Section 5.2 Right Triangle Trigonometry 497

Critical Thinking Exercises

Make Sense? In Exercises 117–120, determine whethereach statement makes sense or does not make sense, and explainyour reasoning.

117. I made an error because the angle I drew in standardposition exceeded a straight angle.

118. When an angle’s measure is given in terms of I know thatit’s measured using radians.

119. When I convert degrees to radians, I multiply by 1, choosingfor 1.

120. Using radian measure, I can always find a positive angleless than coterminal with a given angle by adding orsubtracting

121. If is this angle larger or smaller than a right angle?

122. A railroad curve is laid out on a circle.What radius should beused if the track is to change direction by 20° in a distance of100 miles? Round your answer to the nearest mile.

123. Assuming Earth to be a sphere of radius 4000 miles, howmany miles north of the Equator is Miami, Florida, if it is26° north from the Equator? Round your answer to thenearest mile.

u =32 ,

2p.2p

p180°

p,

Preview ExercisesExercises 124–126 will help you prepare for the material coveredin the next section. In each exercise, let be an acute angle in a righttriangle, as shown in the figure. These exercises require the use ofthe Pythagorean Theorem.

124. If and find the ratio of the length of the sideopposite to the length of the hypotenuse.

125. If and find the ratio of the length of the sideopposite to the length of the hypotenuse. Simplify the ratioby rationalizing the denominator.

126. Simplify: aacb

2

+ abcb

2

.

u

b = 1,a = 1

u

b = 12,a = 5

A

B

C

a

b

c

Length of the side adjacent to u

Length of thehypotenuse

u

Length of theside opposite u

u

5.2 Right Triangle Trigonometry

Mountain climbers haveforever been fascinated

by reaching the top ofMount Everest, sometimeswith tragic results. Themountain, on Asia’s Tibet-

Nepal border, is Earth’s highest, peaking at an incredible 29,035 feet. The heightsof mountains can be found using trigonometry. The word “trigonometry” means“measurement of triangles.” Trigonometry is used in navigation, building, andengineering. For centuries, Muslims used trigonometry and the stars to navigateacross the Arabian desert to Mecca, the birthplace of the prophet Muhammad,the founder of Islam. The ancient Greeks used trigonometry to record thelocations of thousands of stars and worked out the motion of the Moon relativeto Earth. Today, trigonometry is used to study the structure of DNA, the mastermolecule that determines how we grow from a single cell to a complex, fullydeveloped adult.

Objectives

� Use right triangles to evaluatetrigonometric functions.

� Find function values for

and

� Recognize and usefundamental identities.

� Use equal cofunctions ofcomplements.

� Evaluate trigonometricfunctions with a calculator.

� Use right triangletrigonometry to solve appliedproblems.

60° ap

3b .

30° ap

6b , 45° a

p

4b ,

Sec t i on

In the last century, Ang Rita Sherpaclimbed Mount Everest ten times, allwithout the use of bottled oxygen.

A-BLTZMC05_481-604-hr4 13-10-2008 15:53 Page 497

498 Chapter 5 Trigonometric Functions

Side adjacent to u

Hypotenuse

u

Side opposite u

Figure 5.19 Naming a right triangle’ssides from the point of view of an acuteangle u

� Use right triangles to evaluatetrigonometric functions.

The Six Trigonometric FunctionsWe begin the study of trigonometry by defining six functions, the six trigonometricfunctions. The inputs for these functions are measures of acute angles in righttriangles. The outputs are the ratios of the lengths of the sides of right triangles.

Figure 5.19 shows a right triangle with one of its acute angles labeled The side opposite the right angle is known as the hypotenuse. The other sides of thetriangle are described by their position relative to the acute angle One side isopposite and one is adjacent to

The trigonometric functions have names that are words, rather than singleletters such as and For example, the sine of is the length of the sideopposite divided by the length of the hypotenuse:

The ratio of lengths depends on angle and thus is a function of The expression really means where sine is the name of the function and the measure of anacute angle, is the input.

Here are the names of the six trigonometric functions, along with theirabbreviations:

u,sin1u2,sin uu.u

length of side opposite ulength of hypotenuse

sin u= .

Output is the ratio ofthe lengths of the sides.

