1. Law of Sines 8.1

2. 2  Use the Law of Sines to solve oblique triangles (AAS or ASA).  Use the Law of Sines to solve oblique triangles (SSA).  Find the areas of oblique triangles.  Use the Law of Sines to model and solve real-life problems. Objectives

3. 3 Introduction

4. 4 In this section, you will solve oblique triangles—triangles that have no right angles. As standard notation, the angles of a triangle are labeled A, B, and C, and their opposite sides are labeled a, b, and c, as shown on the right. To solve an oblique triangle, you need to know the measure of at least one side and any two other measures of the triangle—either two sides, two angles, or one angle and one side.

5. 5 Introduction This breaks down into the following four cases. 1. Two angles and any side (AAS or ASA) 2. Two sides and an angle opposite one of them (SSA) 3. Three sides (SSS) 4. Two sides and their included angle (SAS) The first two cases can be solved using the Law of Sines, whereas the last two cases require the Law of Cosines.

6. 6 Derive: Law of Sines Step 1: Draw ΔABC as shown with height h. Notice that side a is opposite ∠ A, b is opposite ∠ B, and c is opposite ∠ C.

7. 7 Derive: Law of Sines Step 2: Write an equation for h using sin( ∠ A). Step 3: Write an equation for h using sin( ∠ B).

8. 8 Derive: Law of Sines Step 4: Use substitution to combine the two previous equations and write your final equation as a proportion.

9. 9 Derive: Law of Sines Step 5: Write an equation for k using sin( ∠ B). Step 6: Write an equation for k using sin( ∠ C).

10. 10 Derive: Law of Sines Step 7: Use substitution to combine the two previous equations and write your final equation as a proportion.

11. 11 Derive: Law of Sines Step 8: Finally, use the transitive property to combine the proportion from S4 and S7. This is the Law of Sines.

12. 12 Law of Sines If ΔABC has side lengths a, b, and c, representing the lengths of the sides opposite the angles with measures A, B, and C, then; Looking for an ANGLE: Looking for a SIDE:

13. 13 Solving a Triangle The Law of Sines can be used to “solve a triangle,” which means to find the measures of all of the angles and all of the sides of a triangle. You can use the Law of Sines to solve a triangle if you are given: two angle measures and any side length (ASA or AAS) or two side lengths and a non-included angle measure (SSA). Minimum Requirements: one angle and the side opposite it.

14. 14 Example Law of Sines (AAS) Find p. Round to the nearest tenth. We are given measures of two angles and a nonincluded side, so use the Law of Sines to write a proportion.

15. 15 Example Law of Sines (AAS) Law of Sines Use a calculator. Divide each side by sin Cross Products Property Answer: p ≈ 4.8

16. 16 Your Turn – Given Two Angles and One Side—AAS For the triangle in Figure 3.1, C = 102 , B = 29 , and b = 28 feet. Find the remaining angle and sides. Figure 3.1

17. 17 Solution The third angle of the triangle is A = 180  – B – C = 180  – 29  – 102  = 49 . By the Law of Sines, you have.

18. 18 Solution Using b = 28 produces and cont’d

19. 19 A.A B.B C.C D.D Your Turn: A.4.6 B.29.9 C.7.8 D.8.5 Find c to the nearest tenth.

20. 20 Example Find AC in Δ ABC. Law of Sines (ASA)

21. 21 Example Law of Sines (ASA) First, use the Triangle Angle Sum Theorem to find the measure of the third angle. Find AC. Law of Sines Cross Products Property Divide each side by sin102. Use a calculator. Answer: AC ≈ 11.2 ft.

22. 22 The Ambiguous Case (SSA)

23. 23 The Ambiguous Case (SSA) Earlier, you saw that two angles and one side determine a unique triangle. However, if two sides and one opposite angle are given, three possible situations can occur: (1) no such triangle exists, (2) one such triangle exists, or (3) two distinct triangles may satisfy the conditions.

24. 24 The Ambiguous Case (SSA)

25. 25 Example – Single-Solution Case —SSA For the triangle in Figure 3.4, a = 22 inches, b = 12 inches, and A = 42 . Find the remaining side and angles. One solution: a  b Figure 3.4

26. 26 Example – Solution By the Law of Sines, you have Reciprocal form Multiply each side by b. Substitute for A, a, and b.

27. 27 Example – Solution Now, you can determine that C  180  – 42  – 21.41  = 116.59 . Then, the remaining side is cont’d

28. 28 Example Law of Sines (SSA) Find x. Round to the nearest degree.

29. 29 Example Law of Sines (SSA) Law of Sines m  B = 50, b = 10, a = 11 Cross Products Property Divide each side by 10. Use a calculator. Use the inverse sine ratio. Answer: x ≈ 57.4 ˚

30. 30 A.A B.B C.C D.D Your Turn: A.39 B.43 C.46 D.49 Find x. Round to the nearest degree.

