additional topics in trigonometry 6sciannamath.weebly.com/uploads/1/6/9/2/16922340/chapter... ·...

The work done by a force, such aspushing and pulling objects, can becalculated using vector operations.

SELECTED APPLICATIONSTriangles and vectors have many real-life applications. The applications listed below represent a smallsample of the applications in this chapter.

• Bridge Design,Exercise 39, page 437

• Glide Path,Exercise 41, page 437

• Surveying,Exercise 31, page 444

• Paper Manufacturing,Exercise 45, page 445

• Cable Tension,Exercises 79 and 80, page 458

• Navigation,Exercise 84, page 459

• Revenue,Exercise 65, page 468

• Work,Exercise 73, page 469

6.1 Law of Sines

6.2 Law of Cosines

6.3 Vectors in the Plane

6.4 Vectors and Dot Products

6.5 Trigonometric Form of a Complex Number

Additional Topics in Trigonometry 66

© A

my

Etra

/Ph

oto

Edit

429

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IntroductionIn Chapter 4, you studied techniques for solving right triangles. In this sectionand the next, you will solve oblique triangles—triangles that have no rightangles. As standard notation, the angles of a triangle are labeled and andtheir opposite sides are labeled and as shown in Figure 6.1.

FIGURE 6.1

To solve an oblique triangle, you need to know the measure of at least oneside and any two other parts of the triangle—either two sides, two angles, or oneangle and one side. 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 twocases require the Law of Cosines (see Section 6.2).

The Law of Sines can also be written in the reciprocal form

For a proof of the Law of Sines, see Proofs in Mathematics on page 489.

sin A

a�

sin B

b�

sin C

c.

A B

a

c

b

C

c,b,a,C,B,A,

430 Chapter 6 Additional Topics in Trigonometry

What you should learn• Use the Law of Sines to solve

oblique triangles (AAS or ASA).

• Use the Law of Sines to solveoblique triangles (SSA).

• Find the areas of oblique triangles.

• Use the Law of Sines to modeland solve real-life problems.

Why you should learn itYou can use the Law of Sines tosolve real-life problems involvingoblique triangles. For instance,in Exercise 44 on page 438, youcan use the Law of Sines todetermine the length of theshadow of the Leaning Tower of Pisa.

Law of Sines

Hideo Kurihara/Getty Images

6.1

Law of SinesIf is a triangle with sides and then

is acute. is obtuse.AA

A B

ab

c

h

C

A B

ab

c

h

C

a

sin A�

b

sin B�

c

sin C.

c,b,a,ABC

The HM mathSpace® CD-ROM andEduspace® for this text contain additional resources related to the concepts discussed in this chapter.

333202_0601.qxd 12/5/05 10:40 AM Page 430

Given Two Angles and One Side—AAS

For the triangle in Figure 6.2, and feet. Findthe remaining angle and sides.

SolutionThe third angle of the triangle is

By the Law of Sines, you have

Using produces

and

Now try Exercise 1.

Given Two Angles and One Side—ASA

A pole tilts toward the sun at an angle from the vertical, and it casts a 22-footshadow. The angle of elevation from the tip of the shadow to the top of the poleis How tall is the pole?

SolutionFrom Figure 6.3, note that and So, the third angleis

By the Law of Sines, you have

Because feet, the length of the pole is

Now try Exercise 35.

For practice, try reworking Example 2 for a pole that tilts away from the sununder the same conditions.

a �c

sin C�sin A� �

22

sin 39��sin 43�� 23.84 feet.

c � 22

a

sin A�

c

sin C.

� 39�.

� 180� � 43� � 98�

C � 180� � A � B

B � 90� � 8� � 98�.A � 43�

43�.

8�

c �b

sin B�sin C� �

27.4

sin 28.7��sin 102.3�� 55.75 feet.

a �b

sin B�sin A� �

27.4

sin 28.7��sin 49.0�� 43.06 feet

b � 27.4

a

sin A�

b

sin B�

c

sin C.

� 49.0�.

� 180� � 28.7� � 102.3�

A � 180� � B � C

b � 27.4C � 102.3�, B � 28.7�,

Section 6.1 Law of Sines 431

Example 1

Example 2

A B

a

c

102.3°

28.7°

b = 27.4 ftC

FIGURE 6.2

AB

43°

8°a

b

c = 22 ft

C

FIGURE 6.3

When solving triangles, acareful sketch is useful as aquick test for the feasibility ofan answer. Remember that thelongest side lies opposite thelargest angle, and the shortestside lies opposite the smallestangle.

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

Single-Solution Case—SSA

For the triangle in Figure 6.4, inches, inches, and Findthe remaining side and angles.

SolutionBy the Law of Sines, you have

Reciprocal form

Multiply each side by b.

Substitute for A, a, and b.

B is acute.

Now, you can determine that

Then, the remaining side is


c �a

sin A �sin C� �

22

sin 42��sin 116.59�� 29.40 inches.

c

sin C�

a

sin A

� 116.59�.C � 180� � 42� � 21.41�

B � 21.41�.

sin B � 12�sin 42�

22 �

sin B � b�sin A

a �

sin B

b�

sin Aa

A � 42�.b � 12a � 22


The Ambiguous Case (SSA)Consider a triangle in which you are given and

is acute. is acute. is acute. is acute. is obtuse. is obtuse.

Sketch

Necessarycondition

Triangles None One One Two None Onepossible

a > ba ≤ bh < a < ba > ba � ha < h

ab

A

a

b

A

a ah

b

AA

ab hahb

AA

b ah

AAAAAA

�h � b sin A�A.b,a,

Example 3

a = 22 in.b = 12 in.

A B42°

c

C

One solution:FIGURE 6.4

a > b

333202_0601.qxd 12/5/05 10:40 AM 32

No-Solution Case—SSA

Show that there is no triangle for which and

SolutionBegin by making the sketch shown in Figure 6.5. From this figure it appears thatno triangle is formed. You can verify this using the Law of Sines.

Reciprocal form

Multiply each side by b.

This contradicts the fact that So, no triangle can be formed havingsides and and an angle of


Two-Solution Case—SSA

Find two triangles for which meters, meters, and

SolutionBy the Law of Sines, you have

Reciprocal form

There are two angles and between and whose sine is 0.9047. For you obtain

For you obtain

The resulting triangles are shown in Figure 6.6.

FIGURE 6.6


A

a = 12 mb = 31 m

20.5° 115.2°

B2A

a = 12 mb = 31 m

20.5° 64.8°B1

c �a

sin A�sin C� �

12

sin 20.5��sin 44.3�� 23.93 meters.

C � 180� � 20.5� � 115.2� � 44.3�

B2 � 115.2�,

c �a

sin A�sin C� �

12

sin 20.5��sin 94.7�� 34.15 meters.

C � 180� � 20.5� � 64.8� � 94.7�

B1 � 64.8�,180�0�B2 � 180� � 64.8� � 115.2�B1 � 64.8�

sin B � b�sin A

a � � 31�sin 20.5�

12 � � 0.9047.

sin B

b�

sin Aa

A � 20.5�.b � 31a � 12

A � 85�.b � 25a � 15�sin B� ≤ 1.

sin B � 25�sin 85�

15 � � 1.660 > 1

sin B � b�sin Aa �

sin B

b�

sin Aa

A � 85�.a � 15, b � 25,


Example 4

Example 5

A

a = 15

b = 25

85°

h

No solution:FIGURE 6.5

a < h

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Area of an Oblique TriangleThe procedure used to prove the Law of Sines leads to a simple formula for thearea of an oblique triangle. Referring to Figure 6.7, note that each triangle has aheight of Consequently, the area of each triangle is

By similar arguments, you can develop the formulas

A is acute A is obtuseFIGURE 6.7

Note that if angle is the formula gives the area for a right triangle:

Similar results are obtained for angles and equal to

Finding the Area of a Triangular Lot

Find the area of a triangular lot having two sides of lengths 90 meters and 52meters and an included angle of

SolutionConsider meters, meters, and angle as shown inFigure 6.8. Then, the area of the triangle is


Area �1

2ab sin C �

1

2�90��52��sin 102�� 2289 square meters.

C � 102�,b � 52a � 90

102�.

90�.BC

sin 90� � 1Area �1

2 bc�sin 90��

1

2 bc �

1

2�base��height�.

90�,A

AA

B

a ab b

c

h

Bc

h

C C

Area �1

2ab sin C �

1

2ac sin B.

Area �1

2�base��height� �

1

2�c��b sin A� �

1

2bc sin A.

h � b sin A.


Area of an Oblique TriangleThe area of any triangle is one-half the product of the lengths of two sidestimes the sine of their included angle. That is,

Area �1

2bc sin A �

1

2ab sin C �

1

2ac sin B.

Example 6

a = 90 m

b = 52 m

C

102°

FIGURE 6.8

To see how to obtain the height ofthe obtuse triangle in Figure 6.7,notice the use of the referenceangle and the differ-ence formula for sine, as follows.

� b sin A

� b�0 � cos A � ��1� � sin A�

� cos 180� sin A�

� b�sin 180� cos A

h � b sin�180� � A�

180� � A

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Application

An Application of the Law of Sines

The course for a boat race starts at point in Figure 6.9 and proceeds in thedirection S W to point then in the direction S E to point and finallyback to Point lies 8 kilometers directly south of point Approximate thetotal distance of the race course.

SolutionBecause lines and are parallel, it follows that Consequently, triangle has the measures shown in Figure 6.10. For angle you have Using the Law of Sines

you can let and obtain

and

The total length of the course is approximately


� 19.453 kilometers.

Length � 8 � 6.308 � 5.145

c �8

sin 88��sin 40�� 5.145.

a �8

sin 88��sin 52�� 6.308

b � 8

a

sin 52��

b

sin 88��

c

sin 40�

B � 180� � 52� � 40� � 88�.B,ABC

�BCA �DBC.ACBD

A.CA.C,40�B,52�

A


Example 7

W RITING ABOUT MATHEMATICS

Using the Law of Sines In this section, you have been using the Law of Sines tosolve 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 eachtriangle. Is there an easier way to solve these triangles?

a. b.

50°

AC

a = 10

B

50°

AC

c = 20

B

�ASA��AAS�

A

B

CD

40°

52°

8 km

S

EW

N

FIGURE 6.9

40°

52°B

C

a

c

b = 8 km

A

FIGURE 6.10

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In Exercises 1–18, use the Law of Sines to solve the triangle.Round your answers to two decimal places.

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

In Exercises 19–24, use the Law of Sines to solve (if possible)the triangle. If two solutions exist, find both. Round youranswers to two decimal places.

19.

20.

21.

22.

23.

24.

In Exercises 25–28, find values for b such that the trianglehas (a) one solution, (b) two solutions, and (c) no solution.

25.

26.

27.

28.

In Exercises 29–34, find the area of the triangle having theindicated angle and sides.

29.

30.

31.

32.

33.

34. b � 20a � 16,C � 84� 30�,

c � 64a � 105,B � 72� 30�,

c � 22b � 4.5,A � 5� 15�,

c � 85b � 57,A � 43� 45�,

c � 20a � 62,B � 130�,

b � 6a � 4,C � 120�,

a � 315.6A � 88�,

a � 10.8A � 10�,

a � 10A � 60�,

a � 5A � 36�,

b � 12.8a � 4.5,A � 58�,

b � 12.8a � 11.4,A � 58�,

b � 21a � 34,A � 76�,

b � 20a � 18,A � 76�,

b � 200a � 125,A � 110�,

b � 100a � 125,A � 110�,

a � 358C � 104�,B � 28�,

c �34B � 42�,A � 55�,

c � 50a � 35,C � 85� 20�,

b � 16a � 48,A � 110� 15�,

c � 10a � 125,A � 100�,

c � 14b � 4,C � 145�,

c � 5.8b � 6.2,B � 2� 45�,

b � 6.8a � 4.5,B � 15� 30�,

b � 4.8B � 8� 15�,A � 5� 40�,

c � 18.1 C � 54.6�, A � 83� 20�,

c � 2.68C � 54.6�,A � 24.3�,

a � 21.6C � 16.7�,A � 102.4�,

c � 10a � 9,A � 60�,

b � 5a � 8,A � 36�,

A B135°

10°b a

c = 45

C

A B

b a = 3.5

c25° 35°

C

A B

105°

40°

ab

c = 20

C

A B

a = 20b

c30° 45°

C


Exercises 6.1 The HM mathSpace® CD-ROM and Eduspace® for this text contain step-by-step solutions to all odd-numbered exercises. They also provide Tutorial Exercises for additional help.

VOCABULARY CHECK: Fill in the blanks.

1. An ________ triangle is a triangle that has no right angle.

2. For triangle the Law of Sines is given by

3. The area of an oblique triangle is given by

PREREQUISITE SKILLS REVIEW: Practice and review algebra skills needed for this section at www.Eduspace.com.

12 bc sin A �

12ab sin C � ________ .

asin A

� ________ �c

sin C.ABC,

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www.Eduspace.com


35. Height Because of prevailing winds, a tree grew so thatit was leaning 4 from the vertical. At a point 35 metersfrom the tree, the angle of elevation to the top of the tree is

(see figure). Find the height of the tree.

36. Height A flagpole at a right angle to the horizontal islocated on a slope that makes an angle of with thehorizontal. The flagpole’s shadow is 16 meters long andpoints directly up the slope. The angle of elevation fromthe tip of the shadow to the sun is

(a) Draw a triangle that represents the problem. Show theknown quantities on the triangle and use a variable toindicate 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 a17-meter shadow directly down a slope when the angle ofelevation of the sun is (see figure). Find the angle ofelevation of the ground.

38. Flight Path A plane flies 500 kilometers with a bearingof from Naples to Elgin (see figure). The plane thenflies 720 kilometers from Elgin to Canton. Find the bearingof the flight from Elgin to Canton.

39. Bridge Design A bridge is to be built across a small lakefrom a gazebo to a dock (see figure). The bearing from thegazebo to the dock is S W. From a tree 100 meters fromthe gazebo, the bearings to the gazebo and the dock are S

E and S E, respectively. Find the distance from thegazebo to the dock.

40. Railroad Track Design The circular arc of a railroadcurve has a chord of length 3000 feet and a central angle of

(a) Draw a diagram that visually represents the problem.Show the known quantities on the diagram and use thevariables and to represent the radius of the arc andthe length of the arc, respectively.

(b) Find the radius of the circular arc.

(c) Find the length of the circular arc.

41. Glide Path A pilot has just started on the glide path forlanding at an airport with a runway of length 9000 feet.The angles of depression from the plane to the ends of therunway are and

(a) Draw a diagram that visually represents the problem.

(b) Find the air distance the plane must travel untiltouching down on the near end of the runway.

(c) Find the ground distance the plane must travel untiltouching down.

(d) Find the altitude of the plane when the pilot begins thedescent.

42. Locating a Fire The bearing from the Pine Knob firetower to the Colt Station fire tower is N E, and the twotowers are 30 kilometers apart. A fire spotted by rangers ineach tower has a bearing of from Pine Knob and

from Colt Station (see figure). Find the distance ofthe fire from each tower.

65°Fire

Pine Knob

Colt Station

80°70°

30 kmS

EW

N

Not drawn to scale

S 70� EN 80� E

65�

18.8�.17.5�

s

r

sr

40�.

74°Tree

Dock

Gazebo

100 m

28°

41°

S

EW

N

28�74�

41�

44°720 km 500 km

Naples

N

Canton

Elgin

S

EW

N

Not drawn to scale

316�

42°

B

A

17 m

C

10 m

θ

θ

42° −

�,42�

20�.

12�

23°

h

94°

35 m

h23�

�

333202_0601.qxd 12/5/05 10:40 AM Page 437

43. Distance A boat is sailing due east parallel to theshoreline at a speed of 10 miles per hour. At a given time,the bearing to the lighthouse is S E, and 15 minuteslater the bearing is S E (see figure). The lighthouse islocated at the shoreline. What is the distance from the boatto the shoreline?

Synthesis

True or False? In Exercises 45 and 46, determine whetherthe statement is true or false. Justify your answer.

45. If a triangle contains an obtuse angle, then it must beoblique.

46. Two angles and one side of a triangle do not necessarilydetermine a unique triangle.

47. Graphical and Numerical Analysis In the figure, andare positive angles.

(a) Write as a function of

(b) Use a graphing utility to graph the function. Determineits domain and range.

(c) Use the result of part (a) to write as a function of

(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?

FIGURE FOR 47 FIGURE FOR 48

48. Graphical Analysis

(a) Write the area of the shaded region in the figure as afunction of

(b) Use a graphing utility to graph the area function.

(c) Determine the domain of the area function. Explainhow the area of the region and the domain of thefunction would change if the eight-centimeter linesegment were decreased in length.

Skills Review

In Exercises 49–52, use the fundamental trigonometricidentities to simplify the expression.

49. 50.

51. 52. 1 � cot2��

2� x�1 � sin2��

2� x�

tan x cos x sec xsin x cot x

�.A

30 cm

8 cm

20 cm

2

θ

θ

.c

.

70°63°

d S

EW

N

63�70�


44. Shadow Length The Leaning Tower of Pisa in Italy ischaracterized by its tilt. The tower leans because it wasbuilt on a layer of unstable soil—clay, sand, and water.The tower is approximately 58.36 meters tall from itsfoundation (see figure). The top of the tower leansabout 5.45 meters off center.

(a) Find the angle of lean of the tower.

(b) Write as a function of and where is theangle of elevation to the sun.

(c) Use the Law of Sines to write an equation for thelength of the shadow cast by the tower.