Input is the measureof an acute angle.

u

Uh.f, g,

u.u

u.

u.

Name Abbreviation Name Abbreviation

sine sin cosecant csc

cosine cos secant sec

tangent tan cotangent cot

Now, let be an acute angle in a right triangle, as shown in Figure 5.20. Thelength of the side opposite is the length of the side adjacent to is and thelength of the hypotenuse is c.

b,ua,u

u

A

B

C

a

b

c

Length of the side adjacent to u

Length of thehypotenuse

u

Length of theside opposite u

Figure 5.20

Right Triangle Definitions of Trigonometric FunctionsSee Figure 5.20. The six trigonometric functions of the acute angle are defined as follows:

tan u =

length of side opposite angle ulength of side adjacent to angle u

=

a

b cot u =

length of side adjacent to angle ulength of side opposite angle u

=

ba

.

cos u =

length of side adjacent to angle ulength of hypotenuse

=

bc sec u =

length of hypotenuselength of side adjacent to angle u

=

c

b

sin u =

length of side opposite angle ulength of hypotenuse

=

ac csc u =

length of hypotenuselength of side opposite angle u

=

ca

U

Each of the trigonometric functions of the acute angle is positive. Observe that theratios in the second column in the box are the reciprocals of the correspondingratios in the first column.

u

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Section 5.2 Right Triangle Trigonometry 499

Figure 5.21 shows four right triangles of varying sizes. In each of the triangles,is the same acute angle, measuring approximately 56.3°. All four of these similar

triangles have the same shape and the lengths of corresponding sides are in the sameratio. In each triangle, the tangent function has the same value for the angleu: tan u =

32 .

u

Study TipThe word

is a way to remember the right triangle definitions of the three basic trigonometric functions,sine, cosine, and tangent.

“Some Old Hog Came Around Here and Took Our Apples.”

S O H()*

C A H()*

T O A()*

æopp

hypæ

adj

hypæ

opp

adjSine Cosine Tangent

SOHCAHTOA 1pronounced: so-cah-tow-ah2

b � 2

a � 3

u

4

6

u

1

1.5u

3

4.5

u

tan u � �32

ab tan u � �

32

64 tan u � �

32

1.51 tan u � �

32

4.53

Figure 5.21 A particular acuteangle always gives the same ratio ofopposite to adjacent sides.

In general, the trigonometric function values of depend only on the sizeof angle and not on the size of the triangle.

Evaluating Trigonometric Functions

Find the value of each of the six trigonometric functions of in Figure 5.22.

Solution We need to find the values of the six trigonometric functions of However, we must know the lengths of all three sides of the triangle ( and ) toevaluate all six functions.The values of and are given.We can use the PythagoreanTheorem, to find

Now that we know the lengths of the three sides of the triangle, we apply thedefinitions of the six trigonometric functions of Referring to these lengths asopposite, adjacent, and hypotenuse, we have

tan u =

oppositeadjacent

=

512 cot u =

adjacentopposite

=

125

.

cos u =

adjacenthypotenuse

=

1213 sec u =

hypotenuseadjacent

=

1312

sin u =

oppositehypotenuse

=

513 csc u =

hypotenuseopposite

=

135

u.

c = 2169 = 13

c2=a2+b2=52+122=25+144=169

a = 5 b = 12

c.c2= a2

+ b2,ba

ca, b,u.

u

EXAMPLE 1

U

U

b � 12

a � 5c

u

B

CA

Figure 5.22

Study TipThe function values in the secondcolumn are reciprocals of those inthe first column. You can obtain eachof these values by exchanging thenumerator and denominator ofthe corresponding ratio in the firstcolumn.

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500 Chapter 5 Trigonometric Functions

Check Point 1 Find the value of each of the sixtrigonometric functions of in the figure.u

b � 4

a � 3c

u

B

CA

Evaluating Trigonometric Functions

Find the value of each of the six trigonometric functions of in Figure 5.23.

Solution We begin by finding

Use the Pythagorean Theorem.

Figure 5.23 shows that and

and

Subtract 1 from both sides.

