Law of Sines

Chapter 6 6.1 Additional Topics in Trigonometry Law of Sines What you should learn • 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. Introduction In Chapter 4, you studied techniques for solving right triangles. In this section and the next, 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 in Figure 6.1. C Why you should learn it You can use the Law of Sines to solve real-life problems involving oblique triangles. For instance, in Exercise 44 on page 438, you can use the Law of Sines to determine the length of the shadow of the Leaning Tower of Pisa. a b A B c FIGURE 6.1 To solve an oblique triangle, you need to know the measure of at least one side and any two other parts of the triangle—either two sides, two angles, or one angle and one side. This breaks down into the following four cases. 1. 2. 3. 4. Two angles and any side (AAS or ASA) Two sides and an angle opposite one of them (SSA) Three sides (SSS) 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 (see Section 6.2). Law of Sines If ABC is a triangle with sides a, b, and c, then a b c

. sin A sin B sin C Hideo Kurihara/Getty Images C b A C a h c A is acute. h B b A a c B A is obtuse. The Law of Sines can also be written in the reciprocal form sin A sin B sin C

. a b c For a proof of the Law of Sines, see Proofs in Mathematics on page 489. Section 6.1 C Example 1 b = 27.4 ft a 102.3° 28.7° FIGURE Given Two Angles and One Side—AAS For the triangle in Figure 6.2, C
102.3, B
28.7, and b
27.4 feet. Find the remaining angle and sides. c A Law of Sines B Solution The third angle of the triangle is 6.2 A
180  B  C
180  28.7  102.3
49.0. By the Law of Sines, you have a b c

. sin A sin B sin C Using b
27.4 produces When solving triangles, a careful sketch is useful as a quick test for the feasibility of an answer. Remember that the longest side lies opposite the largest angle, and the shortest side lies opposite the smallest angle. a
b 27.4
sin A

sin 49.0  43.06 feet sin B sin 28.7 c
b 27.4
sin C

sin 102.3  55.75 feet. sin B sin 28.7 and Now try Exercise 1. Example 2 Given Two Angles and One Side—ASA A pole tilts toward the sun at an 8 angle from the vertical, and it casts a 22-foot shadow. The angle of elevation from the tip of the shadow to the top of the pole is 43. How tall is the pole? Solution From Figure 6.3, note that A
43 and B
90  8
98. So, the third angle is C
180  A  B C
180  43  98
39. b a By the Law of Sines, you have a c
. sin A sin C 8° 43° B FIGURE 6.3 c = 22 ft A Because c
22 feet, the length of the pole is c 22 a

sin A

sin 43  23.84 feet. sin C sin 39 Now try Exercise 35. For practice, try reworking Example 2 for a pole that tilts away from the sun under the same conditions. Chapter 6 Additional Topics in Trigonometry The Ambiguous Case (SSA) In Examples 1 and 2 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. The Ambiguous Case (SSA) Consider a triangle in which you are given a, b, and A. A is acute. A is acute.
h
b sin A A is acute. A is acute. A is obtuse. Sketch b h b a h a A A b b a a h a a h b A A A is obtuse. a b A A Necessary condition a < h a
h a > b h < a < b a ≤ b a > b Triangles possible None One One Two None One Example 3 For the triangle in Figure 6.4, a
22 inches, b
12 inches, and A
42. Find the remaining side and angles. C a = 22 in. b = 12 in. Solution 42° A Single-Solution Case—SSA c One solution: a > b FIGURE 6.4 B By the Law of Sines, you have sin B sin A
b a sin B
b  sin B
12 sin A a  Reciprocal form  sin 42 22 Multiply each side by b.  B  21.41. Substitute for A, a, and b. B is acute. Now, you can determine that C  180  42  21.41
116.59. Then, the remaining side is c a
sin C sin A Encourage your students to sketch the triangle, keeping in mind that the longest side lies opposite the largest angle of the triangle. For practice, suggest that students also find h. c
a 22
sin C