31. 31 Your Turn: Find the indicated measure. 1. 2.

32. 32 Area of an Oblique Triangle

33. 33 Area of an Oblique Triangle The procedure used to prove the Law of Sines leads to a simple formula for the area of an oblique triangle. Referring to triangles below, note that each triangle has a height of h = b sin A. A is acute.A is obtuse.

34. 34 Area of an Oblique Triangle Consequently, the area of each triangle is Area = (base)(height) = (c)(b sin A) = bc sin A. By similar arguments, you can develop the formulas Area = ab sin C = ac sin B.

35. 35 Area of an Oblique Triangle Note that when angle A is 90 , the formula gives the area of a right triangle: Area = bc(sin 90  ) = bc = (base)(height). Similar results are obtained for angles C and B equal to 90 . sin 90  = 1

36. 36 c ab Area of triangle C AB ba c Must be the included angle

37. 37 Example – Finding the Area of a Triangle Find the area of triangle ABC shown in Figure 17. Solution: The triangle has sides of length 10 cm and 3 cm, with included angle 120 . Figure 17

38. 38 Solution Therefore sin  (3) sin 120  = 15 sin 60   13 cm 2 Reference angle

39. 39 Your Turn: Calculate the area of the triangle shown. Give your answer correct to one decimal place. Area of triangle = absin C 1212 Area = (3)(4) sin 55  1212 = 4·9149… 4 cm 3 cm C must be the included angle = 4·9 cm 2 55º

40. 40 Find the area of triangle abc, correct to the nearest whole number. Area of triangle = absin C 1212 Area = (14)(18·4) sin 70  1212 = 121·0324… 18·4 14 c ab 44º 66º C must be the included angle |  abc | = 180  – 44  – 66  = 70  70º = 121units 2

41. 41 Your Turn – Finding the Area of a Triangular Lot Find the area of a triangular lot having two sides of lengths 90 meters and 52 meters and an included angle of 102 . Solution: Consider a = 90 meters, b = 52 meters, and angle C = 102 , as shown in Figure 3.6. Then, the area of the triangle is Area = ab sin C = (90)(52)(sin 102  )  2289 square meters. Figure 3.6

42. 42 Application

43. 43 Example – An Application of the Law of Sines The course for a boat race starts at point A and proceeds in the direction S 52  W to point B, then in the direction S 40  E to point C, and finally back to A, as shown in Figure 3.7. Point C lies 8 kilometers directly south of point A. Approximate the total distance of the race course. Figure 3.7

44. 44 Solution Because lines BD and AC are parallel, it follows that BCA  CBD. Consequently, triangle ABC has the measures shown in Figure 3.8. The measure of angle B is 180  – 52  – 40  = 88 . Using the Law of Sines, Figure 3.8

45. 45 Solution Because b = 8, and The total distance of the course is approximately Length  8 + 6.31 + 5.15 =19.46 kilometers. cont’d

46. 46 The Sine Rule 25 o 15 m A D The angle of elevation of the top of a building measured from point A is 25 o. At point D which is 15m closer to the building, the angle of elevation is 35 o Calculate the height of the building. T B Angle TDA = 145 o Angle DTA = 10 o 35 o 36.5 180 – 35 = 145 o 180 – 170 = 10 o Example:

47. 47 A The angle of elevation of the top of a column measured from point A, is 20 o. The angle of elevation of the top of the statue is 25 o. Find the height of the statue when the measurements are taken 50 m from its base 50 m Angle BCA = 70 o Angle ACT = Angle ATC = 110 o 65 o 53.21 m B T C 180 – 110 = 70 o 180 – 70 = 110 o 180 – 115 = 65 o 20 o 25 o 5o5o Your Turn:

48. 48 Example A civil engineer wants to determine the distances from points A and B to an inaccessible point C, as shown. From direct measurement, the engineer knows that AB = 25m,  A = 110 o, and  B = 20 o. Find AC and BC. A B C

49. 49 Example 1.Mark the drawing with known info 2.  C = 180 – (110 + 20) = 50 o 3. 4. A B C 25 110 o 20 o

50. 50 Your Turn: A forest ranger at an observation point (A) sights a fire in the direction 32° east of north. Another ranger at a second observation point (B), 10 miles due east of A, sights the same fire 48° west of north. Find the distance from each observation point to the fire. 10 32 o 48 o 58 o 42 o A B 80 o 8.611 6.795

51. 51 Section 8.1 Notes Worksheet Section 8.1 pg. 598 – 600; #1 – 23 odd, 29 – 43 odd Homework