Take the principal square root and simplify:

Now that we know the lengths of the three sides of the triangle, we apply thedefinitions of the six trigonometric functions of

Because fractional expressions are usually written without radicals in the denomi-nators, we simplify the values of and by rationalizing the denominators:sec utan u

tan u =

oppositeadjacent

=

1

222 cot u =

adjacentopposite

=

2221

= 222

cos u =

adjacenthypotenuse

=

2223 sec u =

hypotenuseadjacent

=

3

222

sin u =

oppositehypotenuse

=

13 csc u =

hypotenuseopposite

=

31

= 3

u.

28 = 24 # 2 = 24 22 = 222 .

b = 28 = 222

b2= 8

32= 9.12

= 1 1 + b2= 9

c = 3.a = 1 12+ b2

= 32

a2+ b2

= c2

b.

u

EXAMPLE 2

We are multiplying by 1 and

not changing the value of .

sec u= �2�2

3

2�2

3= = .=

�2

�22 � 23�2

43�2

32�2

We are multiplying by 1 and

not changing the value of .

tan u= �2�2

1

2�2

1= = =

�2

�22 � 2�2

4�2

12�2

Check Point 2 Find the value of eachof the six trigonometric functions of inthe figure. Express each value in simpli-fied form.

Function Values for Some Special Angles

A 45°, or radian, angle occurs frequently in trigonometry. How do we find the

values of the trigonometric functions of 45°? We construct a right triangle with a45° angle, as shown in Figure 5.24. The triangle actually has two 45° angles. Thus, thetriangle is isosceles—that is, it has two sides of the same length. Assume that each

p

4

u

a = 1c = 3

u

B

CAb = 2�2

Figure 5.23

b

a � 1c � 5

u

B

CA

� Find function values for

and

60°ap

3b .

30°ap

6b , 45°a

p

4b ,

a = 1c = 3

u

B

CAb

Figure 5.23 (repeated,

showing b = 222)

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Section 5.2 Right Triangle Trigonometry 501

leg of the triangle has a length equal to 1. We can find the length of the hypotenuseusing the Pythagorean Theorem.

With Figure 5.24, we can determine the trigonometric function values for 45°.

Evaluating Trigonometric Functions of 45°

Use Figure 5.24 to find sin 45°, cos 45°, and tan 45°.

Solution We apply the definitions of these three trigonometric functions. Whereappropriate, we simplify by rationalizing denominators.

length of side opposite 45�

length of hypotenusesin 45�= = �

Rationalize denominators

�2

1= =

�2

1

�2�2

2�2

length of side adjacent to 45�

length of hypotenusecos 45�= = �

�2

1= =

�2

1

�2

�2

2

�2

length of side opposite 45�

length of side adjacent to 45�

11

tan 45�= = =1

EXAMPLE 3

length of hypotenuse = 22

1length of hypotenuse22 = 12+ 12

= 2

1

1�2

45�

Figure 5.24 Anisosceles right triangle

1

2

�3

60�

30�

Figure 5.25 30°–60°–90° triangle

Check Point 3 Use Figure 5.24 to find csc 45°, sec 45°, and cot 45°.

When you worked Check Point 3, did you actually use Figure 5.24 or did youuse reciprocals to find the values?

Notice that if you use reciprocals, you should take the reciprocal of a function valuebefore the denominator is rationalized. In this way, the reciprocal value will notcontain a radical in the denominator.

Two other angles that occur frequently in trigonometry are 30°, or radian,

and 60°, or radian, angles.We can find the values of the trigonometric functions of

30° and 60° by using a right triangle. To form this right triangle, draw an equilateraltriangle—that is, a triangle with all sides the same length. Assume that each side has alength equal to 2. Now take half of the equilateral triangle.We obtain the right trianglein Figure 5.25.This right triangle has a hypotenuse of length 2 and a leg of length 1.Theother leg has length which can be found using the Pythagorean Theorem.

With the right triangle in Figure 5.25, we can determine the trigonometricfunctions for 30° and 60°.

a = 23

a2= 3

a2+ 1 = 4

a2+ 12

= 22

a,

p

3

p

6

Take the reciprocal

csc 45�=�2

of sin 45° = .1�2

Take the reciprocal

sec 45�=�2

of cos 45° = .1�2

Take the reciprocal

cot 45�=1

of tan 45° = .11

A-BLTZMC05_481-604-hr 18-09-2008 11:34 Page 501

502 Chapter 5 Trigonometric Functions

Evaluating Trigonometric Functions of 30° and 60°

Use Figure 5.25 to find sin 60°, cos 60°, sin 30°, and cos 30°.