(d) Use a graphing utility to complete the table.

d

��,d

Model It

58.36 m

d

5.45 m

θ

α

β

Not drawn to scale

d

60�50�40�30�20�10��

0.4 0.8 1.2 1.6 2.0 2.4 2.8

c

18 9

cα β

γ

333202_0601.qxd 12/5/05 10:40 AM Page 438

Section 6.2 Law of Cosines 439

IntroductionTwo cases remain in the list of conditions needed to solve an oblique triangle—SSS and SAS. If you are given three sides (SSS), or two sides and their includedangle (SAS), none of the ratios in the Law of Sines would be complete. In suchcases, you can use the Law of Cosines.

For a proof of the Law of Cosines, see Proofs in Mathematics on page 490.

Three Sides of a Triangle—SSS

Find the three angles of the triangle in Figure 6.11.

FIGURE 6.11

SolutionIt is a good idea first to find the angle opposite the longest side—side in thiscase. Using the alternative form of the Law of Cosines, you find that

Because is negative, you know that is an obtuse angle given by 116.80 . At this point, it is simpler to use the Law of Sines to determine

Because is obtuse, must be acute, because a triangle can have, at most, oneobtuse angle. So, and

Now try Exercise 1.

41.12�. C � 180� � 22.08� � 116.80� �A � 22.08�AB

sin A � a�sin B

b � � 8�sin 116.80�

19 � � 0.37583

A.�B �Bcos B

� �0.45089.�82 � 142 � 192

2�8��14�cos B �

a2 � c2 � b2

2ac

b

AC

a = 8 ft c = 14 ft

b = 19 ft

B

What you should learn• Use the Law of Cosines to

solve oblique triangles (SSS or SAS).

• Use the Law of Cosines tomodel and solve real-life problems.

• Use Heron’s Area Formula tofind the area of a triangle.

Why you should learn itYou can use the Law of Cosinesto solve real-life problemsinvolving oblique triangles.For instance, in Exercise 31 onpage 444, you can use the Lawof Cosines to approximate thelength of a marsh.

Law of Cosines

© Roger Ressmeyer/Corbis

6.2

Law of CosinesStandard Form Alternative Form

cos C �a2 � b2 � c2

2abc2 � a2 � b2 � 2ab cos C

cos B �a2 � c2 � b2

2acb2 � a2 � c2 � 2ac cos B

cos A �b2 � c2 � a2

2bca2 � b2 � c2 � 2bc cos A

Example 1

333202_0602.qxd 12/5/05 10:41 AM Page 439

Do you see why it was wise to find the largest angle first in Example 1?Knowing the cosine of an angle, you can determine whether the angle is acute orobtuse. That is,

for Acute

for Obtuse

So, in Example 1, once you found that angle was obtuse, you knew that anglesand were both acute. If the largest angle is acute, the remaining two angles

are acute also.

Two Sides and the Included Angle—SAS

Find the remaining angles and side of the triangle in Figure 6.12.

FIGURE 6.12

SolutionUse the Law of Cosines to find the unknown side in the figure.

Because centimeters, you now know the ratio and you can usethe reciprocal form of the Law of Sines to solve for

So, and

Now try Exercise 3.

C � 180� � 115� � 39.75� � 25.25�.B � arcsin 0.63945 � 39.75�

� 0.63945

� 15�sin 115�

21.26 �

sin B � b�sin Aa �

sin Bb

�sin A

a

B.sin A�aa � 21.26

a � 21.26

a2 � 451.79

a2 � 152 � 102 � 2�15��10� cos 115�

a2 � b2 � c2 � 2bc cos A

a

A B

b = 15 cma

c = 10 cm

115°

C

CAB

90� < � < 180�.cos � < 0

0� < � < 90�cos � > 0


Example 2

What familiar formula do youobtain when you use the thirdform of the Law of Cosines

and you let What isthe relationship between the Law of Cosines and thisformula?

C � 90�?

c2 � a2 � b2 � 2ab cos C

Exploration

When solving an oblique triangle given three sides, youuse the alternative form of theLaw of Cosines to solve for anangle. When solving an obliquetriangle given two sides and theirincluded angle, you use the standard form of the Law ofCosines to solve for an unknown.

333202_0602.qxd 12/5/05 10:41 AM Page 440

Applications

An Application of the Law of Cosines

The pitcher’s mound on a women’s softball field is 43 feet from home plate andthe distance between the bases is 60 feet, as shown in Figure 6.13. (The pitcher’smound is not halfway between home plate and second base.) How far is thepitcher’s mound from first base?

SolutionIn triangle (line bisects the right angle at ), and

Using the Law of Cosines for this SAS case, you have

So, the approximate distance from the pitcher’s mound to first base is

feet.


An Application of the Law of Cosines

A ship travels 60 miles due east, then adjusts its course northward, as shown inFigure 6.14. After traveling 80 miles in that direction, the ship is 139 miles fromits point of departure. Describe the bearing from point to point

SolutionYou have and so, using the alternative form of the Lawof Cosines, you have

So, and thus the bearing measured from duenorth from point to point is or


N 76.15� E.166.15� � 90� � 76.15�,CBB � arccos��0.97094� � 166.15�,

� �0.97094.

�802 � 602 � 1392

2�80��60�

cos B �a2 � c2 � b2

2ac

c � 60;b � 139,a � 80,

C.B

h � �1800.3 � 42.43

� 1800.3 � 432 � 602 � 2�43��60� cos 45º

h2 � f 2 � p2 � 2fp cos H

p � 60.f � 43,HHPH � 45�HPF,


Example 3

Example 4

BA

c = 60 mi

C

a = 80 miS

EW

N

b = 139 mi

FIGURE 6.14

H

PF

60 ft 60 ft

60 ft p = 60 ft

f = 43 ft

h

45°

FIGURE 6.13

333202_0602.qxd 12/5/05 10:41 AM Page 441

Heron’s Area FormulaThe Law of Cosines can be used to establish the following formula for the area of a triangle. This formula is called Heron’s Area Formula after the Greekmathematician Heron (c. 100 B.C.).

For a proof of Heron’s Area Formula, see Proofs in Mathematics on page 491.

Using Heron’s Area Formula

Find the area of a triangle having sides of lengths meters, meters,and meters.

SolutionBecause , Heron’s Area Formula yields


You have now studied three different formulas for the area of a triangle.

Standard Formula

Oblique Triangle

Heron’s Area Formula Area � �s�s � a��s � b��s � c�

Area �12 bc sin A �

12 ab sin C �

12 ac sin B

Area �12 bh

� 1131.89 square meters. � �84�41��31��12�

Area � �s�s � a��s � b��s � c�s � �a � b � c��2 � 168�2 � 84

c � 72b � 53a � 43


Heron’s Area FormulaGiven any triangle with sides of lengths and the area of the triangle is

Area

where s � �a � b � c��2.

� �s�s � a��s � b��s � c�

c,b,a,

Example 5


The Area of a Triangle Use the most appropriate formula to find the area of each triangle below. Show your work and give your reasons for choosing each formula.

a. b.

c. d.

4 ft

5 ft

3 ft

4 ft

2 ft

4 ft

2 ft 3 ft

4 ft

2 ft

50°

Historical NoteHeron of Alexandria (c. 100 B.C.)was a Greek geometer andinventor. His works describehow to find the areas of triangles, quadrilaterals,regular polygons having 3 to12 sides, and circles as well asthe surface areas and volumesof three-dimensional objects.

333202_0602.qxd 12/5/05 10:41 AM Page 442


In Exercises 1–16, use the Law of Cosines to solve thetriangle. Round your answers to two decimal places.

1. 2.

3. 4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

In Exercises 17–22, complete the table by solving theparallelogram shown in the figure. (The lengths of thediagonals are given by and )

17. 5 8 45

18. 25 35 120

19. 10 14 20

20. 40 60 80

21. 15 25 20

22. 25 50 35

In Exercises 23–28, use Heron’s Area Formula to find thearea of the triangle.

23.

24.

25.

26.

27.

28.

29. Navigation A boat race runs along a triangular coursemarked by buoys and The race starts with the boatsheaded west for 3700 meters. The other two sides of thecourse lie to the north of the first side, and their lengths are1700 meters and 3000 meters. Draw a figure that gives avisual representation of the problem, and find the bearingsfor the last two legs of the race.

30. Navigation A plane flies 810 miles from Franklin toCenterville with a bearing of Then it flies 648 milesfrom Centerville to Rosemount with a bearing of Draw a figure that visually represents the problem, andfind the straight-line distance and bearing from Franklin toRosemount.

32�. 75�.

C.B,A,

c � 2.45b � 0.75,a � 3.05,

c � 15.05b � 8.46,a � 12.32,

c � 52b � 52,a � 75.4,

c � 9b � 10.2,a � 2.5,

c � 9b � 15,a � 12,

c � 10b � 7,a � 5,

��

��

��dcba

φ

θa

c

b

d

d.c

b �34a �

38,C � 103�,

b �79a �

49,C � 43�,

b � 2.15a � 6.25,C � 15� 15�,

c � 32a � 32,B � 125� 40�,

c � 9.5a � 6.2,B � 75� 20�,

c � 30a � 40,B � 10� 35�,

c � 10b � 3,A � 55�,

c � 9b � 4,A � 135�,

c � 1.25b � 0.75,a � 1.42,

c � 52b � 52,a � 75.4,

c � 72b � 25,a � 55,

c � 20b � 14,a � 11,

BA

a = 10

c

b = 4.5105°

C

A B

b = 15

c = 30

a30°

C

A B

b = 3 a = 8

c = 9

C

A B

b = 10 a = 7

c = 15

C

Exercises 6.2


1. If you are given three sides of a triangle, you would use the Law of ________ to find the three angles of the triangle.

2. The standard form of the Law of Cosines for is ________ .

3. The Law of Cosines can be used to establish a formula for finding the area of a triangle called ________ ________ Formula.


cos B �a2 � c2 � b2

2ac

333202_0602.qxd 12/5/05 10:41 AM Page 443

www.Eduspace.com

31. Surveying To approximate the length of a marsh, asurveyor walks 250 meters from point to point , thenturns and walks 220 meters to point (see figure).Approximate the length of the marsh.

32. Surveying A triangular parcel of land has 115 meters of frontage, and the other boundaries have lengths of 76 meters and 92 meters. What angles does the frontagemake with the two other boundaries?

33. Surveying A triangular parcel of ground has sides oflengths 725 feet, 650 feet, and 575 feet. Find the measureof the largest angle.

34. Streetlight Design Determine the angle in the design ofthe streetlight shown in the figure.

35. Distance Two ships leave a port at 9 A.M. Onetravels at a bearing of N W at 12 miles per hour, andthe other travels at a bearing of S W at 16 miles perhour. Approximate how far apart they are at noon that day.

36. Length A 100-foot vertical tower is to be erected on theside of a hill that makes a angle with the horizontal (seefigure). Find the length of each of the two guy wires thatwill be anchored 75 feet uphill and downhill from the baseof the tower.

37. Navigation On a map, Orlando is 178 millimeters duesouth of Niagara Falls, Denver is 273 millimeters fromOrlando, and Denver is 235 millimeters from Niagara Falls(see figure).

(a) Find the bearing of Denver from Orlando.

(b) Find the bearing of Denver from Niagara Falls.

38. Navigation On a map, Minneapolis is 165 millimetersdue west of Albany, Phoenix is 216 millimeters fromMinneapolis, and Phoenix is 368 millimeters from Albany(see figure).

(a) Find the bearing of Minneapolis from Phoenix.

(b) Find the bearing of Albany from Phoenix.

39. Baseball On a baseball diamond with 90-foot sides, thepitcher’s mound is 60.5 feet from home plate. How far is itfrom the pitcher’s mound to third base?

40. Baseball The baseball player in center field is playingapproximately 330 feet from the television camera that isbehind home plate. A batter hits a fly ball that goes to thewall 420 feet from the camera (see figure). The cameraturns to follow the play. Approximately how far does thecenter fielder have to run to make the catch?

330 ft

420 ft

8°

8�

MinneapolisMinneapolisAlbanyAlbany

PhoenixPhoenix

216 mm216 mm

165 mm165 mm

368 mm368 mm

Denver

Niagara Falls

Orlando

273 mm

235 mm

178 mm

Denver

Niagara Falls

Orlando

273 mm

235 mm

178 mm

100 ft

75 ft75 ft

6°

6�

67�53�

4 21

3

2θ

�

C A

75°

220 m 250 m

B

ACC75�

BA


333202_0602.qxd 12/5/05 10:41 AM Page 444


41. Aircraft Tracking To determine the distance betweentwo aircraft, a tracking station continuously determines thedistance to each aircraft and the angle between them (seefigure). Determine the distance between the planes when

miles, and miles.

FIGURE FOR 41

42. Aircraft Tracking Use the figure for Exercise 41 todetermine the distance between the planes when

miles, and miles.

43. Trusses is the midpoint of the line segment in thetruss rafter shown in the figure. What are the lengths of theline segments and

45. Paper Manufacturing In a process with continuouspaper, the paper passes across three rollers of radii 3 inches,4 inches, and 6 inches (see figure). The centers of the three-inch and six-inch rollers are inches apart, and thelength of the arc in contact with the paper on the four-inchroller is inches. Complete the table.

46. Awning Design A retractable awning above a patio doorlowers at an angle of from the exterior wall at a heightof 10 feet above the ground (see figure). No direct sunlightis to enter the door when the angle of elevation of the sunis greater than What is the length of the awning?

47. Geometry The lengths of the sides of a triangular parcelof land are approximately 200 feet, 500 feet, and 600 feet.Approximate the area of the parcel.

48. Geometry A parking lot has the shape of a parallelogram(see figure). The lengths of two adjacent sides are 70meters and 100 meters. The angle between the two sides is70 What is the area of the parking lot?

70°

70 m

100 m

�.

70°

50°

10 ft

Sun’s raysx

x70�.

50�

ds

4 in.

6 in.

θ

3 in.

s

d

PS

Q

8 8 8 8

10

R

RS?QS,PQ,

PRQ

c � 20b � 20A � 11�,a

A

BC

bc

a

c � 20b � 35A � 42�,a

A

44. Engine Design An engine has a seven-inch connectingrod fastened to a crank (see figure).

(a) Use the Law of Cosines to write an equation givingthe relationship between and

(b) Write as a function of (Select the sign thatyields positive values of )

(c) Use a graphing utility to graph the function in part (b).

(d) Use the graph in part (c) to determine the maxi-mum distance the piston moves in one cycle.

x.�.x

�.x

7 in.1.5 in.

x

θ

Model It

d (inches) 9 10 12 13 14 15 16

(degrees)

s (inches)

�

333202_0602.qxd 12/8/05 9:22 AM Page 445

49. Geometry You want to buy a triangular lot measuring510 yards by 840 yards by 1120 yards. The price of theland is $2000 per acre. How much does the land cost?(Hint:

50. Geometry You want to buy a triangular lot measuring1350 feet by 1860 feet by 2490 feet. The price of the landis $2200 per acre. How much does the land cost? (Hint:

Synthesis

True or False? In Exercises 51–53, determine whether thestatement is true or false. Justify your answer.

51. In Heron’s Area Formula, is the average of the lengths ofthe three sides of the triangle.

52. In addition to SSS and SAS, the Law of Cosines can beused to solve triangles with SSA conditions.

53. A triangle with side lengths of 10 centimeters, 16 centime-ters, and 5 centimeters can be solved using the Law ofCosines.

54. Circumscribed and Inscribed Circles Let and be the radii of the circumscribed and inscribed circles of atriangle respectively (see figure), and let

(a) Prove that

(b) Prove that

Circumscribed and Inscribed Circles In Exercises 55 and56, use the results of Exercise 54.

55. Given a triangle with

and

find the areas of (a) the triangle, (b) the circumscribedcircle, and (c) the inscribed circle.

56. Find the length of the largest circular running track that canbe built on a triangular piece of property with sides oflengths 200 feet, 250 feet, and 325 feet.

57. Proof Use the Law of Cosines to prove that

58. Proof Use the Law of Cosines to prove that

Skills Review

In Exercises 59– 64, evaluate the expression without usinga calculator.

59.

60.

61.

62.

63.

64.

In Exercises 65– 68, write an algebraic expression that isequivalent to the expression.

65.

66.

67.

68.

In Exercises 69–72, use trigonometric substitution to writethe algebraic equation as a trigonometric function of where Then find and

69.

70.

71.

72.

In Exercises 73 and 74, write the sum or difference as aproduct.

73.

74. sin�x ��

2� � sin�x ��

2�

cos 5�

6� cos

�

3

x � 6 tan �12 � �36 � x2,

x � 3 sec ��3 � �x2 � 9,

x � 2 cos ��2 � �4 � x2,

x � 5 sin �5 � �25 � x2,

csc �.sec ��/2 < � < �/2.�,

cos�arcsin x � 1

2 �cot�arctan�x � 2�tan�arccos 3x�sec�arcsin 2x�

arccos��32 �

arcsin��32 �

arctan��3 �arctan �3

arccos 0

arcsin��1�

1

2 bc �1 � cos A� �

a � b � c

2

a � b � c

2.

1

2 bc �1 � cos A� �

a � b � c

2

�a � b � c

2.

c � 72b � 55,a � 25,

r ��s � a��s � b��s � c�s

.

2R �a

sin A�

b

sin B�

c

sin C.

BC a

b c

R

r

A

s �a � b � c

2.