sin 116.59  29.40 inches. sin A sin 42 Now try Exercise 19. Section 6.1 No-Solution Case—SSA Example 4 a = 15 b = 25 Law of Sines Show that there is no triangle for which a
15, b
25, and A
85. h Solution 85° Begin by making the sketch shown in Figure 6.5. From this figure it appears that no triangle is formed. You can verify this using the Law of Sines. A No solution: a < h FIGURE 6.5 sin B sin A
b a Reciprocal form sina A sin 85 sin B
25  1.660 > 1 15  sin B
b  Multiply each side by b.  This contradicts the fact that sin B ≤ 1. So, no triangle can be formed having sides a
15 and b
25 and an angle of A
85. Now try Exercise 21. Activities Have your students determine the number of triangles possible in each of the following cases. 1. A
62, a
10, b
12 (0 triangles) 2. A
98, a
10, b
3 (1 triangle) 3. A
54, a
7, b
10 (0 triangles) Discuss several examples of the twosolution case. Additional Example Find two triangles for which c
29, b
46, and C
31. Solution B
54.8, A
94.2, a
56.2 B
125.2, A
23.8, a
22.7 Example 5 Two-Solution Case—SSA Find two triangles for which a
12 meters, b
31 meters, and A
20.5. Solution By the Law of Sines, you have sin B sin A
b a sin B
b  Reciprocal form    sin A sin 20.5
31  0.9047. a 12 There are two angles B1  64.8 and B2  180  64.8
115.2 between 0 and 180 whose sine is 0.9047. For B1  64.8, you obtain C  180  20.5  64.8
94.7 c
a 12
sin C

sin 94.7  34.15 meters. sin A sin 20.5 For B2  115.2, you obtain C  180  20.5  115.2
44.3 c
a 12
sin C

sin 44.3  23.93 meters. sin A sin 20.5 The resulting triangles are shown in Figure 6.6. b = 31 m 20.5° A FIGURE b = 31 m a = 12 m 64.8° B1 6.6 Now try Exercise 23. A 20.5° 115.2° B2 a = 12 m Chapter 6 Additional Topics in Trigonometry Area of an Oblique Triangle To see how to obtain the height of the obtuse triangle in Figure 6.7, notice the use of the reference angle 180  A and the difference formula for sine, as follows. h
b sin
180  A The procedure used to prove the Law of Sines leads to a simple formula for the area of an oblique triangle. Referring to Figure 6.7, note that each triangle has a height of h
b sin A. Consequently, the area of each triangle is Area
By similar arguments, you can develop the formulas
b
sin 180 cos A 1 1 Area
ab sin C
ac sin B. 2 2  cos 180 sin A
b0  cos A 
1 1 1 1
base
height

c
b sin A
bc sin A. 2 2 2 C  sin A C
b sin A a b Activities 1. Use the given information to find (if possible) the remaining side and angles of the oblique triangle. If two solutions exist, find both. A
58, a
20, c
10 Answer: B
97, C
25, b
23.4 2. Use the given information to find (if possible) the remaining side and angles of the oblique triangle. If two solutions exist, find both. B
78, b
207, c
210 Answer: Two solutions A
19.1, a
69.2, C
82.9 A
4.9, a
18.1, C
97.1 3. Find the area of the triangle with B
120, a
32, and c
50. Answer: Area
692.8 square units h A c A c B A is obtuse Area of an Oblique Triangle The area of any triangle is one-half the product of the lengths of two sides times the sine of their included angle. That is, 1 1 1 Area
bc sin A
ab sin C
ac sin B. 2 2 2 Note that if angle A is 90, the formula gives the area for a right triangle: Area
1 1 1 bc
sin 90
bc

base
height. 2 2 2 sin 90
1 Similar results are obtained for angles C and B equal to 90. 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. b = 52 m 102° C 6.8 a b B A is acute FIGURE 6.7 Example 6 FIGURE h Solution a = 90 m Consider a
90 meters, b
52 meters, and angle C
102, as shown in Figure 6.8. Then, the area of the triangle is 1 1 Area
ab sin C