Solution We begin with 60°. Use the angle on the lower left in Figure 5.25.

To find sin 30° and cos 30°, use the angle on the upper right in Figure 5.25.

Check Point 4 Use Figure 5.25 to find tan 60° and tan 30°. If a radical appearsin a denominator, rationalize the denominator.

Because we will often use the function values of 30°, 45°, and 60°, you shouldlearn to construct the right triangles in Figure 5.24, shown on the previous page, andFigure 5.25. With sufficient practice, you will memorize the values in Table 5.2.

cos 30° =

length of side adjacent to 30°length of hypotenuse

=

232

sin 30° =

length of side opposite 30°length of hypotenuse

=

12

cos 60° =

length of side adjacent to 60°length of hypotenuse

=

12

sin 60° =

length of side opposite 60°length of hypotenuse

=

232

EXAMPLE 4

Table 5.2 Trigonometric Functions of Special Angles

U30° �

P

645° �

P

460° �

P

3

sin U12

222

232

cos U232

222

12

tan U233

1 23

Fundamental IdentitiesMany relationships exist among the six trigonometric functions. These relationshipsare described using trigonometric identities. For example, is defined as thereciprocal of This relationship can be expressed by the identity

This identity is one of six reciprocal identities.

csc u =

1sin u

.

sin u.csc u

Reciprocal Identities

tan u =

1cot u cot u =

1tan u

cos u =

1sec u sec u =

1cos u

sin u =

1csc u csc u =

1sin u

� Recognize and usefundamental identities.

1

2

�3

60�

30�

Figure 5.25 (repeated)

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Section 5.2 Right Triangle Trigonometry 503

Two other relationships that follow from the definitions of the trigonometricfunctions are called the quotient identities.

Quotient Identities

tan u =

sin ucos u cot u =

cos usin u

If and are known, a quotient identity and three reciprocal identitiesmake it possible to find the value of each of the four remaining trigonometricfunctions.

Using Quotient and Reciprocal Identities

Given and find the value of each of the four remaining

trigonometric functions.

Solution We can find by using the quotient identity that describes asthe quotient of and

We use the reciprocal identities to find the value of each of the remaining threefunctions.

We found We could use

but then we would have to rationalize the denominator.

Check Point 5 Given and find the value of each of the

four remaining trigonometric functions.

cos u =

253

,sin u =

23

tan u =

222121

,tan u =

2

221.cot u =

1tan u

=

12

221

=

2212

sec u= ��21

5= = ==

�21

5

�21

�21

21

5�21

5�21

1cos u

1

Rationalize the denominator.

csc u =

1sin u

=

125

=

52

tan u= � ��21

5= = ==

�21

2=

�21

2

�21

�2121

2�21

5�21

sin ucos u

25 2

5

Rationalize the denominator.

cos u.sin utan utan u

cos u =

2215

,sin u =

25

EXAMPLE 5

cos usin u

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504 Chapter 5 Trigonometric Functions

Other relationships among trigonometric functions follow from thePythagorean Theorem. Using Figure 5.26, the Pythagorean Theorem states that

To obtain ratios that correspond to trigonometric functions, divide both sides of thisequation by

Based on the observations in the voice balloons, we see that

We will use the notation for and for With thisnotation, we can write the identity as

Two additional identities can be obtained from by dividing bothsides by and respectively. The three identities are called the Pythagoreanidentities.

a2,b2a2

+ b2= c2

sin2 u + cos2 u = 1.

1cos u22.cos2 u1sin u22sin2 u

1sin u22 + 1cos u22 = 1.

In Figure 5.26, sin u = so this is (sin u)2.

,

ac

b2

c2a2

c2 a b2 b

ca b

2

=1 or =1+ +

ac In Figure 5.26, cos u =

so this is (cos u)2.,b

c

c2.

a2+ b2

= c2.

Pythagorean Identities

sin2 u + cos2 u = 1 1 + tan2 u = sec2 u 1 + cot2 u = csc2 u

Using a Pythagorean Identity

Given that and is an acute angle, find the value of using atrigonometric identity.