ABC,

rR

s

1 acre � 43,560 square feet)

1 acre � 4840 square yards)


333202_0602.qxd 12/5/05 10:41 AM Page 446

Section 6.3 Vectors in the Plane 447

IntroductionQuantities such as force and velocity involve both magnitude and direction andcannot be completely characterized by a single real number. To represent such aquantity, you can use a directed line segment, as shown in Figure 6.15. The directed line segment has initial point and terminal point Its magnitude(or length) is denoted by and can be found using the Distance Formula.

FIGURE 6.15 FIGURE 6.16

Two directed line segments that have the same magnitude and direction areequivalent. For example, the directed line segments in Figure 6.16 are allequivalent. The set of all directed line segments that are equivalent to the directedline segment is a vector v in the plane, written Vectors are denotedby lowercase, boldface letters such as , , and .

Vector Representation by Directed Line Segments

Let be represented by the directed line segment from to and let be represented by the directed line segment from to

as shown in Figure 6.17. Show that

FIGURE 6.17

SolutionFrom the Distance Formula, it follows that and have the same magnitude.

Moreover, both line segments have the same direction because they are bothdirected toward the upper right on lines having a slope of So, and havethe same magnitude and direction, and it follows that

Now try Exercise 1.

u � v.RS

\

PQ\2

3.

�RS\

� � ��4 � 1�2 � �4 � 2�2 � �13

�PQ\

� � ��3 � 0�2 � �2 � 0�2 � �13

RS\

PQ\

x

(1, 2)

(4, 4)

2 3 41(0, 0)

1

2

3

4

R

S(3, 2)

Q

v

uP

5

y

u � v.S � �4, 4�,R � �1, 2�v

�3, 2�,Q �P � �0, 0�u

wvu� PQ

\

.vPQ\

Initial pointP

Q

PQ

Terminal point

�PQ\

�Q.PPQ

\

What you should learn• Represent vectors as directed

line segments.

• Write the component forms of vectors.

• Perform basic vector operations and representthem graphically.

• Write vectors as linear combinations of unit vectors.

• Find the direction angles of vectors.

• Use vectors to model andsolve real-life problems.

Why you should learn itYou can use vectors to modeland solve real-life problemsinvolving magnitude and direc-tion. For instance, in Exercise 84on page 459, you can use vectorsto determine the true directionof a commercial jet.

Vectors in the Plane

Bill Bachman/Photo Researchers, Inc.

6.3

Example 1

333202_0603.qxd 12/5/05 10:42 AM Page 447

Component Form of a VectorThe directed line segment whose initial point is the origin is often the mostconvenient representative of a set of equivalent directed line segments. Thisrepresentative of the vector v is in standard position.

A vector whose initial point is the origin can be uniquely represented bythe coordinates of its terminal point This is the component form of avector v, written as

The coordinates and are the components of . If both the initial point andthe terminal point lie at the origin, is the zero vector and is denoted by

Two vectors and are equal if and only if andFor instance, in Example 1, the vector from to is

and the vector from to is

Finding the Component Form of a Vector

Find the component form and magnitude of the vector that has initial pointand terminal point

SolutionLet and let as shown in Figure6.18. Then, the components of are

So, and the magnitude of is

Now try Exercise 9.

� �169 � 13.

�v � � ��5�2 � 122

vv � ��5, 12�

v2 � q2 � p2 � 5 � ��7� � 12.

v1 � q1 � p1 � �1 � 4 � �5

v � �v1, v2�Q � ��1, 5� � �q1, q2�,P � �4, �7� � � p1, p2�

��1, 5�.�4, �7�v

� �3, 2�.�4 � 1, 4 � 2�v � RS\

�

S � �4, 4�R � �1, 2�v

u � PQ\

� �3 � 0, 2 � 0� � �3, 2�

Q � �3, 2�P � �0, 0�uu2 � v2.u1 � v1v � �v1, v2�u � �u1, u2�

0 � �0, 0�.v

vv2v1

v � �v1, v2�.

�v1, v2�.�0, 0�


Component Form of a VectorThe component form of the vector with initial point and termi-nal point is given by

The magnitude (or length) of is given by

If is a unit vector. Moreover, if and only if is the zerovector 0.

v�v � � 0v�v � � 1,

�v � � ��q1 � p1�2 � �q2 � p2�2 � �v12 � v2

2.

v

PQ\

� �q1 � p1, q2 � p2� � �v1, v2� � v.

Q � �q1, q2�P � �p1, p2�

x

−4

−6

642−2−4−6−8−2

−8

2

6

P = (4, −7)

v

Q = (−1, 5)

y

FIGURE 6.18

You can graph vectors with agraphing utility by graphingdirected line segments. Consultthe user’s guide for your graphingutility for specific instructions.

Techno logy

Example 2

333202_0603.qxd 12/8/05 9:23 AM Page 448

Vector OperationsThe two basic vector operations are scalar multiplication and vector addition.In operations with vectors, numbers are usually referred to as scalars. In this text,scalars will always be real numbers. Geometrically, the product of a vector anda scalar is the vector that is times as long as . If is positive, has thesame direction as , and if is negative, has the direction opposite that of ,as shown in Figure 6.19.

To add two vectors geometrically, position them (without changing theirlengths or directions) so that the initial point of one coincides with the terminalpoint of the other. The sum is formed by joining the initial point of the sec-ond vector with the terminal point of the first vector , as shown in Figure 6.20.This technique is called the parallelogram law for vector addition because the vector , often called the resultant of vector addition, is the diagonal of aparallelogram having and as its adjacent sides.

FIGURE 6.20

The negative of is

Negative

and the difference of and is

Add See Figure 8.21.

Difference

To represent geometrically, you can use directed line segments with thesame initial point. The difference is the vector from the terminal point of

to the terminal point of , which is equal to as shown in Figure 6.21.u � ��v�,uvu � v

u � v

� �u1 � v1, u2 � v2�.

��v�.u � v � u � ��v�

vu

� ��v1, �v2�

�v � ��1�v

v � �v1, v2�

v

u

x

u + v

y

v

u

x

y

vuu � v

uvu � v

vkvkvkvkv�k�k

v


Definitions of Vector Addition and Scalar MultiplicationLet and be vectors and let be a scalar (a realnumber). Then the sum of and is the vector

Sum

and the scalar multiple of times is the vector

Scalar multipleku � k�u1, u2� � �ku1, ku2�.

uk

u � v � �u1 � v1, u2 � v2�

vukv � �v1, v2�u � �u1, u2�

v v v3122 2v −v −

FIGURE 6.19

x

−v u − v

u

vu + (−v)

y

FIGURE 6.21u � v � u � ��v�

333202_0603.qxd 12/5/05 10:42 AM Page 449

The component definitions of vector addition and scalar multiplication areillustrated in Example 3. In this example, notice that each of the vector opera-tions can be interpreted geometrically.

Vector Operations

Let and and find each of the following vectors.

a. 2 b. c.

Solutiona. Because you have

A sketch of 2 is shown in Figure 6.22.

b. The difference of and is

A sketch of is shown in Figure 6.23. Note that the figure shows thevector difference as the sum

c. The sum of and 2 is

A sketch of is shown in Figure 6.24.


v � 2w

� �4, 13�.

� ��2 � 6, 5 � 8�

� ��2, 5� � �6, 8�

� ��2, 5� � �2�3�, 2�4��

v � 2w � ��2, 5� � 2�3, 4�

wv

w � ��v�.w � vw � v

� �5, �1�.

w � v � �3 � ��2�, 4 � 5�

vw

v

� ��4, 10�.

� �2��2�, 2�5��

2v � 2��2, 5�

v � ��2, 5�,

v � 2ww � vv

w � �3, 4�,v � ��2, 5�


Example 3

x−2 2−4−6−8

( 4, 10)−

( 2, 5)−

2v

v

4

6

8

10

y

FIGURE 6.22

x3 4 5

(3, 4)

(5, −1)

4

3

2

1

−1

w

w − v

−v

y

FIGURE 6.23

x−2−4−6 2 4 6 8

(−2, 5)

(4, 13)

2w

v

v + 2w

8

10

12

14

y

FIGURE 6.24

333202_0603.qxd 12/5/05 10:42 AM Page 450

Vector addition and scalar multiplication share many of the properties ofordinary arithmetic.

Property 9 can be stated as follows: the magnitude of the vector is theabsolute value of times the magnitude of

Unit VectorsIn many applications of vectors, it is useful to find a unit vector that has the samedirection as a given nonzero vector To do this, you can divide by its magni-tude to obtain

Unit vector in direction of

Note that is a scalar multiple of The vector has a magnitude of 1 and thesame direction as The vector is called a unit vector in the direction of

Finding a Unit Vector

Find a unit vector in the direction of and verify that the result has amagnitude of 1.

SolutionThe unit vector in the direction of is

This vector has a magnitude of 1 because


�� 2

�292

� � 5

�292

�� 4

29�

25

29��29

29� 1.

� �2

�29,

5

�29�.

�1

�29 ��2, 5�

v

�v ��

��2, 5��2�2 � �5�2

v

v � ��2, 5�

v.uv.uv.u

vu � unit vector �v

�v �� 1

�v �v.

vv.

v.ccv


Properties of Vector Addition and Scalar MultiplicationLet , , and be vectors and let and be scalars. Then the followingproperties are true.

1. 2.

3. 4.

5. 6.

7. 8.

9. �cv � � �c� � v �

1�u� � u, 0�u� � 0c�u � v� � cu � cv

�c � d�u � cu � duc�du� � �cd �u

u � ��u� � 0u � 0 � u

�u � v� � w � u � �v � w�u � v � v � u

dcwvu

Example 4

Historical NoteWilliam Rowan Hamilton(1805–1865), an Irish mathe-matician, did some of the earliest work with vectors.Hamilton spent many yearsdeveloping a system of vector-like quantities calledquaternions. AlthoughHamilton was convinced of the benefits of quaternions,the operations he defined didnot produce good models forphysical phenomena. It wasn’tuntil the latter half of the nineteenth century that theScottish physicist JamesMaxwell (1831–1879) restructured Hamilton’squaternions in a form usefulfor representing physicalquantities such as force,velocity, and acceleration.

The

Gra

ng

er C

olle

ctio

n

333202_0603.qxd 12/5/05 10:42 AM Page 451

The unit vectors and are called the standard unit vectors andare denoted by

and

as shown in Figure 6.25. (Note that the lowercase letter is written in boldface todistinguish it from the imaginary number ) These vectors can be usedto represent any vector as follows.

The scalars and are called the horizontal and vertical components ofrespectively. The vector sum

is called a linear combination of the vectors and . Any vector in the plane canbe written as a linear combination of the standard unit vectors and

Writing a Linear Combination of Unit Vectors

Let be the vector with initial point and terminal point Write as a linear combination of the standard unit vectors and

SolutionBegin by writing the component form of the vector

This result is shown graphically in Figure 6.26.


Vector Operations

Let and let Find

SolutionYou could solve this problem by converting and to component form. This,however, is not necessary. It is just as easy to perform the operations in unitvector form.


� �12i � 19j

� �6i � 16j � 6i � 3j

2u � 3v � 2��3i � 8j� � 3�2i � j�

vu

2u � 3v.v � 2i � j.u � �3i � 8j

� �3i � 8j

� ��3, 8�

u � ��1 � 2, 3 � ��5��

u.

j.iu��1, 3�.�2, �5� u

j.iji

v1i � v2 j

v,v2v1

� v1i � v2 j

� v1�1, 0� � v2�0, 1�

v � �v1, v2�

v � �v1, v2�,i � ��1.

i

j � �0, 1�i � �1, 0�

�0, 1��1, 0�


x

j = ⟨0, 1⟩

i = ⟨1, 0⟩

1 2

1

2

y

FIGURE 6.25

x−8 −6 −4 −2 2 4 6

−2

−4

−6

4

6

8

(−1, 3)

(2, −5)

u

y

FIGURE 6.26

Example 5

Example 6

333202_0603.qxd 12/5/05 10:42 AM Page 452

Direction AnglesIf is a unit vector such that is the angle (measured counterclockwise) fromthe positive -axis to , the terminal point of lies on the unit circle and you have

as shown in Figure 6.27. The angle is the direction angle of the vector Suppose that is a unit vector with direction angle If is any

vector that makes an angle with the positive -axis, it has the same direction asand you can write

Because it follows that the directionangle for is determined from

Quotient identity

Multiply numerator and denominator by

Simplify.

Finding Direction Angles of Vectors

Find the direction angle of each vector.

a.

b.

Solutiona. The direction angle is

So, as shown in Figure 6.28.

b. The direction angle is

Moreover, because lies in Quadrant IV, lies in Quadrant IV andits reference angle is

So, it follows that as shown in Figure 6.29.


� � 360� � 53.13� � 306.87�,

� � �arctan��4

3�� 53.13�� 53.13�.

�v � 3i � 4j

tan � �b

a�

�4

3.

� � 45�,

tan � �b

a�

3

3� 1.

v � 3i � 4j

u � 3i � 3j

�b

a.

� v �. ��v � sin �

�v � cos �

tan � �sin �

cos �

v�v � a i � bj � �v � �cos ��i � �v � �sin ��j,

� � v � �cos ��i � � v � �sin ��j.

v � � v � �cos �, sin �

ux�

v � a i � bj�.uu.�

u � �x, y � �cos �, sin � � �cos ��i � �sin ��j

uux�u


xθ

u

−1

−1

1

1

θx = cos

θy = sin

( , )x y

y

FIGURE 6.27 �u� � 1

x1 2 3

1

2

3(3, 3)

u

= 45°θ

y

FIGURE 6.28

x−1 2 3 41

1

−2

−1

−3

−4 (3, −4)

v

306.87°

y

FIGURE 6.29

Example 7

333202_0603.qxd 12/8/05 9:26 AM Page 453

Applications of Vectors

Finding the Component Form of a Vector

Find the component form of the vector that represents the velocity of an airplanedescending at a speed of 100 miles per hour at an angle 30 below the horizon-tal, as shown in Figure 6.30.

SolutionThe velocity vector has a magnitude of 100 and a direction angle of

You can check that has a magnitude of 100, as follows.


Using Vectors to Determine Weight

A force of 600 pounds is required to pull a boat and trailer up a ramp inclined at15 from the horizontal. Find the combined weight of the boat and trailer.

SolutionBased on Figure 6.31, you can make the following observations.

By construction, triangles and are similar. So, angle is 15 andso in triangle you have

Consequently, the combined weight is approximately 2318 pounds. (In Figure6.31, note that is parallel to the ramp.)


AC\

�BA\

� �600

sin 15�� 2318.

sin 15� �� AC

\

�� BA

\

��

600

�BA\

�

ABC�,ABCABCBWD

�AC\

� � force required to move boat up ramp � 600 pounds

�BC\

� � force against ramp

�BA\

� � force of gravity � combined weight of boat and trailer

�

� 100 � �10,000

� �7500 � 2500

�v � � ��50�3�2� ��50�2

v

� ��50�3, �50�

� �50�3 i � 50j

� 100��3

2 i � 100��1

2j

� 100�cos 210��i � 100�sin 210��j

v � �v � �cos ��i � �v � �sin ��j

� � 210�.v

�


x

210°

100

y

−50

−50

−75

−75−100

FIGURE 6.30

AC

B

D15°15°

W

FIGURE 6.31

Example 8

Example 9

333202_0603.qxd 12/5/05 10:42 AM Page 454

Using Vectors to Find Speed and Direction

An airplane is traveling at a speed of 500 miles per hour with a bearing of at a fixed altitude with a negligible wind velocity as shown in Figure 6.32(a).When the airplane reaches a certain point, it encounters a wind with a velocity of70 miles per hour in the direction as shown in Figure 6.32(b).What arethe resultant speed and direction of the airplane?

(a) (b)

FIGURE 6.32

SolutionUsing Figure 6.32, the velocity of the airplane (alone) is

and the velocity of the wind is

So, the velocity of the airplane (in the wind) is

and the resultant speed of the airplane is

miles per hour.

Finally, if is the direction angle of the flight path, you have

which implies that

So, the true direction of the airplane is


337.4�.

� 112.6�.� 180� � 67.4�� 180� � arctan��2.4065�

� �2.4065

tan � �482.5

�200.5

�

� 522.5

�v � � ��200.5�2 � �482.5�2

� ��200.5, 482.5�

� ��250 � 35�2, 250�3 � 35�2�v � v1 � v2

� �35�2, 35�2�.

v2 � 70�cos 45�, sin 45��

� ��250, 250�3�v1 � 500�cos 120�, sin 120��

xx

v1

v2

vWind

y

θx

120°v1

y

N 45� E,

330�


Example 10

Recall from Section 4.8 that inair navigation, bearings aremeasured in degrees clockwisefrom north.

333202_0603.qxd 12/5/05 10:42 AM Page 455

In Exercises 1 and 2, show that

1. 2.

In Exercises 3–14, find the component form and themagnitude of the vector

3. 4.

5. 6.

7. 8.

Initial Point Terminal Point

9.

10.

11.

12.

13.

14.

In Exercises 15–20, use the figure to sketch a graph of thespecified vector. To print an enlarged copy of the graph, goto the website, www.mathgraphs.com.

15. 16. 5

17. 18.

19. 20. v �12uu � 2v

u � vu � v

v�v

x

u v

y

�5, �17��2, 7��8, �9��1, 3��9, 40��3, 11��5, 1��3, �5��9, 3��1, 11��15, 12��1, 5�

x

(3, −1)(−4, −1)4−5

−2−3−4−5

v

23

1

4

y

x

(3, 3)

(3, −2)

2 51 4−2 −1

−2−3

v23

1

4

y

x

(−1, −1)

(3, 5)

42−2−4

v2

6

4

y

x

(−1, 4)

(2, 2)

21 3−2 −1−3

v

2

3

5

1

y

x

( 4, 2)− −

1

−3

−2

−1v−2−4 −3

y

x

(3, 2)

1 2 3 4

1

2

3

4

v

y

v.