90
52
sin 102  2289 square meters. 2 2 Now try Exercise 29. Section 6.1 N A W Law of Sines Application E S Example 7 52° B 8 km 40° An Application of the Law of Sines The course for a boat race starts at point A in Figure 6.9 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. Point C lies 8 kilometers directly south of point A. Approximate the total distance of the race course. Solution C D FIGURE 6.9 a b c

sin 52 sin 88 sin 40 A c b = 8 km a you can let b
8 and obtain 52° B 40° C FIGURE Because lines BD and AC are parallel, it follows that
BCA
DBC. Consequently, triangle ABC has the measures shown in Figure 6.10. For angle B, you have B
180  52  40
88. Using the Law of Sines 6.10 a
8
sin 52  6.308 sin 88 c
8
sin 40  5.145. sin 88 and The total length of the course is approximately Length  8  6.308  5.145
19.453 kilometers. Now try Exercise 39. Alternative Writing About Mathematics: Error Analysis You are a math instructor, and one of your students hands in the following solution. Discuss what is wrong with your student’s solution. How could you help the student avoid making a similar mistake in the future? Use a diagram to illustrate your explanation. Find side c in a triangle that has a
5.8, b
7, and C
82. Student’s Solution Because this is an SSA situation, I can use the Law of Sines. 5.8 c
7 sin 82 5.8
sin 82
0.821 7 W RITING ABOUT MATHEMATICS Using the Law of Sines In this section, you have been using the Law of Sines to solve oblique triangles. Can the Law of Sines also be used to solve a right triangle? If so, write a short paragraph explaining how to use the Law of Sines to solve each triangle. Is there an easier way to solve these triangles? a.
AAS b.
ASA B B 50° C 50° c = 20 a = 10 A C A Chapter 6 6.1 Additional Topics in Trigonometry Exercises Exercises with no solution: 20, 21, 24 Exercises with two solutions: 6, 23 VOCABULARY CHECK: Fill in the blanks. 1. An ________ triangle is a triangle that has no right angle. 2. For triangle ABC, the Law of Sines is given by a c
________
. sin A sin C 1 1 3. The area of an oblique triangle is given by 2 bc sin A
2ab sin C
________ . In Exercises 1–18, use the Law of Sines to solve the triangle. Round your answers to two decimal places. 1. b a = 20 45° 30° c 2. a B c = 20 3. C 25° 35° c 4. B a 10° A B c = 45 5. A
36, a
8, b
5 9. A
83 20, 10. A
5 40, 11. B
15 30, 12. B
2 45, a
125, b
100 a
125, b
200 21. A
76, a
18, b
20 22. A
76, a
34, b
21 23. A
58, a
11.4, 24. A
58, a
4.5, b
12.8 b
12.8 25. A
36, a
5 C
16.7, a
21.6 C
54.6, c
2.68 C
54.6, B
8 15, a
4.5, b
6.2, 13. C
145, b
4, 27. A
10, a
10.8 28. A
88, a
315.6 In Exercises 29–34, find the area of the triangle having the indicated angle and sides. 6. A
60, a
9, c
10 8. A
24.3, 5 a
38 26. A
60, a
10 135° 7. A
102.4, C
104, c
50 c
34 In Exercises 25–28, find values for b such that the triangle has (a) one solution, (b) two solutions, and (c) no solution. C b 18. B
28, b
16 a = 3.5 b A B
42, 20. A
110, 40° A a
35, 17. A
55, 19. A
110, 105° c
10 a
48, In Exercises 19–24, use the Law of Sines to solve (if possible) the triangle. If two solutions exist, find both. Round your answers to two decimal places. B C b a
125, 15. A
110 15, 16. C
85 20, C A 14. A
100, c
18.1 b
4.8 b
6.8 c
5.8 c
14 29. C
120, a
4, 30. B
130, a
62, 31. A
43 45, 32. A
5 15, b
6 c
20 b
57, c
85 b
4.5, c
22 33. B
72 30, a
105, c
64 34. C
84 30, a
16, b
20 Section 6.1 35. Height Because of prevailing winds, a tree grew so that it was leaning 4 from the vertical. At a point 35 meters from the tree, the angle of elevation to the top of the tree is 23 (see figure). Find the height h of the tree. Law of Sines 39. Bridge Design A bridge is to be built across a small lake from a gazebo to a dock (see figure). The bearing from the gazebo to the dock is S 41 W. From a tree 100 meters from the gazebo, the bearings to the gazebo and the dock are S 74 E and S 28 E, respectively. Find the distance from the gazebo to the dock. N h 94° Tree 100 m 74° Gazebo 41° 35 m 36. Height A flagpole at a right angle to the horizontal is located on a slope that makes an angle of 12 with the horizontal. The flagpole’s shadow is 16 meters long and points directly up the slope. The angle of elevation from the tip of the shadow to the sun is 20. (a) Draw a triangle that represents the problem. Show the known quantities on the triangle and use a variable to indicate the height of the flagpole. (b) Write an equation involving the unknown quantity. (c) Find the height of the flagpole. 