Solution We can find the value of by using the Pythagorean identity

We are given that

Square

Subtract from both sides.

Simplify:

Because is an acute angle, is positive.

Thus,

Check Point 6 Given that and is an acute angle, find the value ofusing a trigonometric identity.

Trigonometric Functions and ComplementsTwo positive angles are complements if their sum is 90° or For example, angles of70° and 20° are complements because 70° + 20° = 90°.

p

2.

cos uusin u =

12

cos u =

45

.

cos uu cos u =

A1625

=

45

1 -

925

=

2525

-

925

=

1625

. cos2 u =

1625

925

cos2 u = 1 -

925

35

: a35b

2

=

32

52 =

925

. 925

+ cos2 u = 1

sin u =

35

. a35b

2

+ cos2 u = 1

sin2 u + cos2 u = 1.

cos u

cos uusin u =35

EXAMPLE 6

� Use equal cofunctionsof complements.

b

ac

u

Figure 5.26

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Section 5.2 Right Triangle Trigonometry 505

Another relationship among trigonometric functions is based on angles thatare complements. Refer to Figure 5.27. Because the sum of the angles of any triangleis 180°, in a right triangle the sum of the acute angles is 90°. Thus, the acute anglesare complements. If the degree measure of one acute angle is then the degreemeasure of the other acute angle is This angle is shown on the upper rightin Figure 5.27.

Let’s use Figure 5.27 to compare and

Thus, If two angles are complements, the sine of one equalsthe cosine of the other. Because of this relationship, the sine and cosine are calledcofunctions of each other. The name cosine is a shortened form of the phrasecomplement’s sine.

Any pair of trigonometric functions and for which

are called cofunctions. Using Figure 5.27, we can show that the tangent andcotangent are also cofunctions of each other. So are the secant and cosecant.

f1u2 = g190° - u2 and g1u2 = f190° - u2

gf

sin u = cos190° - u2.

cos190° - u2 =

length of side adjacent to 190° - u2

length of hypotenuse=

ac

sin u =

length of side opposite ulength of hypotenuse

=

ac

cos190° - u2.sin u

190° - u2.u,

b

90� − ua

c

u

Figure 5.27

Cofunction IdentitiesThe value of a trigonometric function of is equal to the cofunction of thecomplement of Cofunctions of complementary angles are equal.

If is in radians, replace 90° with p

2.u

sec u = csc190° - u2 csc u = sec190° - u2

tan u = cot190° - u2 cot u = tan190° - u2

sin u = cos190° - u2 cos u = sin190° - u2

u.u

Using Cofunction Identities

Find a cofunction with the same value as the given expression:

a. sin 72° b.

Solution Because the value of a trigonometric function of is equal to thecofunction of the complement of we need to find the complement of each angle.

We do this by subtracting the angle’s measure from 90° or its radian equivalent,

Check Point 7 Find a cofunction with the same value as the given expression:

a. sin 46° b. cot p

12.

-=secb. cscp

2p

3p

6p

3 a b -=sec =sec3p6

2p6a b

We have a cofunctionand its function.

Perform the subtraction using theleast common denominator, 6.

a. sin 72�=cos(90�-72�)=cos 18�

We have a function andits cofunction.

p

2.

u,u

csc p

3.

EXAMPLE 7

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506 Chapter 5 Trigonometric Functions

Using a Calculator to Evaluate Trigonometric FunctionsThe values of the trigonometric functions obtained with the special triangles areexact values. For most acute angles other than 30°, 45°, and 60°, we approximate thevalue of each of the trigonometric functions using a calculator.The first step is to setthe calculator to the correct mode, degrees or radians, depending on how the acuteangle is measured.

Most calculators have keys marked and For example, to

find the value of sin 30°, set the calculator to the degree mode and enter 30 on

most scientific calculators and 30 on most graphing calculators.Consult the manual for your calculator.

To evaluate the cosecant, secant, and cotangent functions, use the key for the

respective reciprocal function, or and then use the reciprocal

key. The reciprocal key is on many scientific calculators and on many

graphing calculators. For example, we can evaluate using the followingreciprocal relationship:

Using the radian mode, enter one of the following keystroke sequences:

Many Scientific Calculators

Many Graphing Calculators

Rounding the display to four decimal places, we obtain

Evaluating Trigonometric Functions with a Calculator

Use a calculator to find the value to four decimal places:

a. cos 48.2° b. cot 1.2.