(0, −5)

(0, 4)

(3, 3)

42−4−2

−4

vu

4

x

y

(−3, −4)

(4, 1)(0, 0)

(6, 5)

(2, 4)

−2−2 2 4 6

v

u4

2

6

x

y

u � v.


Exercises 6.3


1. A ________ ________ ________ can be used to represent a quantity that involves both magnitude and direction.

2. The directed line segment has ________ point and ________ point

3. The ________ of the directed line segment is denoted by

4. The set of all directed line segments that are equivalent to a given directed line segment is a ________ in the plane.

5. The directed line segment whose initial point is the origin is said to be in ________ ________ .

6. A vector that has a magnitude of 1 is called a ________ ________ .

7. The two basic vector operations are scalar ________ and vector ________ .

8. The vector is called the ________ of vector addition.

9. The vector sum is called a ________ ________ of the vectors and and the scalars and are called the ________ and ________ components of respectively.


v,v2v1

j,iv1i � v2 j

u � v

vPQ

\

�PQ�.PQ\

Q.PPQ\

333202_0603.qxd 12/5/05 10:42 AM Page 456

www.Eduspace.com

www.mathgraphs.com


In Exercises 21–28, find (a) , (b) , and (c)Then sketch the resultant vector.

21.

22.

23.

24.

25.

26.

27.

28.

In Exercises 29–38, find a unit vector in the direction of thegiven vector.

29. 30.

31. 32.

33. 34.

35. 36.

37. 38.

In Exercises 39– 42, find the vector v with the givenmagnitude and the same direction as u.

Magnitude Direction

39.

40.

41.

42.

In Exercises 43–46, the initial and terminal points of a vectorare given. Write a linear combination of the standard unitvectors i and j.

Initial Point Terminal Point

43.

44.

45.

46.

In Exercises 47–52, find the component form of andsketch the specified vector operations geometrically,where and

47.

48.

49.

50.

51.

52.

In Exercises 53–56, find the magnitude and direction angleof the vector v.

53.

54.

55.

56.

In Exercises 57–64, find the component form of v given itsmagnitude and the angle it makes with the positive -axis.Sketch v.

Magnitude Angle

57.

58.

59.

60.

61.

62.

63. in the direction

64. in the direction

In Exercises 65–68, find the component form of the sum ofu and v with direction angles and

Magnitude Angle

65.

66.

67.

68.

In Exercises 69 and 70, use the Law of Cosines to find theangle between the vectors. ( Assume )

69.

70.

Resultant Force In Exercises 71 and 72, find the anglebetween the forces given the magnitude of their resultant.(Hint: Write force 1 as a vector in the direction of thepositive -axis and force 2 as a vector at an angle with thepositive -axis.)

Force 1 Force 2 Resultant Force

71. 45 pounds 60 pounds 90 pounds

72. 3000 pounds 1000 pounds 3750 pounds

x�x

w � 2i � jv � i � 2j,

w � 2i � 2jv � i � j,

0� ≤ � ≤ 180�.�

� 110��v�v � � 30

� 30��u�u � � 50

� 180��v�v � � 50

� 45��u�u � � 20

� 90��v�v � � 4

� 60��u�u � � 4

� 90��v�v � � 5

� 0��u�u � � 5

�v .�u

3i � 4jv�v � � 3

i � 3jv�v � � 2

� � 90��v � � 4�3

� � 150��v � � 3�2

� � 45��v � �52

� � 150��v � �72

� � 45��v � � 1

� � 0��v � � 3

x

v � �5i � 4j

v � 6i � 6j

v � 8�cos 135�i � sin 135�j �v � 3�cos 60�i � sin 60�j �

v � u � 2w

v �12�3u � w�

v � �u � w

v � u � 2w

v �34 w

v �32u

w � i � 2j.u � 2i � j

v

�0, 1��6, 4��2, 3��1, �5��3, 6��0, �2��4, 5��3, 1�

u � ��10, 0��v � � 10

u � �2, 5��v � � 9

u � ��3, 3��v � � 6

u � �3, 3��v � � 5

w � 7j � 3iw � i � 2j

w � �6iw � 4j

v � i � jv � 6i � 2j

v � �5, �12�v � ��2, 2�u � �0, �2�u � �3, 0�

v � 2iu � 3j,

v � ju � 2i,

v � �i � 2ju � �2i � j,

v � 2i � 3ju � i � j,

v � �2, 1�u � �0, 0�,v � �0, 0�u � ��5, 3�,

v � �4, 0�u � �2, 3�,v � �1, 3�u � �2, 1�,

2u � 3v.u � vu � v

333202_0603.qxd 12/5/05 10:42 AM Page 457

73. Resultant Force Forces with magnitudes of 125 newtonsand 300 newtons act on a hook (see figure). The anglebetween the two forces is Find the direction andmagnitude of the resultant of these forces.

74. Resultant Force Forces with magnitudes of 2000newtons and 900 newtons act on a machine part at anglesof and respectively, with the -axis (see figure).Find the direction and magnitude of the resultant of theseforces.

75. Resultant Force Three forces with magnitudes of 75 pounds, 100 pounds, and 125 pounds act on an object atangles of and respectively, with the positive-axis. Find the direction and magnitude of the resultant of

these forces.

76. Resultant Force Three forces with magnitudes of 70 pounds, 40 pounds, and 60 pounds act on an object atangles of 4 and respectively, with thepositive -axis. Find the direction and magnitude of theresultant of these forces.

77. Velocity A ball is thrown with an initial velocity of 70 feet per second, at an angle of with the horizontal(see figure). Find the vertical and horizontal components ofthe velocity.

78. Velocity A gun with a muzzle velocity of 1200 feet persecond is fired at an angle of with the horizontal. Findthe vertical and horizontal components of the velocity.

Cable Tension In Exercises 79 and 80, use the figure todetermine the tension in each cable supporting the load.

79. 80.

81. Tow Line Tension A loaded barge is being towed by twotugboats, and the magnitude of the resultant is 6000 poundsdirected along the axis of the barge (see figure). Find thetension in the tow lines if they each make an angle withthe axis of the barge.

82. Rope Tension To carry a 100-pound cylindrical weight,two people lift on the ends of short ropes that are tied to aneyelet on the top center of the cylinder. Each rope makes a

angle with the vertical. Draw a figure that gives a visualrepresentation of the problem, and find the tension in theropes.

83. Navigation An airplane is flying in the direction of with an airspeed of 875 kilometers per hour.Because of the wind, its groundspeed and direction are 800 kilometers per hour and respectively (see figure).Find the direction and speed of the wind.

x

140°148°

875 kilometers per hour

800 kilometers per hour

y

Win

d

S

EW

N

140�,

148�,

20�

18°

18°

18�

A

C

B

10 in. 20 in.

24 in.

5000 lb

A

C

B50° 30°

2000 lb

6�

35˚

70 ftsec

35�

x135�,45�,�30�,

x120�,45�,30�,

30°

−45°

900 newtons

x

2000 newtons

x�45�,30�

x300 newtons

125 newtons

45°

y

45�.


333202_0603.qxd 12/5/05 10:42 AM Page 458


85. Work A heavy implement is pulled 30 feet across a floor,using a force of 100 pounds. The force is exerted at anangle of above the horizontal (see figure). Find thework done. (Use the formula for work, where is the component of the force in the direction of motion and

is the distance.)

FIGURE FOR 85 FIGURE FOR 86

86. Rope Tension A tetherball weighing 1 pound is pulledoutward from the pole by a horizontal force until the ropemakes a angle with the pole (see figure). Determine theresulting tension in the rope and the magnitude of .

Synthesis

True or False? In Exercises 87 and 88, decide whether thestatement is true or false. Justify your answer.

87. If and have the same magnitude and direction, then

88. If is a unit vector, then

89. Think About It Consider two forces of equal magnitudeacting on a point.

(a) If the magnitude of the resultant is the sum of themagnitudes of the two forces, make a conjecture aboutthe angle between the forces.

(b) If the resultant of the forces is make a conjectureabout the angle between the forces.

(c) Can the magnitude of the resultant be greater than thesum of the magnitudes of the two forces? Explain.

90. Graphical Reasoning Consider two forces

(a) Find as a function of

(b) Use a graphing utility to graph the function in part (a)for

(c) Use the graph in part (b) to determine the range of thefunction. What is its maximum, and for what value of

does it occur? What is its minimum, and for whatvalue of does it occur?

(d) Explain why the magnitude of the resultant is never 0.

91. Proof Prove that is a unit vector forany value of

92. Technology Write a program for your graphing utilitythat graphs two vectors and their difference given thevectors in component form.

In Exercises 93 and 94, use the program in Exercise 92 tofind the difference of the vectors shown in the figure.

93. 94.

Skills Review

In Exercises 95–98, use the trigonometric substitution towrite the algebraic expression as a trigonometric functionof where

95.

96.

97.

98.

In Exercises 99–102, solve the equation.

99.

100.

101.

102. cos x csc x � cos x�2 � 0

3 sec x sin x � 2�3 sin x � 0

sin x�2 sin x � �2� � 0

cos x�cos x � 1� � 0

x � 5 sec ��x 2 � 25�3,

x � 6 tan ��x 2 � 36,

x � 8 sin ��64 � x 2,

x � 8 sec ��x 2 � 64,

0 < � < �/2.�,

x

(80, 80)

(10, 60)

(−20, 70)

(−100, 0) 50−50

125

y

x2 4 6 8

2

4

6

8(1, 6)

(5, 2)

(4, 5)

(9, 4)

y

�.�cos ��i � �sin ��j

��

0 ≤ � < 2�.

�.�F1 � F2 �

F1 � �10, 0� and F2 � 5�cos �, sin ��.

0,

a 2 � b2 � 1.u � ai � bj

u � v.vu

u45�

u

u

1 lb

Tension

45°50°

30 ft

100 lb

D

FW � FD,50�

84. Navigation A commercial jet is flying from Miami toSeattle. The jet’s velocity with respect to the air is 580miles per hour, and its bearing is The wind, at thealtitude of the plane, is blowing from the southwestwith a velocity of 60 miles per hour.

(a) Draw a figure that gives a visual representation ofthe problem.

(b) Write the velocity of the wind as a vector incomponent form.

(c) Write the velocity of the jet relative to the air incomponent form.

(d) What is the speed of the jet with respect to theground?

(e) What is the true direction of the jet?

332�.

Model It

333202_0603.qxd 12/5/05 10:42 AM Page 459

The Dot Product of Two VectorsSo far you have studied two vector operations—vector addition and multiplica-tion by a scalar—each of which yields another vector. In this section, you willstudy a third vector operation, the dot product. This product yields a scalar,rather than a vector.

For proofs of the properties of the dot product, see Proofs in Mathematics onpage 492.

Finding Dot Products

Find each dot product.

a. b. c.

Solutiona.

b.

c.

Now try Exercise 1.

In Example 1, be sure you see that the dot product of two vectors is a scalar(a real number), not a vector. Moreover, notice that the dot product can bepositive, zero, or negative.

�0, 3� � �4, �2� � 0�4� � 3��2� � 0 � 6 � �6

�2, �1� � �1, 2� � 2�1� � ��1��2� � 2 � 2 � 0

� 23

� 8 � 15

�4, 5� � �2, 3� � 4�2� � 5�3�

�0, 3� � �4, �2��2, �1� � �1, 2��4, 5� � �2, 3�


What you should learn• Find the dot product of two

vectors and use the Propertiesof the Dot Product.

• Find the angle between two vectors and determinewhether two vectors areorthogonal.

• Write a vector as the sum oftwo vector components.

• Use vectors to find the workdone by a force.

Why you should learn itYou can use the dot product of two vectors to solve real-lifeproblems involving two vectorquantities. For instance, inExercise 68 on page 468, you can use the dot product to findthe force necessary to keep asport utility vehicle from rollingdown a hill.

Vectors and Dot Products

Edward Ewert

6.4

Definition of the Dot ProductThe dot product of and is

u � v � u1v1 � u2v2.

v � �v1, v2�u � �u1, u2�

Properties of the Dot ProductLet , , and be vectors in the plane or in space and let be a scalar.

1.

2.

3.

4.

5. c�u � v� � cu � v � u � cv

v � v � �v �2

u � �v � w� � u � v � u � w

0 � v � 0

u � v � v � u

cwvu

Example 1

333202_0604.qxd 12/5/05 10:44 AM Page 460

Using Properties of Dot Products

Let and Find each dot product.

a. b.

SolutionBegin by finding the dot product of and

a.

b.

Notice that the product in part (a) is a vector, whereas the product in part (b) is ascalar. Can you see why?


Dot Product and Magnitude

The dot product of with itself is 5. What is the magnitude of ?

SolutionBecause and it follows that


The Angle Between Two VectorsThe angle between two nonzero vectors is the angle betweentheir respective standard position vectors, as shown in Figure 6.33. This anglecan be found using the dot product. (Note that the angle between the zero vectorand another vector is not defined.)

For a proof of the angle between two vectors, see Proofs in Mathematics on page492.

0 ≤ � ≤ �,�,

��5.

�u� ��u � u

u � u � 5,�u �2 � u � u

uu

� �28

� 2��14�u � 2v � 2�u � v�

� ��14, 28��u � v�w � �14�1, �2�

� �14

� ��1��2� � 3��4�

u � v � ��1, 3� � �2, �4�

v.u

u � 2v�u � v�w

w � �1, �2�.v � �2, �4�,u � ��1, 3�,

Section 6.4 Vectors and Dot Products 461

Example 2

Example 3

Angle Between Two VectorsIf is the angle between two nonzero vectors and then

cos � �u � v

� u� �v �.

v,u�

−

vu

Origin

θ

v u

FIGURE 6.33

333202_0604.qxd 12/5/05 10:44 AM Page 461

Finding the Angle Between Two Vectors

Find the angle between and

Solution

This implies that the angle between the two vectors is

as shown in Figure 6.34.


Rewriting the expression for the angle between two vectors in the form

Alternative form of dot product

produces an alternative way to calculate the dot product. From this form, you cansee that because and are always positive, and will alwayshave the same sign. Figure 6.35 shows the five possible orientations of twovectors.

The terms orthogonal and perpendicular mean essentially the samething—meeting at right angles. Even though the angle between the zero vectorand another vector is not defined, it is convenient to extend the definition oforthogonality to include the zero vector. In other words, the zero vector isorthogonal to every vector u, because 0 � u � 0.

cos �u � v�v �� u�

u � v � � u� �v � cos �

� � arccos 27

5�34� 22.2�

�27

5�34

��4, 3� � �3, 5�

� �4, 3�� 3, 5��

cos � �u � v

� u� �v �

v � �3, 5�.u � �4, 3�


Example 4

u v

θ

Opposite DirectionFIGURE 6.35

cos � � �1� � �

u

v

θ

Obtuse Angle�1 < cos � < 0

�

2< � < �

u

v

θ

Angle90�

cos � � 0

� ��

2

u

vθ

Acute Angle0 < cos � < 1

0 < � <�

2

u

v

Same Directioncos � � 1� � 0

Definition of Orthogonal VectorsThe vectors and are orthogonal if u � v � 0.vu

x1 2 3 4 5 6

1

2

3

4

5

6v = ⟨3, 5⟩

u = ⟨4, 3⟩

θ

y

FIGURE 6.34

333202_0604.qxd 12/8/05 9:34 AM Page 462

Determining Orthogonal Vectors

Are the vectors and orthogonal?

SolutionBegin by finding the dot product of the two vectors.

Because the dot product is 0, the two vectors are orthogonal (see Figure 6.36).

FIGURE 6.36


Finding Vector ComponentsYou have already seen applications in which two vectors are added to produce aresultant vector. Many applications in physics and engineering pose the reverseproblem—decomposing a given vector into the sum of two vector components.

Consider a boat on an inclined ramp, as shown in Figure 6.37. The force due to gravity pulls the boat down the ramp and against the ramp. These twoorthogonal forces, and are vector components of . That is,

Vector components of

The negative of component represents the force needed to keep the boat fromrolling down the ramp, whereas represents the force that the tires mustwithstand against the ramp. A procedure for finding and is shown on thefollowing page.

FIGURE 6.37

Fw2

w1

w2w1

w2

w1

FF � w1 � w2.

Fw2,w1

F

v = ⟨6, 4⟩

u = ⟨2, −3⟩

x1 2 3 4 5 6 7

1

2

3

4

−1

−2

−3

y

� 0� 2�6� � ��3��4� u � v � �2, �3� � �6, 4�

v � �6, 4�u � �2, �3�


Example 5

The graphing utility programFinding the Angle Between TwoVectors, found on our website college.hmco.com, graphs two vectors and in standard position and finds themeasure of the angle betweenthem. Use the program to verifythe solutions for Examples 4 and 5.

v � �c , d�u � �a, b�

Techno logy

333202_0604.qxd 12/5/05 10:44 AM Page 463

From the definition of vector components, you can see that it is easy to findthe component once you have found the projection of onto . To find theprojection, you can use the dot product, as follows.

is a scalar multiple of

Take dot product of each side with

and are orthogonal.