37. Angle of Elevation A 10-meter telephone pole casts a 17-meter shadow directly down a slope when the angle of elevation of the sun is 42 (see figure). Find , the angle of elevation of the ground. A 10 m B 42° − θ m θ 17 E S Canton 40. Railroad Track Design The circular arc of a railroad curve has a chord of length 3000 feet and a central angle of 40. (a) Draw a diagram that visually represents the problem. Show the known quantities on the diagram and use the variables r and s to represent the radius of the arc and the length of the arc, respectively. (b) Find the radius r of the circular arc. (c) Find the length s of the circular arc. 41. Glide Path A pilot has just started on the glide path for landing at an airport with a runway of length 9000 feet. The angles of depression from the plane to the ends of the runway are 17.5 and 18.8. (b) Find the air distance the plane must travel until touching down on the near end of the runway. (c) Find the ground distance the plane must travel until touching down. N Elgin N Dock (a) Draw a diagram that visually represents the problem. C 38. Flight Path A plane flies 500 kilometers with a bearing of 316 from Naples to Elgin (see figure). The plane then flies 720 kilometers from Elgin to Canton. Find the bearing of the flight from Elgin to Canton. W E S 28° 23° 42° W 720 km 500 km (d) Find the altitude of the plane when the pilot begins the descent. 42. Locating a Fire The bearing from the Pine Knob fire tower to the Colt Station fire tower is N 65 E, and the two towers are 30 kilometers apart. A fire spotted by rangers in each tower has a bearing of N 80 E from Pine Knob and S 70 E from Colt Station (see figure). Find the distance of the fire from each tower. 44° Not drawn to scale Naples N W E Colt Station S 80° 65° 30 km Pine Knob 70° Fire Not drawn to scale Chapter 6 Additional Topics in Trigonometry 43. Distance A boat is sailing due east parallel to the shoreline at a speed of 10 miles per hour. At a given time, the bearing to the lighthouse is S 70 E, and 15 minutes later the bearing is S 63 E (see figure). The lighthouse is located at the shoreline. What is the distance from the boat to the shoreline? N 63° W 70° d E S Synthesis True or False? In Exercises 45 and 46, determine whether the statement is true or false. Justify your answer. 45. If a triangle contains an obtuse angle, then it must be oblique. 46. Two angles and one side of a triangle do not necessarily determine a unique triangle. 47. Graphical and Numerical Analysis are positive angles. (a) Write In the figure, and as a function of . (b) Use a graphing utility to graph the function. Determine its domain and range. (c) Use the result of part (a) to write c as a function of . Model It 44. Shadow Length The Leaning Tower of Pisa in Italy is characterized by its tilt. The tower leans because it was built on a layer of unstable soil—clay, sand, and water. The tower is approximately 58.36 meters tall from its foundation (see figure). The top of the tower leans about 5.45 meters off center. (d) Use a graphing utility to graph the function in part (c). Determine its domain and range. (e) Complete the table. What can you infer? 0.4 0.8 1.2 1.6 2.0 2.4 2.8 c 5.45 m 20 cm β α θ 2 58.36 m 18 α θ FIGURE FOR d Not drawn to scale (a) Find the angle of lean of the tower. (b) Write as a function of d and , where  is the angle of elevation to the sun. (c) Use the Law of Sines to write an equation for the length d of the shadow cast by the tower. (d) Use a graphing utility to complete the table.  10 20 30 40 50 60 8 cm γ c 47 9 β θ 30 cm FIGURE FOR 48 48. Graphical Analysis (a) Write the area A of the shaded region in the figure as a function of . (b) Use a graphing utility to graph the area function. (c) Determine the domain of the area function. Explain how the area of the region and the domain of the function would change if the eight-centimeter line segment were decreased in length. Skills Review d In Exercises 49–52, use the fundamental trigonometric identities to simplify the expression. 49. sin x cot x 51. 1  sin2 2  x 50. tan x cos x sec x 52. 1  cot2 2  x