Solution

EXAMPLE 8

sec p

12L 1.0353.

� 1 � � COS � � 1 � � p � � , � 12 � 2 � � 2 � � x-1 � � ENTER �.

� p � � , � 12 � = � � COS � � 1>x �

sec p

12=

1

cos p

12

.

sec p

12

� x-1 �� 1>x �

� TAN �,� SIN �, � COS �,

� ENTER �� SIN �� SIN �

� TAN �.� SIN �, � COS �,

� Evaluate trigonometric functionswith a calculator.

Scientific Calculator Solution

Function Mode Keystrokes Display, Rounded to Four Decimal Places

a. cos 48.2° Degree 48.2 � COS � 0.6665

b. cot 1.2 Radian 1.2 � TAN � � 1>x � 0.3888

Graphing Calculator Solution

Function Mode Keystrokes Display, Rounded to Four Decimal Places

a. cos 48.2° Degree � COS � 48.2 � ENTER � 0.6665

b. cot 1.2 Radian � 1 � � TAN � 1.2 � 2 � � x-1 � � ENTER � 0.3888

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Section 5.2 Right Triangle Trigonometry 507

Check Point 8 Use a calculator to find the value to four decimal places:

a. sin 72.8° b. csc 1.5.

ApplicationsMany applications of right triangle trigonometry involve the angle made with animaginary horizontal line. As shown in Figure 5.28, an angle formed by a horizontalline and the line of sight to an object that is above the horizontal line is called theangle of elevation. The angle formed by a horizontal line and the line of sight to anobject that is below the horizontal line is called the angle of depression.Transits andsextants are instruments used to measure such angles.

� Use right triangle trigonometryto solve applied problems.

Line of sight above observer

Angle of elevation

Angle of depressionLine of sight below observer

Observerlocatedhere

Horizontal

Figure 5.28

Problem Solving Using an Angle of Elevation

Sighting the top of a building, a surveyor measured the angle of elevation to be 22°.The transit is 5 feet above the ground and 300 feet from the building. Find thebuilding’s height.

Solution The situation is illustrated in Figure 5.29. Let be the height of theportion of the building that lies above the transit. The height of the building is thetransit’s height, 5 feet, plus Thus, we need to identify a trigonometric function thatwill make it possible to find In terms of the 22° angle, we are looking for the sideopposite the angle. The transit is 300 feet from the building, so the side adjacent tothe 22° angle is 300 feet. Because we have a known angle, an unknown opposite side,and a known adjacent side, we select the tangent function.

a.a.

a

EXAMPLE 9

ha

22°300 feet

5 feetLine of sightTransit

Figure 5.29

Multiply both sides of the equation by 300.

Use a calculator in the degree mode.

The height of the part of the building above the transit is approximately 121 feet. Thus,the height of the building is determined by adding the transit’s height, 5 feet, to 121 feet.

The building’s height is approximately 126 feet.

h L 5 + 121 = 126

a L 121

a = 300 tan 22°

a

300tan 22�=

Length of side opposite the 22° angle

Length of side adjacent to the 22° angle

A-BLTZMC05_481-604-hr 18-09-2008 11:34 Page 507

508 Chapter 5 Trigonometric Functions

Figure 5.30

25 m

21 m

u

Angle of elevation

Figure 5.31

Check Point 9 The irregular blueshape in Figure 5.30 represents alake. The distance across the lake,is unknown. To find this distance, asurveyor took the measurementsshown in the figure. What is thedistance across the lake?

If two sides of a right triangle are known, an appropriate trigonometric functioncan be used to find an acute angle in the triangle.You will also need to use an inversetrigonometric key on a calculator.These keys use a function value to display the acuteangle For example, suppose that We can find in the degree mode by

using the secondary inverse sine key, usually labeled The key is not a

button you will actually press. It is the secondary function for the button labeled

The display should show approximately 59.99, which can be rounded to 60. Thus, ifthen

Determining the Angle of Elevation

A building that is 21 meters tall casts a shadow 25 meters long. Find the angle ofelevation of the sun to the nearest degree.

Solution The situation is illustrated in Figure 5.31. We are asked to find u.

