1. Trigonometric Functions of Angles 6

2. The Law of Sines 6.4

3. Introduction In Section 6-2, we used the trigonometric ratios to solve right triangles. The trigonometric functions can also be used to solve oblique triangles—triangles with no right angles. To do this, we first study the Law of Sines here and then the Law of Cosines in the next section.

4. Introduction To state these laws (or formulas) more easily, we follow the convention of labeling: The angles of a triangle as A, B, C. The lengths of the corresponding opposite sides as a, b, c.

5. Determining a Triangle To solve a triangle, we need to know certain information about its sides and angles. To decide whether we have enough information, it’s often helpful to make a sketch.

6. Two Angles and Included Side For instance, if we are given two angles and the included side, then it’s clear that one and only one triangle can be formed.

7. Two Sides and Included Angle Similarly, if two sides and the included angle are known, then a unique triangle is determined.

8. Three Angles and No Sides However, if we know all three angles and no sides, we cannot uniquely determine the triangle. Many triangles can have the same three angles. All these triangles would be similar, of course. So, we won’t consider this case.

9. Determining a Triangle In general, a triangle is determined by three of its six parts (angles and sides) as long as at least one of these three parts is a side. So, the possibilities are as follows.

10. Determining a Triangle CaseAngles and SidesAbbrvn. 1One side and two anglesASA/SAA 2 Two sides and the angle opposite one of those sides SSA 3Two sides and the included angleSAS 4Three sidesSSS

11. Determining a Triangle Cases 1 and 2 are solved using the Law of Sines. Cases 3 and 4 require the Law of Cosines.

12. The Law of Sines

13. Law of Sines The law says that, in any triangle, the lengths of the sides are proportional to the sines of the corresponding opposite angles. In triangle ABC, we have:

14. Law of Sines—Proof To see why the Law of Sines is true, refer to the figure. By the formula in Section 6-3, the area of triangle ABC is ½ab sin C. By the same formula, the area of this triangle is also ½ac sin B and ½bc sin A.

15. Law of Sines—Proof Thus, ½bc sin A = ½ac sin B = ½ab sin C Multiplying by 2/(abc) gives the Law of Sines.

16. E.g. 1—Tracking a Satellite (ASA) A satellite orbiting the earth passes directly overhead at observation stations in Phoenix and Los Angeles, 340 mi apart. At an instant when it is between these two stations, its angle of elevation is simultaneously observed to be 60° at Phoenix and 75° at Los Angeles.

17. E.g. 1—Tracking a Satellite (ASA) How far is the satellite from Los Angeles? In other words, find the distance AC in the figure.

18. E.g. 1—Tracking a Satellite (ASA) Whenever two angles in a triangle are known, the third can be determined immediately as the sum of the angles of a triangle is 180°. In this case,

19. E.g. 1—Tracking a Satellite (ASA) So, we have: The distance of the satellite from Los Angeles is approximately 416 mi.

20. E.g. 2—Solving a Triangle (SAA) Solve the triangle in the figure. First,

21. E.g. 2—Solving a Triangle (SAA) Since side c is known, to find side a, we use the relation:

22. E.g. 2—Solving a Triangle (SAA) Similarly, to find b, we use:

23. The Ambiguous Case

24. In Examples 1 and 2, a unique triangle was determined by the information given. This is always true of Case 1 (ASA or SAA).

25. The Ambiguous Case However, in Case 2 (SSA), there may be two triangles, one triangle, or no triangle with the given properties. For this reason, Case 2 is sometimes called the ambiguous case.

26. The Ambiguous Case To see why this is so, here we show the possibilities when angle A and sides a and b are given.

27. The Ambiguous Case In part (a), no solution is possible. Side a is too short to complete the triangle. In part (b), the solution is a right triangle.

28. The Ambiguous Case In part (c), two solutions are possible. In part (d), there is a unique triangle with the given properties. We illustrate the possibilities of Case 2 in the following examples.

29. E.g. 3—SSA, the One-Solution Case Solve triangle ABC, where: b = 7

30. E.g. 3—SSA, the One-Solution Case We first sketch the triangle with the information we have. Our sketch is necessarily tentative—since we don’t yet know the other angles. Nevertheless, we can now see the possibilities.

31. E.g. 3—SSA, the One-Solution Case We first find.

32. E.g. 3—SSA, the One-Solution Case Which angles B have sin B = ½? From Section 6-3, we know that there are two such angles smaller than 180°. They are 30° and 150°.

33. E.g. 3—SSA, the One-Solution Case Which of these angles is compatible with what we know about triangle ABC? Since, we cannot have, because 45° + 150° > 180°.

34. E.g. 3—SSA, the One-Solution Case Thus, The remaining angle is:

35. E.g. 3—SSA, the One-Solution Case Now, we can find side c.

36. The Ambiguous Case In Example 3, there were two possibilities for angle B. One of these was not compatible with the rest of the information.

37. The Ambiguous Case In general, if sin A < 1, we must check the angle and its supplement as possibilities. This is because any angle smaller than 180° can be in the triangle. To decide whether either possibility works, we check to see whether the resulting sum of the angles exceeds 180°.

38. The Ambiguous Case It can happen that both possibilities are compatible with the given information. In that case, two different triangles are solutions to the problem.

39. E.g. 4—SSA, the Two-Solution Case Solve triangle ABC if: a = 186.2 b = 248.6

40. E.g. 4—SSA, the Two-Solution Case From the given information, we sketch the triangle shown. Note that side a may be drawn in two possible positions to complete the triangle.

41. E.g. 4—SSA, the Two-Solution Case From the Law of Sines,

42. E.g. 4—SSA, the Two-Solution Case There are two possible angles B between 0° and 180° such that sin B = 0.91225. Using the SIN -1 key on a calculator (or INV SIN or ARCSIN ), we find one of these angles is approximately 65.8°. The other is approximately 180° – 65.8° = 114.2°.

43. E.g. 4—SSA, the Two-Solution Case We denote these two angles by B 1 and B 2 so that: Thus, two triangles satisfy the given conditions: triangle A 1 B 1 C 1 and triangle A 2 B 2 C 2.

44. E.g. 4—SSA, the Two-Solution Case Solve triangle A 1 B 1 C 1 : Thus,

45. E.g. 4—SSA, the Two-Solution Case Solve triangle A 2 B 2 C 2 : Thus,

46. The No-Solution Case The next example presents a situation for which no triangle is compatible with the given data.

47. E.g. 5—SSA, the No-Solution Case Solve triangle ABC, where: b = 122

48. E.g. 5—SSA, the No-Solution Case To organize the given information, we sketch this diagram. Let’s try to find.

49. E.g. 5—SSA, the No-Solution Case The sine of an angle is never greater than 1. Thus, we conclude that no triangle satisfies the conditions given in this problem.