So,

and

w1 � projv u � cv �u � v�v�2 v.

c �u � v�v�2

vw2 � c�v�2 � 0

� cv � v � w2 � v

v. u � v � �cv � w2� � v

v.w1 u � w1 � w2 � cv � w2

vuw2


Definition of Vector ComponentsLet and be nonzero vectors such that

where and are orthogonal and is parallel to (or a scalar multiple of), as shown in Figure 6.38. The vectors and are called vector compo-

nents of . The vector is the projection of onto and is denoted by

The vector is given by w2 � u � w1.w2

w1 � projvu.

vuw1uw2w1v

w1w2w1

u � w1 � w2

vu

Projection of u onto vLet and be nonzero vectors. The projection of onto is

projvu � �u � v�v �2 v.

vuvu

w1

w2 u

vθ

is acute.FIGURE 6.38�

w2

w1

u

vθ

is obtuse.�

333202_0604.qxd 12/5/05 10:44 AM Page 464

Decomposing a Vector into Components

Find the projection of onto Then write as the sum oftwo orthogonal vectors, one of which is

SolutionThe projection of onto is

as shown in Figure 6.39. The other component, is

So,


Finding a Force

A 200-pound cart sits on a ramp inclined at 30 , as shown in Figure 6.40. Whatforce is required to keep the cart from rolling down the ramp?

SolutionBecause the force due to gravity is vertical and downward, you can represent thegravitational force by the vector

Force due to gravity

To find the force required to keep the cart from rolling down the ramp, project onto a unit vector in the direction of the ramp, as follows.

Unit vector along ramp

Therefore, the projection of onto is

The magnitude of this force is 100, and so a force of 100 pounds is required tokeep the cart from rolling down the ramp.


� �100��3

2i �

1

2j.

� ��200��1

2v

� �F � v�v

� �F � v�v �2 v

w1 � projvF

vF

v � �cos 30��i � �sin 30��j ��3

2i �

1

2j

vF

F � �200j.

�

u � w1 � w2 � 6

5,

2

5� � 9

5, �

27

5 � � �3, �5�.

w2 � u � w1 � �3, �5� � 6

5,

2

5� � 9

5,�

27

5 �.

w2,

w1 � projvu � �u � v�v �2 v � � 8

40�6, 2� � 6

5,

2

5�vu

projvu.uv � �6, 2�.u � �3, �5�


x−1 1 2 3 4 5 6

1

2

−1

−2

−3

−4

−5

v = ⟨6, 2⟩

u = ⟨3, −5⟩

w1

w2

y

FIGURE 6.39

F

v w1

30°

FIGURE 6.40

Example 6

Example 7

333202_0604.qxd 12/5/05 10:44 AM Page 465

WorkThe work done by a constant force acting along the line of motion of anobject is given by

as shown in Figure 6.41. If the constant force is not directed along the line ofmotion, as shown in Figure 6.42, the work done by the force is given by

Projection form for work

Alternative form of dot product

Force acts along the line of motion. Force acts at angle with the line of motion.FIGURE 6.41 FIGURE 6.42

This notion of work is summarized in the following definition.

Finding Work

To close a sliding door, a person pulls on a rope with a constant force of 50pounds at a constant angle of 60 , as shown in Figure 6.43. Find the work donein moving the door 12 feet to its closed position.

SolutionUsing a projection, you can calculate the work as follows.

Projection form for work

foot-pounds

So, the work done is 300 foot-pounds. You can verify this result by finding thevectors and and calculating their dot product.


PQ\

F

� 300 �1

2�50��12�

� �cos 60��F � �PQ\

�

W � �proj PQ\ F � � PQ

\

�

�

�

P Q

F

Fθ

projPQ

P Q

F

� F � PQ\

.

�projPQ\

F � � �cos �� F �� cos �� F � �PQ\

�

W � �projPQ\

F � �PQ\

�

WF

� �F � � PQ\

� W � �magnitude of force��distance�

FW


Definition of WorkThe work done by a constant force as its point of application movesalong the vector is given by either of the following.

1. Projection form

2. Dot product formW � F � PQ\

W � �projPQ\F� �PQ

\

�

PQ\

FW

Example 8

12 ft

12 ft

F

60°P Q

projPQF

FIGURE 6.43

333202_0604.qxd 12/5/05 10:44 AM Page 466


In Exercises 1–8, find the dot product of and

1. 2.

3. 4.

5. 6.

7. 8.

In Exercises 9–18, use the vectors and to find the indicated quantity. Statewhether the result is a vector or a scalar.

9. 10.

11. 12.

13. 14.

15. 16.

17. 18.

In Exercises 19–24, use the dot product to find themagnitude of u.

19. 20.

21. 22.

23. 24.

In Exercises 25 –34, find the angle between the vectors.

25. 26.

27. 28.

29. 30.

31. 32.

33.

34.

In Exercises 35–38, graph the vectors and find the degreemeasure of the angle between the vectors.

35. 36.

37. 38.

In Exercises 39–42, use vectors to find the interior angles ofthe triangle with the given vertices.

39. 40.

41. 42.

In Exercises 43–46, find where is the angle betweenand

43.

44.

45.

46. �u� � 4, �v� � 12, � ��

3

�u� � 9, �v� � 36, � �3�

4

�u � � 100, � v � � 250, � ��

6

�u � � 4, �v � � 10, � �2�

3

v.u�u � v,

�7, 9��1, 9�,��3, 5�,��3, 0�, �2, 2�, �0, 6)

�8, 2��1, 7�,��3, �4�,�1, 2�, �3, 4�, �2, 5�

�v � 8i � 3jv � �8i � 8j

u � 2i � 3ju � 5i � 5j

v � �4i � 4jv � �7i � 5j

u � 6i � 3ju � 3i � 4j

�

v � cos��

2 i � sin��

2 j

u � cos��

4 i � sin��

4 j

v � cos�3�

4 i � sin�3�

4 j

u � cos��

3 i � sin��

3 j

v � 4i � 3jv � �6i � 6j

u � 2i � 3ju � 5i � 5j

v � �8i � 4jv � 6i � 4j

u � �6i � 3ju � 2i � j

v � i � 2jv � �2j

u � 2i � 3ju � 3i � 4j

v � �4, 0�v � �0, �2�u � �3, 2�u � �1, 0�

�

u � �21iu � 6j

u � 12i � 16ju � 20i � 25j

u � �2, �4�u � ��5, 12�

�v � u� � �w � v��u � v� � �u � �w�2 � �u��w� � 1

�u � 2v�w�3w � v�u�v � u�w�u � v�v3u � vu � u

w � <1, �2>v � <�3, 4>,u � <2, 2>,

v � �2i � jv � �2i � 3j

u � i � 2ju � 3i � 2j

v � 7i � 2jv � i � j

u � 3i � 4ju � 4i � 2j

v � ��1, �2�v � �2, �3�u � ��2, 5�u � ��4, 1�v � ��3, 2�v � ��2, 3�u � �5, 12�u � �6, 1�

v.u

Exercises 6.4


1. The ________ ________ of two vectors yields a scalar, rather than a vector.

2. If is the angle between two nonzero vectors and then

3. The vectors and are ________ if

4. The projection of onto is given by

5. The work done by a constant force as its point of application moves along the vector is given by or


W � ________ .W � ________PQ

\

FW

projvu � ________ .vu

u � v � 0.vu

cos � � ________ .v,u�

333202_0604.qxd 12/5/05 10:44 AM Page 467

www.Eduspace.com

In Exercises 47–52, determine whether and are orthog-onal, parallel, or neither.

47. 48.

49. 50.

51. 52.

In Exercises 53–56, find the projection of onto . Thenwrite as the sum of two orthogonal vectors, one of whichis

53. 54.

55. 56.

In Exercises 57 and 58, use the graph to determine mentallythe projection of onto . (The coordinates of the terminalpoints of the vectors in standard position are given.) Usethe formula for the projection of onto to verify yourresult.

57. 58.

In Exercises 59–62, find two vectors in opposite directionsthat are orthogonal to the vector (There are manycorrect answers.)

59.

60.

61.

62.

Work In Exercises 63 and 64, find the work done inmoving a particle from to if the magnitude anddirection of the force are given by

63.

64.

65. Revenue The vector gives the numbersof units of two types of baking pans produced by acompany. The vector gives the prices(in dollars) of the two types of pans, respectively.

(a) Find the dot product and interpret the result in thecontext of the problem.

(b) Identify the vector operation used to increase the pricesby 5%.

66. Revenue The vector gives the numbersof hamburgers and hot dogs, respectively, sold at a fast-foodstand in one month. The vector gives theprices (in dollars) of the food items.

(a) Find the dot product and interpret the result in thecontext of the problem.

(b) Identify the vector operation used to increase the pricesby 2.5%.

68. Braking Load A sport utility vehicle with a gross weightof 5400 pounds is parked on a slope of Assume thatthe only force to overcome is the force of gravity. Find theforce required to keep the vehicle from rolling down thehill. Find the force perpendicular to the hill.

10�.

u � v

v � �1.75, 1.25�

u � �3240, 2450�

u � v

v � �15.25, 10.50�

u � �1650, 3200�

P � �1, 3�, Q � ��3, 5�, v � �2i � 3j

P � �0, 0�, Q � �4, 7�, v � �1, 4�

v.QP

u � �52 i � 3j

u �12 i �

23 j

u � ��8, 3�u � �3, 5�

u.

2−2−2

−4

4 6

2

4(6, 4)

(2, −3)u

v

x

y

2−2−2

4 6

6(6, 4)

(−2, 3)

u

v

x

y

vu

vu

v � ��4, �1�v � �2, 15�u � ��3, �2�u � �0, 3�v � �1, �2�v � �6, 1�u � �4, 2�u � �2, 2�

projv u.u

vu

v � �sin �, �cos ��v � �i � j

u � �cos �, sin ��u � 2i � 2j

v � �2i � 2jv � 5i � 6j

u � iu �14�3i � j�

v � ��1, 5�v � � 12, �5

4�u � �3, 15�u � ��12, 30�

vu


67. Braking Load A truck with a gross weight of 30,000pounds is parked on a slope of (see figure). Assumethat the only force to overcome is the force of gravity.

(a) Find the force required to keep the truck fromrolling down the hill in terms of the slope

(b) Use a graphing utility to complete the table.

(c) Find the force perpendicular to the hill whend � 5�.

d.

d°

Weight = 30,000 lb

d �

Model It

d

Force

5�4�3�2�1�0�

d

Force

10�9�8�7�6�

333202_0604.qxd 12/5/05 10:44 AM Page 468


69. Work Determine the work done by a person lifting a 25-kilogram (245-newton) bag of sugar.

70. Work Determine the work done by a crane lifting a 2400-pound car 5 feet.

71. Work A force of 45 pounds exerted at an angle of above the horizontal is required to slide a table across afloor (see figure). The table is dragged 20 feet. Determinethe work done in sliding the table.

72. Work A tractor pulls a log 800 meters, and the tension inthe cable connecting the tractor and log is approximately1600 kilograms (15,691 newtons). The direction of theforce is above the horizontal. Approximate the workdone in pulling the log.

73. Work One of the events in a local strongman contest is topull a cement block 100 feet. One competitor pulls theblock by exerting a force of 250 pounds on a rope attachedto the block at an angle of with the horizontal (seefigure). Find the work done in pulling the block.

74. Work A toy wagon is pulled by exerting a force of 25 pounds on a handle that makes a angle with the horizontal (see figure). Find the work done in pulling thewagon 50 feet.

Synthesis

True or False? In Exercises 75 and 76, determine whetherthe statement is true or false. Justify your answer.

75. The work done by a constant force acting along theline of motion of an object is represented by a vector.

76. A sliding door moves along the line of vector If aforce is applied to the door along a vector that is orthogo-nal to then no work is done.

77. Think About It What is known about , the anglebetween two nonzero vectors and , under eachcondition?

(a) (b) (c)

78. Think About It What can be said about the vectors andunder each condition?

(a) The projection of onto equals .

(b) The projection of onto equals .

79. Proof Use vectors to prove that the diagonals of arhombus are perpendicular.

80. Proof Prove the following.

Skills Review

In Exercises 81–84, perform the operation and write theresult in standard form.

81.

82.

83.

84.

In Exercises 85–88, find all solutions of the equation in theinterval

85.

86.

87.

88.

In Exercises 89–92, find the exact value of thetrigonometric function given that and

(Both and are in Quadrant IV.)

89.

90.

91.

92. tan�u � v�cos�v � u�sin�u � v�sin�u � v�

vucos v � 2425.

sin u � �1213

cos 2x � 3 sin x � 2

2 tan x � tan 2x

sin 2x ��2 cos x � 0

sin 2x ��3 sin x � 0

[0, 2��.

��12 � ��96

��3 � ��8

�18 � �112

�42 � �24

�u � v �2 � � u�2 � �v �2 � 2u � v

0vu

uvu

vu

u � v < 0u � v > 0u � v � 0

vu�

PQ\

,

PQ\

.

FW

20°

20�

30˚

100 ftNot drawn to scale

30�

35�

45 lb

30°

20 ft

30�

333202_0604.qxd 12/8/05 10:08 AM Page 469

The Complex PlaneJust as real numbers can be represented by points on the real number line, youcan represent a complex number

as the point in a coordinate plane (the complex plane). The horizontal axisis called the real axis and the vertical axis is called the imaginary axis, as shownin Figure 6.44.

FIGURE 6.44

The absolute value of the complex number is defined as the distancebetween the origin and the point

If the complex number is a real number (that is, if ), then thisdefinition agrees with that given for the absolute value of a real number

Finding the Absolute Value of a Complex Number

Plot and find its absolute value.

SolutionThe number is plotted in Figure 6.45. It has an absolute value of

Now try Exercise 3.

� �29.

�z� � ��2�2 � 52

z � �2 � 5i

�a � 0i� � �a2 � 02 � �a�.

b � 0a � bi

�a, b�.�0, 0�a � bi

Realaxis1

3

2

1

−1

−2

2 3−2 −1−3

(3, 1)or3 + i

−2 − i(−2, −1) or

Imaginaryaxis

�a, b�

z � a � bi


What you should learn• Plot complex numbers in

the complex plane and findabsolute values of complexnumbers.

• Write the trigonometric forms of complex numbers.

• Multiply and divide complex numbers written in trigonometric form.

• Use DeMoivre’s Theorem to find powers of complexnumbers.

• Find th roots of complexnumbers.

Why you should learn itYou can use the trigonometricform of a complex number to perform operations withcomplex numbers. For instance,in Exercises 105–112 on page 480,you can use the trigonometricforms of complex numbers tohelp you solve polynomialequations.

n

Trigonometric Form of a Complex Number6.5

1

3

4

5

2 3−2 −1−3−4 4

29

Realaxis

(−2, 5)

Imaginaryaxis

FIGURE 6.45

Definition of the Absolute Value of a Complex NumberThe absolute value of the complex number is

�a � bi� � �a2 � b2.

z � a � bi

Example 1

333202_0605.qxd 12/5/05 10:45 AM Page 470

Trigonometric Form of a Complex NumberIn Section 2.4, you learned how to add, subtract, multiply, and divide complexnumbers. To work effectively with powers and roots of complex numbers, it ishelpful to write complex numbers in trigonometric form. In Figure 6.46, considerthe nonzero complex number By letting be the angle from the positivereal axis (measured counterclockwise) to the line segment connecting the originand the point you can write

and

where Consequently, you have

from which you can obtain the trigonometric form of a complex number.

The trigonometric form of a complex number is also called the polar form.Because there are infinitely many choices for the trigonometric form of acomplex number is not unique. Normally, is restricted to the interval

although on occasion it is convenient to use

Writing a Complex Number in Trigonometric Form

Write the complex number in trigonometric form.

SolutionThe absolute value of is

and the reference angle is given by

Because and because lies in Quadrant III, youchoose to be So, the trigonometric form is

See Figure 6.47.


� 4�cos 4�

3� i sin

4�

3 �.

z � r�cos � � i sin ��

� � � � ��3 � 4��3.�z � �2 � 2�3 itan��3� � �3

tan �� b

a�

�2�3

�2� �3.

��

r � ��2 � 2�3 i� � ��2�2 � ��2�3 �2� �16 � 4

z

z � �2 � 2�3 i

� < 0.0 ≤ � < 2�,��,

a � bi � �r cos �� r sin ��i

r � �a2 � b2.

b � r sin �a � r cos �

�a, b�,

�a � bi.

Section 6.5 Trigonometric Form of a Complex Number 471

Trigonometric Form of a Complex NumberThe trigonometric form of the complex number is

where and Thenumber is the modulus of and is called an argument of z.�z,r

tan � � b�a.a � r cos �, b � r sin �, r � �a2 � b2,

z � r�cos � � i sin ��

z � a � bi

θ

rb

a

(a ,b)

Realaxis

Imaginaryaxis

FIGURE 6.46

1

−3

−2

−4

−2−33

4

z = 4

πRealaxis

z = −2 − 2 3 i

Imaginaryaxis

FIGURE 6.47

Example 2

333202_0605.qxd 12/5/05 10:45 AM Page 471

Writing a Complex Number in Standard Form

Write the complex number in standard form

SolutionBecause and you can write


Multiplication and Division of Complex NumbersThe trigonometric form adapts nicely to multiplication and division of complexnumbers. Suppose you are given two complex numbers

and

The product of and is given by

Using the sum and difference formulas for cosine and sine, you can rewrite thisequation as

This establishes the first part of the following rule. The second part is left for youto verify (see Exercise 117).

Note that this rule says that to multiply two complex numbers you multiplymoduli and add arguments, whereas to divide two complex numbers you dividemoduli and subtract arguments.

z1z2 � r1r2�cos��1 � �2� � i sin��1 � �2�.

� r1r2��cos �1 cos �2 � sin �1 sin �2� � i�sin �1 cos �2 � cos �1 sin �2�.

z1z2 � r1r2�cos �1 � i sin �1��cos �2 � i sin �2�

z2z1

z2 � r2�cos �2 � i sin �2�.z1 � r1�cos �1 � i sin �1�

� �2 � �6 i.