EXAMPLE 10

u L 60°.sin u = 0.866,

.866 2nd SIN

Many Scientific Calculators:

Pressing 2nd SINaccesses the inversesine key, SIN−1 .

2nd SIN .866 ENTER

Many Graphing Calculators:

� SIN �.� SIN-1

�� SIN-1 �.usin u = 0.866.u.

u

a,

750 ydA C

B

a

24°

We begin with the tangent function.

tan u =

side opposite uside adjacent to u

=

2125

A-BLTZMC05_481-604-hr 18-09-2008 11:34 Page 508

Section 5.2 Right Triangle Trigonometry 509

We use a calculator in the degree mode to find

The display should show approximately 40. Thus, the angle of elevation of the sun isapproximately 40°.

Check Point 10 A flagpole that is 14 meters tall casts a shadow 10 meters long.Find the angle of elevation of the sun to the nearest degree.

2521÷ TAN2nd)(

Many Scientific Calculators:

Pressing 2nd TANaccesses the inversetangent key, TAN−1 .

2nd TAN ENTER

Many Graphing Calculators:

2521÷ )(

u.

The Mountain ManIn the 1930s, a National Geographic team headed by Brad Washburnused trigonometry to create a map of the 5000-square-mile region ofthe Yukon, near the Canadian border. The team started with aerialphotography. By drawing a network of angles on the photographs, theapproximate locations of the major mountains and their rough heightswere determined. The expedition then spent three months on foot tofind the exact heights. Team members established two base points aknown distance apart, one directly under the mountain’s peak. Bymeasuring the angle of elevation from one of the base points to thepeak, the tangent function was used to determine the peak’s height.The Yukon expedition was a major advance in the way maps are made.

Exercise Set 5.2

Practice ExercisesIn Exercises 1–8, use the Pythagorean Theorem to find the lengthof the missing side of each right triangle. Then find the value ofeach of the six trigonometric functions of

1. 2.

3. 4.

15

17

u

B

C A

21

29

u

B

C A

8

6

u

B

CA

12

9

u

B

CA

u.

5.

1026

u

B

C A

6.

40

41

u

B

C A

7.

8.24

25

uB

C

A

21

35

u

B

C

A

A-BLTZMC05_481-604-hr 18-09-2008 11:34 Page 509

510 Chapter 5 Trigonometric Functions

In Exercises 9–16, use the given triangles to evaluate eachexpression. If necessary, express the value without a square root inthe denominator by rationalizing the denominator.

9. cos 30° 10. tan 30°

11. sec 45° 12. csc 45°

13. 14.

15. 16.

In Exercises 17–20, is an acute angle and and aregiven. Use identities to find and Wherenecessary, rationalize denominators.

17.

18.

19.

20.

In Exercises 21–24, is an acute angle and is given. Use thePythagorean identity to find

21. 22.

23. 24.

In Exercises 25–30, use an identity to find the value of eachexpression. Do not use a calculator.

25. sin 37° csc 37° 26. cos 53° sec 53°

27. 28.

29. 30.

In Exercises 31–38, find a cofunction with the same value as thegiven expression.

31. sin 7° 32. sin 19°

33. csc 25° 34. csc 35°

35. 36.

37. 38. cos 3p8

cos 2p5

tan p

7tan p

9

csc2 63° - cot2 63°sec2 23° - tan2 23°

sin2 p

10+ cos2

p

10sin2 p

9+ cos2

p

9

sin u =

2215

sin u =

2398

sin u =

78

sin u =

67

cos u.sin2 u + cos2 u = 1sin uu

sin u =

67

, cos u =

2137

sin u =

13

, cos u =

2223

sin u =

35

, cos u =

45

sin u =

817

, cos u =

1517

cot u.tan u, csc u, sec u,cos usin uu

tan p

4+ csc

p

6sin p

4- cos

p

4

cot p

3tan p

3

1

1�2

45�

45�

1

2�3

60�

30�

In Exercises 39–48, use a calculator to find the value of thetrigonometric function to four decimal places.

39. sin 38° 40. cos 21°

41. tan 32.7° 42. tan 52.6°

43. csc 17° 44. sec 55°

45. 46.

47. 48.

