1. Slide 7.1 - 1 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley

2. OBJECTIVES Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley The Law of Sines Learn the vocabulary and conventions associated with solving triangles. Learn the statement and the derivation of the Law of Sines. Learn to solve AAS and ASA triangles by using the Law of Sines. Learn to solve for possible triangles in the ambiguous SSA case. SECTION 7.1 1 2 3 4

3. Slide 7.1 - 3 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley LAW OF SINES The following diagrams illustrate the Law of Sines.

4. Slide 7.1 - 4 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley LAW OF SINES In any triangle ABC, with sides of length a, b, and c, We can rewrite these relations in compact notation:

5. Slide 7.1 - 5 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley SOLVING AAS AND ASA TRIANGLES Step 1:Find the third angle. Find the measure of the third angle by subtracting the measures of the known angles from 180º. Step 2:Make a chart. Make a chart of the six parts of the triangle, indicating the known parts and the parts to be computed. Make a sketch of the triangle.

6. Slide 7.1 - 6 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley SOLVING AAS AND ASA TRIANGLES Step 3:Apply the Law of Sines. Select two ratios of the Law of Sines in which three of the four quantities are known. Solve for the fourth quantity. Use the form of the Law of Sines in which the unknown quantity is in the numerator. Step 4:Show the solution. Show the solution by completing the chart.

7. Slide 7.1 - 7 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 2 Height of Mount Everest From a point on a level plain at the foot of the mountain, a surveyor finds the angle of elevation of the peak of Mount Everest to be 20º. She walks 3465 meters closer (on a direct line between the first point and the base of the mountain) and finds the angle of elevation to be 23º. Estimate the height of the peak of Mount Everest to the nearest meter.

8. Slide 7.1 - 8 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 2 Height of Mount Everest Solution

9. Slide 7.1 - 9 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution continued EXAMPLE 2 Height of Mount Everest Height of Mount Everest is about 8848 meters.

10. Slide 7.1 - 10 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley SOLVING SSA TRIANGLES (AMBGUOUS CASE) If the lengths of two sides of a triangle and the measure of the angle opposite one of theses sides are given, then depending on the measurements, there may be 1. No such triangle 2. One such triangle 3. Two such triangles For this reason, Case 2 is called the ambiguous case.

11. Slide 7.1 - 11 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley SOLVING SSA TRIANGLES (AMBGUOUS CASE) A is an acute angle.

12. Slide 7.1 - 12 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley SOLVING SSA TRIANGLES (AMBGUOUS CASE) A is an acute angle.

13. Slide 7.1 - 13 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley SOLVING SSA TRIANGLES (AMBGUOUS CASE) A is an obtuse angle.

14. Slide 7.1 - 14 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley SOLVING SSA TRIANGLES Step 1:Make a chart. Make a chart of the six parts of the triangle, indicating the known parts and the parts to be computed. Step 2:Apply the Law of Sines to the two ratios of in which three of the four quantities are known. Solve for the sine of the fourth quantity. Use the form of the Law of Sines in which the unknown quantity is in the numerator.

15. Slide 7.1 - 15 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley SOLVING SSA TRIANGLES Step 3:If the sine of the angle, say , in Step 2 is greater than 1, there is no triangle with the given measurements. If sin  in Step 2 is between 0 and 1, go to Step 4. Step 4:Suppose the sine of the angle in Step 2 is x, with 0 ≤ x ≤ 1. If x ≠ 1, there are two possible values for the angle: (i)   = sin -1 x with 0 <   < 90º and (ii)   = 180º –  

16. Slide 7.1 - 16 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley SOLVING SSA TRIANGLES Step 5:If x ≠ 1, with (known angle) +  1 < 180º and (known angle) +  2 < 180º, then there are two triangles. Otherwise, there is only one triangle, and if x = 1, it is a right triangle. Step 6:Find the third angle of the triangle(s). Step 7:Use the Law of Sines to find the remaining side(s). Step 8:Show solution(s).

17. Slide 7.1 - 17 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 3 Solving an SSA Triangle (No Solution) Solve triangle ABC with A = 50º, a = 8 inches, and b = 15 inches. Solution Step 1 Make a chart. Step 2 Apply the Law of Sines. c = ?C = ? b = 15B = ? a = 8A = 50º

18. Slide 7.1 - 18 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution continued EXAMPLE 3 Solving an SSA Triangle (No Solution) Step 3Since sin B ≈ 1.44 > 1 and the range of the sine function is [–1, 1], we conclude that no triangle exists that is compatible with the given information.

19. Slide 7.1 - 19 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 5 Solving an SSA Triangle (Two Solutions) Solve triangle ABC with A = 47º, a = 11 meters, and c = 13 meters. Round each answer to the nearest tenth. Solution Step 1 Make a chart. Step 2 Apply the Law of Sines. c = 13C = ? b = ?B = ? a = 11A = 47º

20. Slide 7.1 - 20 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution continued EXAMPLE 5 Solving an SSA Triangle (Two Solutions) Step 4Two possible values of C with sin C ≈ 0.864 are C 1 ≈ sin -1 (0.864) = 59.8º and C 2 ≈ 180º – 59.8º = 120.2º

21. Slide 7.1 - 21 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution continued EXAMPLE 5 Solving an SSA Triangle (Two Solutions) Step 5A + C 1 = 47º + 59.8º = 106.8º < 180º, A + C 2 = 47º + 120.2º = 167.2º < 180º. There are two triangles with the given measurements. Step 6

22. Slide 7.1 - 22 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution continued EXAMPLE 5 Solving an SSA Triangle (Two Solutions) Step 7

23. Slide 7.1 - 23 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution continued EXAMPLE 5 Solving an SSA Triangle (Two Solutions) Step 8 c = 13C 2 ≈ 120.2º b ≈ 3.3 ≈ 12.8º a = 11A = 47º c = 13C 1 ≈ 59.8º b 1 ≈ 14.4 ≈ 73.2º a = 11A = 47º

24. Slide 7.1 - 24 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley BEARINGS A bearing is the measure of an acute angle from due north or due south. N 40º EN 60º W

25. Slide 7.1 - 25 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley BEARINGS S 30º WS 75º E

26. Slide 7.1 - 26 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 6 Navigating Using Bearings A ship sailing due west at 20 miles per hour records the bearing of an oil rig at N 55.4º W. An hour and a half later the bearing of the same rig is N 66.8º E. a. How far is the ship from the oil rig the second time? b. How close did the ship pass to the oil rig?

27. Slide 7.1 - 27 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley EXAMPLE 6 Navigating Using Bearings Solution a.In an hour and a half the ship travels (1.5)(20) = 30 miles due west. The oil rig (C), the starting point for the ship (A), and the position of the ship after an hour and a half (B) form the vertices of the triangle ABC.

28. Slide 7.1 - 28 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution continued ThenA = 90º – 55.4º = 34.6º and The ship is approximately 20 miles from the oil rig when the second bearing is taken. EXAMPLE 6 Navigating Using Bearings B = 90º – 66.8º = 23.2º. Thus C = 180º – 34.6º – 23.2º = 122.2º.

29. Slide 7.1 - 29 Copyright © 2008 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Solution continued b.The shortest distance between the ship and the oil rig is the length of the segment h in triangle ABC. EXAMPLE 6 Navigating Using Bearings The ship passes within 7.9 miles of the oil rig.