� 2�2�1

2�

�3

2i�

z � �8cos��

3� � i sin��

3��sin��3� � ��3�2,cos��3� �

12

z � �8cos��

3� � i sin��

3��a � bi.


Product and Quotient of Two Complex NumbersLet and be complexnumbers.

Product

Quotientz2 � 0z1

z2

�r1

r2

�cos��1 � �2� � i sin��1 � �2�,

z1z2 � r1r2�cos��1 � �2� � i sin��1 � �2�

z2 � r2�cos �2 � i sin �2�z1 � r1�cos �1 � i sin �1�

Example 3

A graphing utility can be used to convert a complex number in trigonometric (or polar) form to standard form. For specifickeystrokes, see the user’s manualfor your graphing utility.

Techno logy

333202_0605.qxd 12/5/05 10:45 AM Page 472

Multiplying Complex Numbers

Find the product of the complex numbers.

Solution

You can check this result by first converting the complex numbers to the standardforms and and then multiplying algebraically,as in Section 2.4.


Dividing Complex Numbers

Find the quotient of the complex numbers.

Solution


� �3�2

2�

3�2

2i

� 3��2

2 � � i��2

2 �� 3�cos 225� � i sin 225��

�24

8�cos�300� � 75�� i sin�300� � 75��

z1

z2

�24�cos 300� � i sin 300��

8�cos 75� � i sin 75��

z2 � 8�cos 75� � i sin 75��z1 � 24�cos 300� � i sin 300��

z1�z2

� 16i

� �4�3 � 4i � 12i � 4�3

z1z2 � ��1 � �3 i��4�3 � 4i�

z2 � 4�3 � 4iz1 � �1 � �3 i

� 16i

� 16�0 � i�1�

� 16�cos �

2� i sin

�

2�

� 16�cos 5�

2� i sin

5�

2 �

� 16cos� 2�

3�

11�

6 � � i sin� 2�

3�

11�

6 ��

z1z2 � 2�cos 2�

3� i sin

2�

3 � 8�cos 11�

6� i sin

11�

6 �

z2 � 8�cos 11�

6� i sin

11�

6 �z1 � 2�cos 2�

3� i sin

2�

3 �z1z2


Example 4

Example 5

Some graphing utilities can multiplyand divide complex numbers intrigonometric form. If you haveaccess to such a graphing utility, useit to find and in Examples4 and 5.

z1�z2z1z2

Techno logy

Multiply moduli and add arguments.

Divide moduli andsubtract arguments.

333202_0605.qxd 12/5/05 10:45 AM Page 473

Powers of Complex NumbersThe trigonometric form of a complex number is used to raise a complex numberto a power. To accomplish this, consider repeated use of the multiplication rule.

This pattern leads to DeMoivre’s Theorem, which is named after the Frenchmathematician Abraham DeMoivre (1667–1754).

Finding Powers of a Complex Number

Use DeMoivre’s Theorem to find

SolutionFirst convert the complex number to trigonometric form using

and

So, the trigonometric form is

Then, by DeMoivre’s Theorem, you have


� 4096.

� 4096�1 � 0�

� 4096�cos 8� � i sin 8��

� 212cos 12�2��

3� i sin

12�2��3 �

��1 � �3 i�12 � 2�cos 2�

3� i sin

2�

3 ��12

z � �1 � �3 i � 2�cos 2�

3� i sin

2�

3 �.

� � arctan �3�1

�2�

3.r � ��1�2 � ��3�2

� 2

��1 � �3 i�12.

. . .

z5 � r5�cos 5� � i sin 5��

z4 � r4�cos 4� � i sin 4��

z3 � r2�cos 2� � i sin 2��r�cos � � i sin �� r3�cos 3� � i sin 3��

z2 � r �cos � � i sin ��r�cos � � i sin �� r2�cos 2� � i sin 2��

z � r �cos � � i sin ��


DeMoivre’s TheoremIf is a complex number and is a positive integer,then

� rn �cos n� � i sin n��.

zn � �r�cos � � i sin ��n

nz � r�cos � � i sin ��

Example 6Historical NoteAbraham DeMoivre(1667–1754) is remembered forhis work in probability theoryand DeMoivre’s Theorem. Hisbook The Doctrine of Chances(published in 1718) includesthe theory of recurring seriesand the theory of partialfractions.

The

Gra

ng

er C

olle

ctio

n

333202_0605.qxd 12/5/05 10:45 AM Page 474

Roots of Complex NumbersRecall that a consequence of the Fundamental Theorem of Algebra is that apolynomial equation of degree has solutions in the complex number system.So, the equation has six solutions, and in this particular case you can findthe six solutions by factoring and using the Quadratic Formula.

Consequently, the solutions are

and

Each of these numbers is a sixth root of 1. In general, the nth root of a complexnumber is defined as follows.

To find a formula for an th root of a complex number, let be an th rootof where

and

By DeMoivre’s Theorem and the fact that you have

Taking the absolute value of each side of this equation, it follows that Substituting back into the previous equation and dividing by you get

So, it follows that

and

Because both sine and cosine have a period of these last two equations havesolutions if and only if the angles differ by a multiple of Consequently, theremust exist an integer such that

By substituting this value of into the trigonometric form of you get the resultstated on the following page.

u,

�� 2�k

n.

n � � � 2�k

k2�.

2�,

sin n � sin �.cos n � cos �

cos n � i sin n � cos � � i sin �.

r,sn � r.

sn �cos n � i sin n� � r�cos � � i sin ��.

un � z,

z � r �cos � � i sin ��.

u � s�cos � i sin �

z,nun

x �1 ± �3 i

2.x �

�1 ± �3 i

2,x � ±1,

� �x � 1��x2 � x � 1��x � 1��x2 � x � 1� � 0

x6 � 1 � �x3 � 1��x3 � 1�

x6 � 1nn


Definition of the nth Root of a Complex Number The complex number is an nth root of the complex number if

z � un � �a � bi�n.

zu � a � bi

The th roots of a complexnumber are useful for solvingsome polynomial equations. Forinstance, explain how you canuse DeMoivre’s Theorem tosolve the polynomial equation

[Hint: Write as]16�cos � � i sin ��.

�16

x4 � 16 � 0.

n

Exploration

333202_0605.qxd 12/5/05 10:45 AM 5

When exceeds the roots begin to repeat. For instance, if theangle

is coterminal with which is also obtained when The formula for the th roots of a complex number has a nice geometrical

interpretation, as shown in Figure 6.48. Note that because the th roots of allhave the same magnitude they all lie on a circle of radius with center atthe origin. Furthermore, because successive th roots have arguments that differby the roots are equally spaced around the circle.

You have already found the sixth roots of 1 by factoring and by using theQuadratic Formula. Example 7 shows how you can solve the same problem withthe formula for th roots.

Finding the nth Roots of a Real Number

Find all the sixth roots of 1.

SolutionFirst write 1 in the trigonometric form Then, by the throot formula, with and the roots have the form

So, for 1, 2, 3, 4, and 5, the sixth roots are as follows. (See Figure 6.49.)

Increment by


cos 5�

3� i sin

5�

3�

1

2�

�3

2 i

cos 4�

3� i sin

4�

3� �

1

2�

�3

2 i

cos � � i sin � � �1

cos 2�

3� i sin

2�

3� �

1

2�

�3

2 i

2�

n�

2�

6�

�

3 cos

�

3� i sin

�

3�

1

2�

�3

2 i

cos 0 � i sin 0 � 1

k � 0,

6�1�cos 0 � 2�k

6� i sin

0 � 2�k

6 � � cos �k

3� i sin

�k

3.

r � 1,n � 6n1 � 1�cos 0 � i sin 0�.

n

n2��n,n

n�rn�r,zn

znk � 0.��n,

� � 2�n

n�

�

n� 2�

k � n,n � 1,k


2

2n

nrn

π

πRealaxis

Imaginaryaxis

FIGURE 6.48

+

1−1

Realaxis

−1 + 0i 1 + 0i

12

++ i32

12

− i12

32

− i12

32

−

i32

−

Imaginaryaxis

FIGURE 6.49

Finding nth Roots of a Complex NumberFor a positive integer the complex number hasexactly distinct th roots given by

where 1, 2, . . . , n � 1.k � 0,

n�r�cos � � 2�k

n� i sin

� � 2�kn �

nnz � r�cos � � i sin ��n,

Example 7

333202_0605.qxd 12/5/05 10:45 AM Page 476

In Figure 8.49, notice that the roots obtained in Example 7 all have a mag-nitude of 1 and are equally spaced around the unit circle. Also notice that thecomplex roots occur in conjugate pairs, as discussed in Section 2.5. The dis-tinct th roots of 1 are called the nth roots of unity.

Finding the nth Roots of a Complex Number

Find the three cube roots of

SolutionBecause lies in Quadrant II, the trigonometric form of is

By the formula for th roots, the cube roots have the form

Finally, for you obtain the roots

See Figure 6.50.


� 0.3660 � 1.3660i.

6�8�cos 135� � 360��2�

3� i sin

135� � 360��2�3 � ��2�cos 285� � i sin 285��

� �1.3660 � 0.3660i

6�8�cos 135� � 360��1�

3� i sin

135� � 360��1�3 � ��2�cos 165� � i sin 165��

� 1 � i

6�8�cos 135� � 360��0�

3� i sin

135� � 360��0�3 � ��2�cos 45� � i sin 45��

k � 0, 1, and 2,

6�8 �cos 135� � 360�k

3� i sin

135º � 360�k

3 �.

n

� � arctan�2��2� � 135� � �8 �cos 135� � i sin 135��.

z � �2 � 2i

zz

z � �2 � 2i.

nn


Example 8


A Famous Mathematical Formula The famous formula

is called Euler’s Formula, after the Swiss mathematician Leonhard Euler (1707–1783).Although the interpretation of this formula is beyond the scope of this text, wedecided to include it because it gives rise to one of the most wonderful equations inmathematics.

This elegant equation relates the five most famous numbers in mathematics—0, 1,and —in a single equation. Show how Euler’s Formula can be used to derive

this equation.ie,�,

e� i � 1 � 0

ea �bi � e a�cos b � i sin b�

1 + i−1.3660 + 0.3660i

0.3660 − 1.3660i

Realaxis

Imaginaryaxis

1−2

−2

−1

1

2

FIGURE 6.50

Note in Example 8 that theabsolute value of is

and the angle is given by

tan � �ba

�2

�2� �1.

�

� �8

� ��2�2 � 22

r � ��2 � 2i�z

333202_0605.qxd 12/5/05 10:45 AM Page 477

In Exercises 1–6, plot the complex number and find itsabsolute value.

1. 2.

3. 4.

5. 6.

In Exercises 7–10, write the complex number in trigono-metric form.

7. 8.

9. 10.

In Exercises 11–30, represent the complex numbergraphically, and find the trigonometric form of the number.

11. 12.

13. 14.

15. 16.

17. 18.

19. 20.

21. 7 22. 4

23. 24.

25. 26.

27. 28.

29. 30.

In Exercises 31– 40, represent the complex numbergraphically, and find the standard form of the number.

31.

32.

33.

34.

35.

36.

37.

38.

39.

40.

In Exercises 41– 44, use a graphing utility to represent thecomplex number in standard form.

41. 42.

43.

44.

In Exercises 45 and 46, represent the powers and graphically. Describe the pattern.

45. 46. z �12

�1 � �3 i�z ��22

�1 � i�

z 4z3,z2,z,

9�cos 58º � i sin 58º�3�cos 165.5� � i sin 165.5��

10�cos 2�

5� i sin

2�

5 �5�cos �

9� i sin

�

9�

6�cos�230º 30�� i sin�230º 30��3�cos�18� 45�� i sin�18� 45��7�cos 0 � i sin 0�

8�cos �

2� i sin

�

2�

6�cos 5�

12� i sin

5�

12�

3.75�cos 3�

4� i sin

3�

4 �14�cos 225� � i sin 225��

32�cos 300� � i sin 300��5�cos 135� � i sin 135��3�cos 120� � i sin 120��

�9 � 2�10 i�8 � 5�3 i

8 � 3i5 � 2i

1 � 3i�3 � i

2�2 � i3 � �3 i

3 � i�7 � 4i

4i�5i

52��3 � i��2�1 � �3 i�4 � 4�3 i�3 � i

2 � 2i3 � 3i

Realaxis−1−2

3

−3

z i= 1 + 3−

Imaginaryaxis

Realaxis

−3

z i= 3 −3

Imaginaryaxis

s

Realaxis2

z = 2− 2

4

−2

−4

−4−6

Imaginaryaxiaxis

Realaxis1

1234

2−1−2

Imaginary

z = 3i

�8 � 3i6 � 7i

5 � 12i�4 � 4i

�7�7i


Exercises 6.5


1. The ________ ________ of a complex number is the distance between the origin and the point

2. The ________ ________ of a complex number is given by where is the ________ of and is the ________ of

3. ________ Theorem states that if is a complex number and is a positive integer,then

4. The complex number is an ________ ________ of the complex number if


z � un � �a � bi�n.zu � a � bi

zn � r n�cos n� � i sin n��.nz � r �cos � � i sin ��

z.�zrz � r �cos � � i sin ��,z � a � bi

�a, b�.�0, 0�a � bi

333202_0605.qxd 12/8/05 10:09 AM Page 478

www.Eduspace.com


In Exercises 47–58, perform the operation and leave theresult in trigonometric form.

47.

48.

49.

50.

51.

52.

53.

54.

55.

56.

57.

58.

In Exercises 59–66, (a) write the trigonometric forms of thecomplex numbers, (b) perform the indicated operationusing the trigonometric forms, and (c) perform theindicated operation using the standard forms, and checkyour result with that of part (b).

59. 60.

61. 62.

63. 64.

65.

66.

In Exercises 67–70, sketch the graph of all complex numbersz satisfying the given condition.

67.

68.

69.

70.

In Exercises 71–88, use DeMoivre’s Theorem to find theindicated power of the complex number.Write the result instandard form.

71. 72.

73. 74.

75. 76.

77.

78.

79.

80.

81.

82.

83.

84.

85.

86.

87.

88.

In Exercises 89–104, (a) use the theorem on page 476 to findthe indicated roots of the complex number, (b) representeach of the roots graphically, and (c) write each of the rootsin standard form.

89. Square roots of

90. Square roots of

91. Cube roots of

92. Fifth roots of

93. Square roots of

94. Fourth roots of

95. Cube roots of

96. Cube roots of

97. Fourth roots of 16

98. Fourth roots of

99. Fifth roots of 1

100. Cube roots of 1000

101. Cube roots of

102. Fourth roots of

103. Fifth roots of

104. Sixth roots of 64i

128��1 � i��4

�125

i

�4�2�1 � i��125

2 �1 � �3 i�625i

�25i

32�cos 5�

6� i sin

5�

6 �

8�cos 2�

3� i sin

2�

3 �16�cos 60� � i sin 60��5�cos 120� � i sin 120��

�2�cos �

8� i sin

�

8��6

�2�cos �

10� i sin

�

10��5

�2�cos 10� � i sin 10��8

�3�cos 15� � i sin 15��4

��5 � 4i�3

�3 � 2i�5

�cos 0 � i sin 0�20

�5�cos 3.2 � i sin 3.2�4

�2�cos �

2� i sin

�

2��8

�cos �

4� i sin

�

4�12

�3�cos 150� � i sin 150��4

�5�cos 20� � i sin 20��3

4�1 � �3 i�32��3 � i�7

�3 � 2i�8��1 � i�10

�2 � 2i�6�1 � i�5

� �5�

4

� ��

6

z � 3z � 2

4i

�4 � 2i

5

2 � 3i

1 � �3 i6 � 3i

3 � 4i

1 � �3 i

4�1 � �3 i��2i�1 � i��3 � i��1 � i��2 � 2i��1 � i�

6�cos 40� � i sin 40��7�cos 100� � i sin 100��

12�cos 52� � i sin 52��3�cos 110� � i sin 110��

5�cos 4.3 � i sin 4.3�4�cos 2.1 � i sin 2.1�

cos�5��3� � i sin�5��3�cos � � i sin �

2�cos 120� � i sin 120��4�cos 40� � i sin 40��

cos 50� � i sin 50�

cos 20� � i sin 20�

�cos 5� � i sin 5��cos 20� � i sin 20��0.60�cos 200� � i sin 200��0.45�cos 310��i sin 310��

�0.8�cos 300� � i sin 300��0.5�cos 100� � i sin 100��

�53�cos 140� � i sin 140�� 2

3�cos 60� � i sin 60��3

4�cos �

3� i sin

�

3��4�cos 3�

4� i sin

3�

4 ��2�cos

�

4� i sin

�

4��6�cos �

12� i sin

�

12��

333202_0605.qxd 12/8/05 10:09 AM Page 479

In Exercises 105–112, use the theorem on page 476 to findall the solutions of the equation and represent thesolutions graphically.

105.

106.

107.

108.

109.

110.

111.

112.

Synthesis

True or False? In Exercises 113–116, determine whetherthe statement is true or false. Justify your answer.

113. Although the square of the complex number is given bythe absolute value of the complex number

is defined as

114. Geometrically, the th roots of any complex number areall equally spaced around the unit circle centered at theorigin.

115. The product of two complex numbers

and

is zero only when and/or

116. By DeMoivre’s Theorem,

117. Given two complex numbers andshow that

118. Show that is the complexconjugate of

119. Use the trigonometric forms of and in Exercise 118 tofind (a) and (b)

120. Show that the negative of is

121. Show that is a sixth root of 1.

122. Show that is a fourth root of

Graphical Reasoning In Exercises 123 and 124, use thegraph of the roots of a complex number.

