚. 25˚ 61.4km 84.6km 65˚ 115˚ D2D2 D θ D

36. Airspeed 355 mph Wind 28.5 mph Copyright © 2008 Pearson Addison-Wesley. All rights reserved. 7-36 A plane with an airspeed of 355 mph is headed on a bearing of 62°. A west wind is blowing (from west to east) at 28.5 mph. Find the groundspeed and the actual bearing of the plane. 7.5 Example 5 Applying Vectors to a Navigation Problem (page 347) 28.5 2 = 355 2 + 380.4 2 – 2355380.4cos θ θ ≈ 2.0˚ The plane’s groundspeed is about 380.4 mph. The bearing is about 64°. 62˚ 28˚ 152˚ Groundspeed = G G2G2 G ≈ 380.4 mph θ