(a) Write each of the roots in trigonometric form.

(b) Identify the complex number whose roots are given.

(c) Use a graphing utility to verify the results of part (b).

123.

124.

Skills Review

In Exercises 125–130, solve the right triangle shown in thefigure. Round your answers to two decimal places.

125. 126.

127. 128.

129. 130.

Harmonic Motion In Exercises 131–134, for the simpleharmonic motion described by the trigonometric function,find the maximum displacement and the least positivevalue of for which

131. 132.

133. 134.

In Exercises 135 and 136, write the product as a sum ordifference.

135. 136. 2 cos 5� sin 2�6 sin 8� cos 3�

d �112

sin 60�td �116

sin 54

�t

d �18

cos 12�td � 16 cos �

4t

d � 0.t

c � 6.8B � 81� 30�,c � 11.2A � 42� 15�,

b � 211.2B � 6�,b � 112.6A � 30�,

a � 33.5B � 66�,a � 8A � 22�,

C Ab

a c

B

3

3 3

345° 45°

45°45°

Realaxis

Imaginaryaxis

2

2 230° 30°

1−1

Realaxis

Imaginaryaxis

�2.2�1�4�1 � i��

12�1 � �3 i�

�z � r �cos�� i sin�� .z � r �cos � � i sin ��

z � 0.z�z,zzzz

z � r �cos � � i sin ��.z � r �cos�� i sin��

z1

z 2

�r1

r2

�cos��1 � �2� � i sin��1 � �2��.

z2 � 0,z2 � r2�cos �2 � i sin �2�,z1 � r1�cos �1�i sin �1�

�4 � �6 i�8� cos�32� � i sin�8�6�.

r2 � 0.r1 � 0

z2 � r2�cos �2 � i sin �2�.

z1 � r1�cos �1 � i sin �1�

zn

�a � bi� � �a2 � b2.

z � a � bi�bi�2 � �b2,

bi

x 4 � �1 � i� � 0

x3 � �1 � i� � 0

x6 � 64i � 0

x 4 � 16i � 0

x3 � 27 � 0

x5 � 243 � 0

x3 � 1 � 0

x 4 � i � 0


333202_0605.qxd 12/8/05 10:09 AM Page 480

Chapter Summary 481

Chapter Summary6

What did you learn?

Section 6.1 Review Exercises� Use the Law of Sines to solve oblique triangles (AAS, ASA, or SSA) (p. 430, 432). 1–12

� Find areas of oblique triangles (p. 434). 13–16

� Use the Law of Sines to model and solve real-life problems (p. 435). 17–20

Section 6.2� Use the Law of Cosines to solve oblique triangles (SSS or SAS) (p. 439). 21–28

� Use the Law of Cosines to model and solve real-life problems (p. 441). 29–32

� Use Heron's Area Formula to find areas of triangles (p. 442). 33–36

Section 6.3� Represent vectors as directed line segments (p. 447). 37, 38

� Write the component forms of vectors (p. 448). 39–44

� Perform basic vector operations and represent vectors graphically (p. 449). 45–56

� Write vectors as linear combinations of unit vectors (p. 451). 57–62

� Find the direction angles of vectors (p. 453). 63–68

� Use vectors to model and solve real-life problems (p. 454). 69–72

Section 6.4� Find the dot product of two vectors and use the properties of the 73–80

dot product (p. 460).

� Find the angle between two vectors and determine whether two 81–88vectors are orthogonal (p. 461).

� Write vectors as sums of two vector components (p. 463). 89–92

� Use vectors to find the work done by a force (p. 466). 93–96

Section 6.5� Plot complex numbers in the complex plane and find absolute values 97–100

of complex numbers (p. 470).

� Write the trigonometric forms of complex numbers (p. 471). 101–104

� Multiply and divide complex numbers written in trigonometric form (p. 472). 105, 106

� Use DeMoivre’s Theorem to find powers of complex numbers (p. 474) 107–110

� Find nth roots of complex numbers (p. 475). 111–118

333202_060R.qxd 12/5/05 10:48 AM Page 481

In Exercises 1–12, use the Law of Sines to solve (ifpossible) the triangle. If two solutions exist, find both.Round your answers to two decimal places.

1. 2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

In Exercises 13–16, find the area of the triangle having theindicated angle and sides.

13.

14.

15.

16.

17. Height From a certain distance, the angle of elevation tothe top of a building is At a point 50 meters closer tothe building, the angle of elevation is Approximate theheight of the building.

18. Geometry Find the length of the side of theparallelogram.

19. Height A tree stands on a hillside of slope from thehorizontal. From a point 75 feet down the hill, the angle ofelevation to the top of the tree is (see figure). Find theheight of the tree.

FIGURE FOR 19

20. River Width A surveyor finds that a tree on the oppositebank of a river, flowing due east, has a bearing of N E from a certain point and a bearing of N Wfrom a point 400 feet downstream. Find the width of theriver.

In Exercises 21–28, use the Law of Cosines to solvethe triangle. Round your answers to two decimal places.

21.

22.

23.

24.

25.

26.

27.

28.

29. Geometry The lengths of the diagonals of a parallelogramare 10 feet and 16 feet. Find the lengths of the sides of theparallelogram if the diagonals intersect at an angle of

30. Geometry The lengths of the diagonals of a parallelogramare 30 meters and 40 meters. Find the lengths of the sidesof the parallelogram if the diagonals intersect at an angle of

31. Surveying To approximate the length of a marsh, asurveyor walks 425 meters from point to point Thenthe surveyor turns and walks 300 meters to point (seefigure). Approximate the length of the marsh.

A

65°

300 m 425 m

B

C

ACC65�

B.A

34�.

28�.

c � 19.52b � 11.34,A � 62�,

b � 31.4a � 22.5,C � 43�,

c � 20a � 10,B � 150�,

c � 4a � 4,B � 110�,

c � 12.2b � 8.8,a � 16.4,

c � 4.5b � 5.0,a � 2.5,

c � 100b � 60,a � 80,

c � 10b � 8,a � 5,

6.2

15�22� 30�

45°

28°

75 ft

45�

28�

140°16

w

12

w

31�.17�.

c � 21b � 22,A � 11�,

b � 5a � 16,C � 123�,

c � 8a � 4,B � 80�,

c � 7b � 5,A � 27�,

b � 4a � 6.2,B � 25�,

b � 33.7a � 51.2,A � 75�,

b � 3a � 10,B � 150�,

c � 10b � 30,B � 150�,

a � 367C � 36�,B � 64�,

b � 27.5C � 48�,A � 24�,

c � 104.8B � 45�,A � 95�,

c � 8.4B � 98�,A � 16�,

c � 33C � 20�,B � 10�,

b � 54C � 82�,B � 72�,

A C121°

a = 1722°

b

cB

A C

71° a = 8

35°b

c

B

6.1


Review Exercises6

333202_060R.qxd 12/5/05 10:48 AM Page 482

Review Exercises 483

32. Navigation Two planes leave Raleigh-Durham Airport atapproximately the same time. One is flying 425 miles perhour at a bearing of , and the other is flying 530 milesper hour at a bearing of Draw a figure that gives a visual representation of the problem and determine thedistance between the planes after they have flown for 2 hours.

In Exercises 33–36, use Heron’s Area Formula to find thearea of the triangle.

33.

34.

35.

36.

In Exercises 37 and 38, show that

37. 38.

In Exercises 39– 44, find the component form of the vectorsatisfying the conditions.

39. 40.

41. Initial point: terminal point:

42. Initial point: terminal point:

43.

44.

In Exercises 45–52, find (a) (b) (c) and(d)

45.

46.

47.

48.

49.

50.

51.

52.

In Exercises 53–56, find the component form of andsketch the specified vector operations geometrically,where and

53.

54.

55.

56.

In Exercises 57– 60, write vector as a linear combination ofthe standard unit vectors and

57.

58.

59. has initial point and terminal point

60. has initial point and terminal point

In Exercises 61 and 62, write the vector in the form

61.

62.

In Exercises 63–68, find the magnitude and the directionangle of the vector

63.

64.

65.

66.

67.

68.

69. Resultant Force Forces with magnitudes of 85 poundsand 50 pounds act on a single point. The angle between theforces is Describe the resultant force.

70. Rope Tension A 180-pound weight is supported by tworopes, as shown in the figure. Find the tension in each rope.

30° 30°

180 lb

15�.

v � 8i � j

v � �3i � 3j

v � �4i � 7j

v � 5i � 4j

v � 3�cos 150�i � sin 150�j�v � 7�cos 60�i � sin 60�j�

v.

v � 4i � j

v � �10i � 10j

�v��cos � i � sin �j�.v

�5, �9�.��2, 7�u

�9, 8�.�3, 4�u

u � ��6, �8�u � ��3, 4�

j.iu

w �12v

w � 3v

w � 4u � 5v

w � 2u � v

v � 1 � i � 3j.u � 6i � 5j

w

u � �6j, v � i � j

u � 4i, v � �i � 6j

u � �7i � 3j, v � 4i � j

u � 2i � j, v � 5i � 3j

u � �1, �8�, v � �3, �2�u � ��5, 2�, v � �4, 4�u � �4, 5�, v � �0, �1�u � ��1, �3�, v � ��3, 6�

2v � 5u.3u ,u � v,u � v,

� � 225��v � �12,

� � 120��v � � 8,

�15, 9��1, 5�;�7, 3��0, 10�;

( )4

6

(0, 1)x

v

2 4 6

726,

y

2

−2

(−5, 4)

2

−2−4

4

6

(2, −1)

x

v

y

v

(1, 4)

42−4

vu

4

2

x

y

(−1, −4)

(−3, 2)

(3, −2)

( 2, 1)−

(0, 2)−

(6, 3)

(4, 6)

6−2−2

v

u4

6

x

y

u � v.6.3

c � 19.4b � 26.7,a � 38.1,

c � 3.7b � 15.8,a � 12.3,

c � 10b � 8,a � 15,

c � 7b � 5,a � 4,

67�.355�

333202_060R.qxd 12/5/05 10:48 AM Page 483

71. Navigation An airplane has an airspeed of 430 miles perhour at a bearing of The wind velocity is 35 miles perhour in the direction of N E. Find the resultant speedand direction of the airplane.

72. Navigation An airplane has an airspeed of 724 kilome-ters per hour at a bearing of The wind velocity is 32 kilometers per hour from the west. Find the resultantspeed and direction of the airplane.

In Exercises 73–76, find the dot product of and

73. 74.

75. 76.

In Exercises 77– 80, use the vectors andto find the indicated quantity. State whether

the result is a vector or a scalar.

77.

78.

79.

80.

In Exercises 81– 84, find the angle between the vectors.

81.

82.

83.

84.

In Exercises 85–88, determine whether and areorthogonal, parallel, or neither.

85. 86.

87. 88.

In Exercises 89–92, find the projection of onto Thenwrite as the sum of two orthogonal vectors, one of whichis

89.

90.

91.

92.

Work In Exercises 93 and 94, find the work done inmoving a particle from to if the magnitude anddirection of the force are given by

93.

94.

95. Work Determine the work done by a crane lifting an18,000-pound truck 48 inches.

96. Work A mover exerts a horizontal force of 25 poundson a crate as it is pushed up a ramp that is 12 feet long andinclined at an angle of above the horizontal. Find thework done in pushing the crate.

In Exercises 97–100, plot the complex number andfind its absolute value.

97. 98.

99.

100.

In Exercises 101–104, write the complex number in trigonometric form.

101. 102.

103. 104.

In Exercises 105 and 106, (a) write the two complexnumbers in trigonometric form, and (b) use the trigonometric forms to find and where

105.

106.

In Exercises 107–110, use DeMoivre’s Theorem to find theindicated power of the complex number.Write the result instandard form.

107.

108.

109.

110.

In Exercises 111–114, (a) use the theorem on page 476 to find the indicated roots of the complex number,(b) represent each of the roots graphically, and (c) writeeach of the roots in standard form.

111. Sixth roots of

112. Fourth roots of

113. Cube roots of 8

114. Fifth roots of �1024

256i

�729i

�1 � i �8

�2 � 3i �6

�2�cos 4�

15� i sin

4�

15��5

�5�cos �

12� i sin

�

12��4

z2 � 2��3 � i�z1 � �3�1 � i�,z2 � �10iz1 � 2�3 � 2i,

z2 � 0.z1/z2,z1 z2

�7�3�3 � 3i

5 � 12i5 � 5i

�10 � 4i

5 � 3i

�6i7i

6.5

20�

P � ��2, �9�, Q � ��12, 8�, v � 3i � 6j

P � �5, 3�, Q � �8, 9�, v � �2, 7

v.QP

v � ��5, 2u � ��3, 5,v � �1, �1u � �2, 7,v � �10, 0u � �5, 6,

v � ��8, �2u � ��4, 3,

projvu.u

v.u

v � 3i � 6jv � i � 2j

u � �2i � ju � �i

v � ��2, 4v � �8, 3u � �1

4, �12u � ��3, 8

vu

v � �4, 3�3u � �3, �3,

v � ��2, 1u � �2�2, �4,

v � cos 300�i � sin 300�j

u � cos 45�i � sin 45�j

v � cos 5�

6 i � sin

5�

6 j

u � cos 7�

4 i � sin

7�

4 j

�

3u � v

u�u � v�v2

2u � u

v � <2, 1>u � <�3, 4>

v � 16i � 12jv � 11i � 5j

u � �7i � 2ju � 3i � 7j

v � ��4, �14v � ��3, 9u � ��7, 12u � �6, 7

v.u.6.4

30�.

30�135�.


333202_060R.qxd 12/8/05 10:10 AM Page 484

Review Exercises 485

In Exercises 115–118, use the theorem on page 476 to findall solutions of the equation and represent the solutionsgraphically.

115.

116.

117.

118.

Synthesis

True or False? In Exercises 119–123, determine whetherthe statement is true or false. Justify your answer.

119. The Law of Sines is true if one of the angles in thetriangle is a right angle.

120. When the Law of Sines is used, the solution is alwaysunique.

121. If u is a unit vector in the direction of then

122. If then

123. is a solution of the equation

124. State the Law of Sines from memory.

125. State the Law of Cosines from memory.

126. What characterizes a vector in the plane?

127. Which vectors in the figure appear to be equivalent?

128. The vectors and have the same magnitudes in the twofigures. In which figure will the magnitude of the sum begreater? Give a reason for your answer.

(a) (b)

129. Give a geometric description of the scalar multiple ofthe vector for and for

130. Give a geometric description of the sum of the vectors and

Graphical Reasoning In Exercises 131 and 132, use thegraph of the roots of a complex number.

(a) Write each of the roots in trigonometric form.

(b) Identify the complex number whose roots are given.

(c) Use a graphing utility to verify the results of part (b).

131.

132.

133. The figure shows and Describe and

134. One of the fourth roots of a complex number is shownin the figure.

(a) How many roots are not shown?

(b) Describe the other roots.

Realaxis1−1

1z 30°

Imaginaryaxis

z

Realaxis1−1

1

z1z2

θθ

Imaginaryaxis

z1�z2.z1z2z2.z1

60°

60° 30°

30°4

4

44

3

3

Realaxis

Imaginaryaxis

60°

60°4

4

4−2−2

2

Realaxis

Imaginaryaxis

v.u

k < 0.k > 0u,ku

x

uv

y

xuv

y

vu

x

C BA

ED

y

x2 � 8i � 0.x � �3 � i

a � �b.v � a i � bj � 0,

v � �v � u.v,

�x3 � 1��x2 � 1� � 0

x3 � 8i � 0

x5 � 32 � 0

x4 � 81 � 0

333202_060R.qxd 12/8/05 10:10 AM Page 485


Chapter Test6

Take this test as you would take a test in class. When you are finished, check yourwork against the answers given in the back of the book.

In Exercises 1–6, use the information to solve the triangle. If two solutions exist, findboth solutions. Round your answers to two decimal places.

1. 2.

3. 4.

5. 6.

7. A triangular parcel of land has borders of lengths 60 meters, 70 meters, and 82 meters.Find the area of the parcel of land.

8. An airplane flies 370 miles from point to point with a bearing of It then flies240 miles from point to point with a bearing of (see figure). Find the distanceand bearing from point to point

In Exercises 9 and 10, find the component form of the vector satisfying the givenconditions.

9. Initial point of terminal point of

10. Magnitude of direction of

In Exercises 11–13, and Find the resultant vector and sketchits graph.

11. 12. 13.

14. Find a unit vector in the direction of

15. Forces with magnitudes of 250 pounds and 130 pounds act on an object at angles ofand , respectively, with the -axis. Find the direction and magnitude of the

resultant of these forces.

16. Find the angle between the vectors and

17. Are the vectors and orthogonal?

18. Find the projection of onto Then write as the sum of twoorthogonal vectors.

19. A 500-pound motorcycle is headed up a hill inclined at What force is required tokeep the motorcycle from rolling down the hill when stopped at a red light?

20. Write the complex number in trigonometric form.

21. Write the complex number in standard form.

In Exercises 22 and 23, use DeMoivre’s Theorem to find the indicated power of thecomplex number. Write the result in standard form.

22. 23.

24. Find the fourth roots of

25. Find all solutions of the equation and represent the solutions graphically.x3 � 27i � 0

256�1 � �3 i�.

�3 � 3i�6�3�cos 7�

6� i sin

7�

6 ��8

z � 6�cos 120� � i sin 120��z � 5 � 5i

12�.

uv � ��5, �1.u � �6, 7v � �2, 3u � �6, 10

v � �3, �2.u � ��1, 5

x�60�45�

u � �4, �3.

5u � 3vu � vu � v

v � <�7, 1>.u � <3, 5>u � �3, �5v:v � 12;v:

�11, �16�v:��3, 7�;v:

v

C.A37�CB

24�.BA

b � 57a � 41,C � 123�,b � 23a � 15,B � 100�,

c � 12.4b � 7.3,a � 4.0,b � 13.4a � 11.2,A � 24�,

a � 18.1C � 33�,B � 104�,a � 12.2B � 68�,A � 24�,

A

B

24°

370 mi

37°

240 mi C

FIGURE FOR 8

333202_060R.qxd 12/8/05 10:11 AM Page 486

Cumulative Test for Chapters 4–6 487

Cumulative Test for Chapter 4–66

Take this test to review the material from earlier chapters. When you are finished,check your work against the answers given in the back of the book.

1. Consider the angle

(a) Sketch the angle in standard position.

(b) Determine a coterminal angle in the interval

(c) Convert the angle to radian measure.

(d) Find the reference angle

(e) Find the exact values of the six trigonometric functions of

2. Convert the angle radians to degrees. Round the answer to one decimalplace.

3. Find if and

In Exercises 4 –6, sketch the graph of the function. (Include two full periods.)

4. 5. 6.

7. Find and such that the graph of the function matches thegraph in the figure.

8. Sketch the graph of the function over the interval

In Exercises 9 and 10, find the exact value of the expression without using a calculator.

9. 10.

11. Write an algebraic expression equivalent to

12. Use the fundamental identities to simplify:

13. Subtract and simplify:

In Exercises 14 –16, verify the identity.

14.

15.

16.

In Exercises 17 and 18, find all solutions of the equation in the interval

17.

18.

19. Use the Quadratic Formula to solve the equation in the interval

20. Given that and angles and are both in Quadrant I, find

21. If find the exact value of tan�2��.tan � �12,

tan�u � v�.vucos v �

35,sin u �

1213,

sin2 x � 2 sin x � 1 � 0.�0, 2��:

3 tan � � cot � � 0

2 cos2 � cos � 0

[0, 2��.

sin2 x cos2 x �18�1 � cos 4x�

sin�x � y� sin�x � y� � sin2 x � sin2 y

cot2 �sec2 � 1� � 1

sin � � 1

cos ��

cos �

sin � � 1.

cos�

2� x csc x.

sin�arccos 2x�.

tan�arcsin 35�tan�arctan 6.7�

�3� ≤ x ≤ 3�.f �x� �12x sin x

h�x� � a cos�bx � c�cb,a,

h�x� � �sec�x � ��g�x� �12

tanx ��

2f �x� � 3 � 2 sin �x

sin � < 0.tan � � �43cos �

� � 2.35

�.

��.

�0�, 360��.

� � �120�.

x1 3

4

−3−4

y

FIGURE FOR 7

333202_060R.qxd 12/5/05 10:48 AM Page 487


22. If find the exact value of

23. Write the product as a sum or difference.

24. Write as a product.

In Exercises 25–28, use the information to solve the triangle shown in the figure.Round your answers to two decimal places.

25.

26.

27.

28.

29. Two sides of a triangle have lengths 7 inches and 12 inches. Their included anglemeasures Find the area of the triangle.

30. Find the area of a triangle with sides of lengths 11 inches, 16 inches, and 17 inches.

31. Write the vector as a linear combination of the standard unit vectors and

32. Find a unit vector in the direction of

33. Find for and

34. Find the projection of onto Then write as the sum of twoorthogonal vectors.

35. Write the complex number in trigonometric form.

36. Find the product of Write theanswer in standard form.

37. Find the three cube roots of 1.

38. Find all the solutions of the equation

39. A ceiling fan with 21-inch blades makes 63 revolutions per minute. Find the angularspeed of the fan in radians per minute. Find the linear speed of the tips of the bladesin inches per minute.

40. Find the area of the sector of a circle with a radius of 8 yards and a central angle of

41. From a point 200 feet from a flagpole, the angles of elevation to the bottom and topof the flag are and respectively. Approximate the height of the flag to thenearest foot.

42. To determine the angle of elevation of a star in the sky, you get the star in your line ofvision with the backboard of a basketball hoop that is 5 feet higher than your eyes (seefigure). Your horizontal distance from the backboard is 12 feet. What is the angle ofelevation of the star?

43. Write a model for a particle in simple harmonic motion with a displacement of 4 inches and a period of 8 seconds.

44. An airplane’s velocity with respect to the air is 500 kilometers per hour, with a bearing of The wind at the altitude of the plane has a velocity of 50 kilometersper hour with a bearing of N E. What is the true direction of the plane, and whatis its speed relative to the ground?

45. A force of 85 pounds exerted at an angle of above the horizontal is required toslide an object across a floor. The object is dragged 10 feet. Determine the work donein sliding the object.

60�

60�30�.

18�,16� 45�

114�.

x5 � 243 � 0.

�4�cos 30� � i sin 30��6�cos 120� � i sin 120��.�2 � 2i

uv � �1, 5�.u � �8, �2�v � i � 2j.u � 3i � 4ju � v

v � i � j.

j.iu � �3, 5�

60�.

c � 9b � 8,a � 4,

b � 10C � 90�,A � 30�,

A � 30�, b � 8, c � 10

A � 30�, a � 9, b � 8

cos 8x � cos 4x

5 sin 3�

4� cos

7�

4

sin �

2.tan � �

43

,

A

b a

c B

C

FIGURE FOR 25–28

12 feet

5 feet

FIGURE FOR 42

333202_060R.qxd 12/5/05 10:48 AM Page 488

489

ProofLet h be the altitude of either triangle found in the figure above. Then you have

or

or

Equating these two values of you have

or

Note that and because no angle of a triangle can have ameasure of or In a similar manner, construct an altitude from vertex toside (extended in the obtuse triangle), as shown at the left. Then you have

or

or

Equating these two values of you have

or

By the Transitive Property of Equality you know that

So, the Law of Sines is established.

asin A

�b

sin B�

csin C

.

asin A

�c

sin C.a sin C � c sin A

h,

h � a sin C.sin C �ha

h � c sin Asin A �hc

ACB180�.0�

sin B � 0sin A � 0

asin A

�b

sin B.a sin B � b sin A

h,

h � a sin B.sin B �ha

h � b sin Asin A �hb

Proofs in Mathematics

Law of Sines (p. 430)

If is a triangle with sides and then

a

c BA

b

C

a

c BA

b

C

asin A

�b

sin B�

csin C

.

c,b,a,ABC

A is acute. A is obtuse.

Law of TangentsBesides the Law of Sines and theLaw of Cosines, there is also aLaw of Tangents, which wasdeveloped by Francois Viete(1540–1603). The Law ofTangents follows from the Lawof Sines and the sum-to-productformulas for sine and is definedas follows.

The Law of Tangents can beused to solve a triangle whentwo sides and the included angle are given (SAS). Beforecalculators were invented, theLaw of Tangents was used tosolve the SAS case instead ofthe Law of Cosines, becausecomputation with a table oftangent values was easier.

a � ba � b

�tan��A � B� 2�tan��A � B� 2�

a

c BA

b

C

A is obtuse.

a

c BA

b

C

A is acute.

333202_060R.qxd 12/5/05 10:48 AM Page 489

ProofTo prove the first formula, consider the top triangle at the left, which has threeacute angles. Note that vertex has coordinates Furthermore, hascoordinates where and Because is the distancefrom vertex to vertex it follows that

Distance Formula

Square each side.

Substitute for x and y.

Expand.

Factor out

To prove the second formula, consider the bottom triangle at the left, which alsohas three acute angles. Note that vertex has coordinates Furthermore,has coordinates where and Because is thedistance from vertex to vertex it follows that

Distance Formula

Square each side.

Substitute for x and y.

Expand.

Factor out

A similar argument is used to establish the third formula.

sin2 B � cos2 B � 1 b2 � a2 � c2 � 2ac cos B.

a2. b2 � a2�sin2 B � cos2 B� � c2 � 2ac cos B

b2 � a2 cos2 B � 2ac cos B � c2 � a2 sin2 B

b2 � �a cos B � c�2 � �a sin B�2

b2 � �x � c�2 � �y � 0�2

b � ��x � c�2 � �y � 0�2

A,Cby � a sin B.x � a cos B�x, y�,

C�c, 0�.A

sin2 A � cos2 A � 1 a2 � b2 � c2 � 2bc cos A.

b2. a2 � b2�sin2 A � cos2 A� � c2 � 2bc cos A

a2 � b2 cos2 A � 2bc cos A � c2 � b2 sin2 A

a2 � �b cos A � c�2 � �b sin A�2

a2 � �x � c�2 � �y � 0�2

a � ��x � c�2 � �y � 0�2

B,Cay � b sin A.x � b cos A�x, y�,

C�c, 0�.B

490

x

C x y= ( , )

B c= ( , 0)

y

x

b

A

a

c

y

x

C x y= ( , )

y

x

a

B

b

c

y

A = (c, 0)

Law of Cosines (p. 439)

Standard Form Alternative Form

cos C �a2 � b2 � c2

2abc2 � a2 � b2 � 2ab cos C

cos B �a2 � c2 � b2

2acb2 � a2 � c2 � 2ac cos B

cos A �b2 � c2 � a2

2bca2 � b2 � c2 � 2bc cos A

333202_060R.qxd 12/5/05 10:48 AM Page 490

491

ProofFrom Section 6.1, you know that

Square each side.

Pythagorean Identity

Factor.

Using the Law of Cosines, you can show that

and

Letting these two equations can be rewritten as

and

By substituting into the last formula for area, you can conclude that

Area � �s�s � a��s � b��s � c�.

12

bc�1 � cos A� � �s � b��s � c�.

12

bc�1 � cos A� � s�s � a�

s � �a � b � c� 2,

12

bc�1 � cos A� �a � b � c

2�

a � b � c2

.

12

bc�1 � cos A� �a � b � c

2�

�a � b � c2

��12

bc�1 � cos A��12

bc�1 � cos A��.

��14

b2c2�1 � cos2 A�

Area ��14

b2c2 sin2 A

�Area�2 �14

b2c2 sin2 A

Area �12

bc sin A

Heron’s Area Formula (p. 442)

Given any triangle with sides of lengths and the area of the triangle is

where s ��a � b � c�

2.

Area � �s�s � a��s � b��s � c�

c,b,a,

Formula for the area of an oblique triangle

Take the square root ofeach side.

333202_060R.qxd 12/5/05 10:48 AM Page 491

ProofLet and let c be a scalar.

1.

2.

3.

4.

5.

ProofConsider the triangle determined by vectors u, v, and as shown in thefigure. By the Law of Cosines, you can write

cos � �u � v

�u� �v�.

�v�2 � 2u � v � �u�2 � �u�2 � �v�2 � 2�u� �v� cos �

v � v � u � v � v � u � u � u � �u�2 � �v�2 � 2�u� �v� cos �

�v � u� � v � �v � u� � u � �u�2 � �v�2 � 2�u� �v� cos �

�v � u� � �v � u� � �u�2 � �v�2 � 2�u� �v� cos �

�v � u�2 � �u�2 � �v�2 � 2�u� �v�cos �

v � u,

� cu � v

� �cu1, cu2� � �v1, v2� � �cu1�v1 � �cu2�v2

� c�u1v1 � u2v2� c�u � v� � c��u1, u2� � �v1, v2��

� �v�2� ��v12 � v2

2�2 v � v � v1

2 � v22

� u � v � u � w � �u1v1 � u2v2� � �u1w1 � u2w2� � u1v1 � u1w1 � u2v2 � u2w2

� u1�v1 � w1� � u2�v2 � w2� u � �v � w� � u � �v1 � w1, v2 � w2�

� 0 0 � v � 0 � v1 � 0 � v2

� v � u� v1u1 � v2u2 u � v � u1v1 � u2v2

0 � �0, 0�,w � �w1, w2�,v � �v1, v2�,u � �u1, u2�,

492

Properties of the Dot Product (p. 460)

Let and be vectors in the plane or in space and let be a scalar.

1. 2.

3. 4.

5. c�u � v� � cu � v � u � cv

v � v � �v�2u � �v � w� � u � v � u � w

0 � v � 0u � v � v � u

cwv,u,

Angle Between Two Vectors (p. 461)

If is the angle between two nonzero vectors and then cos � �u � v

�u� �v�.v,u�

−

vu

Origin

θ

v u

333202_060R.qxd 12/8/05 9:34 AM Page 492

1. In the figure, a beam of light is directed at the blue mirror,reflected to the red mirror, and then reflected back to theblue mirror. Find the distance that the light travels fromthe red mirror back to the blue mirror.

2. A triathlete sets a course to swim S E from a point onshore to a buoy mile away. After swimming 300 yardsthrough a strong current, the triathlete is off course at a bear-ing of S E. Find the bearing and distance the triathleteneeds to swim to correct her course.

3. A hiking party is lost in a national park. Two ranger stationshave received an emergency SOS signal from the party.Station B is 75 miles due east of station A. The bearing fromstation A to the signal is S E and the bearing from stationB to the signal is S W.

(a) Draw a diagram that gives a visual representation of theproblem.

(b) Find the distance from each station to the SOS signal.

(c) A rescue party is in the park 20 miles from station A ata bearing of S E. Find the distance and the bearingthe rescue party must travel to reach the lost hikingparty.

4. You are seeding a triangular courtyard. One side of thecourtyard is 52 feet long and another side is 46 feet long.The angle opposite the 52-foot side is

(a) Draw a diagram that gives a visual representation of theproblem.

(b) How long is the third side of the courtyard?

(c) One bag of grass covers an area of 50 square feet. Howmany bags of grass will you need to cover the courtyard?

5. For each pair of vectors, find the following.

(i) (ii) (iii)

(iv) (v) (vi)

(a) (b)

(c) (d)

6. A skydiver is falling at a constant downward velocity of 120 miles per hour. In the figure, vector u represents the skydiver’s velocity. A steady breeze pushes the skydiver tothe east at 40 miles per hour. Vector v represents the windvelocity.

(a) Write the vectors and in component form.

(b) Let Use the figure to sketch To print anenlarged copy of the graph, go to the website,www.mathgraphs.com.

(c) Find the magnitude of What information does themagnitude give you about the skydiver’s fall?

(d) If there were no wind, the skydiver would fall in a pathperpendicular to the ground. At what angle to theground is the path of the skydiver when the skydiver isaffected by the 40 mile per hour wind from due west?

(e) The skydiver is blown to the west at 30 miles per hour.Draw a new figure that gives a visual representation ofthe problem and find the skydiver’s new velocity.

s.

s.s � u � v.

vu

u

v

−20 20 40 60

20

40

60

80

100

120

140

EW

Down

Up

v � �5, 5�v � �2, 3�u � �2, �4�u � �1, 12�v � �3, �3�v � ��1, 2�u � �0, 1�u � �1, �1�

� u � v�u � v�� v

�v�� u�u��

�u � v��v��u�

65�.

80�

75�60�

Buoy

300 yd

25°

35° 34

mi

S

EW

N

35�

34

25�

O T Q

P

25°

6 ft

4.7 ft

Blue mirror

Red mirror

θθ

α α

PT

P.S. Problem Solving

This collection of thought-provoking and challenging exercises further exploresand expands upon concepts learned in this chapter.

493

333202_060R.qxd 12/5/05 10:48 AM Page 493

www.mathgraphs.com

7. Write the vector in terms of and given that theterminal point of bisects the line segment (see figure).

8. Prove that if is orthogonal to and then isorthogonal to

for any scalars and (see figure).

9. Two forces of the same magnitude and act at anglesand respectively. Use a diagram to compare the work

done by with the work done by in moving along thevector if

(a)

(b) and

10. Four basic forces are in action during flight: weight, lift,thrust, and drag. To fly through the air, an object mustovercome its own weight. To do this, it must create anupward force called lift. To generate lift, a forward motioncalled thrust is needed. The thrust must be great enough toovercome air resistance, which is called drag.

For a commercial jet aircraft, a quick climb is important tomaximize efficiency, because the performance of anaircraft at high altitudes is enhanced. In addition, it isnecessary to clear obstacles such as buildings andmountains and reduce noise in residential areas. In thediagram, the angle is called the climb angle. The velocityof the plane can be represented by a vector with a verticalcomponent (called climb speed) and a horizontalcomponent where is the speed of the plane.

When taking off, a pilot must decide how much of the thrust toapply to each component. The more the thrust is applied to thehorizontal component, the faster the airplane will gain speed.The more the thrust is applied to the vertical component, thequicker the airplane will climb.

FIGURE FOR 10

(a) Complete the table for an airplane that has a speed ofmiles per hour.

(b) Does an airplane’s speed equal the sum of the verticaland horizontal components of its velocity? If not, howcould you find the speed of an airplane whose velocitycomponents were known?

(c) Use the result of part (b) to find the speed of anairplane with the given velocity components.

(i) miles per hour

miles per hour

(ii) miles per hour

miles per hour�v� cos � � 149.634

�v� sin � � 10.463

�v� cos � � 149.909

�v� sin � � 5.235

�v� � 100

Thrust

Drag

Velocity

Weight

Lift

Climb angle

θ

θ

�v��v� cos �,�v� sin �

v�

�2 � 30�.�1 � 60�

�1 � ��2

PQF2F1

�2,�1

F2F1

wv

u

dc

cv � dw

uw,vu

wv

u

wv,uw

494

�v� cos �

�v� sin �

3.0�2.5�2.0�1.5�1.0�0.5��

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