In Exercises 49–54, find the measure of the side of the right trianglewhose length is designated by a lowercase letter. Round answers tothe nearest whole number.

49. 50.

51. 52.

53.

16 mc

B

C A23�

a

B

CA34�

13 m

b

B

CA34�

220 in.

10 cm

a

B

CA61�

250 cm

a

B

CA37�

cot p

18cot p

12

sin 3p10

cos p

10

54.

In Exercises 55–58, use a calculator to find the value of the acuteangle to the nearest degree.

55. 56.

57. 58.

In Exercises 59–62, use a calculator to find the value of the acuteangle in radians, rounded to three decimal places.

59. 60.

61. 62. tan u = 0.5117tan u = 0.4169

sin u = 0.9499cos u = 0.4112

u

tan u = 26.0307tan u = 4.6252

cos u = 0.8771sin u = 0.2974

u

b

B

CA44�

23 yd

A-BLTZMC05_481-604-hr 18-09-2008 11:34 Page 510

Section 5.2 Right Triangle Trigonometry 511

Practice PlusIn Exercises 63–68, find the exact value of each expression. Do notuse a calculator.

63. 64.

65. 66.

67.

68.

In Exercises 69–70, express the exact value of each function as asingle fraction. Do not use a calculator.

69. If find

70. If find

71. If is an acute angle and find

72. If is an acute angle and find

Application Exercises73. To find the distance across a lake, a surveyor took the mea-

surements shown in the figure. Use these measurements todetermine how far it is across the lake. Round to the nearestyard.

74. At a certain time of day, the angle of elevation of the sun is 40°.To the nearest foot, find the height of a tree whose shadowis 35 feet long.

35 ft

h

40°

630 ydA C

B

a = ?

40°

cscap

2- ub .cos u =

13

,u

tanap

2- ub .cot u =

14

,u

fap

3b .f1u2 = 2 sin u - sin

u

2,

fap

6b .f1u2 = 2 cos u - cos 2u,

cos 12° sin 78° + cos 78° sin 12°

csc 37° sec 53° - tan 53° cot 37°

1 - tan2 10° + csc2 80°1 + sin2 40° + sin2 50°

1

cot p

4

-

2

csc p

6

tan p

32

-

1

sec p

6

75. A tower that is 125 feet tall casts a shadow 172 feet long. Findthe angle of elevation of the sun to the nearest degree.

76. The Washington Monument is 555 feet high. If you are stand-ing one quarter of a mile, or 1320 feet, from the base of themonument and looking to the top, find the angle of elevationto the nearest degree.

77. A plane rises from take-off and flies at an angle of 10° withthe horizontal runway. When it has gained 500 feet, find thedistance, to the nearest foot, the plane has flown.

78. A road is inclined at an angle of 5°. After driving 5000 feetalong this road, find the driver’s increase in altitude. Roundto the nearest foot.

79. A telephone pole is 60 feet tall. A guy wire 75 feet long isattached from the ground to the top of the pole. Find theangle between the wire and the pole to the nearest degree.

60 ft75 ft

u

5000 ft

C

B

A

a = ?5°

500 ft10°

c = ?

A C

B

1320 ft

WashingtonMonument

555 ft

u

172 ft

125 ft

u

A-BLTZMC05_481-604-hr 18-09-2008 11:34 Page 511

512 Chapter 5 Trigonometric Functions

80. A telephone pole is 55 feet tall. A guy wire 80 feet long isattached from the ground to the top of the pole. Find theangle between the wire and the pole to the nearest degree.

Writing in Mathematics81. If you are given the lengths of the sides of a right triangle,

describe how to find the sine of either acute angle.

82. Describe one similarity and one difference between thedefinitions of and where is an acute angle of aright triangle.

83. Describe the triangle used to find the trigonometricfunctions of 45°.

84. Describe the triangle used to find the trigonometricfunctions of 30° and 60°.

85. What is a trigonometric identity?

86. Use words (not an equation) to describe one of thereciprocal identities.

87. Use words (not an equation) to describe one of the quotientidentities.

88. Use words (not an equation) to describe one of thePythagorean identities.

89. Describe a relationship among trigonometric functions thatis based on angles that are complements.

90. Describe what is meant by an angle of elevation and an angleof depression.

91. Stonehenge, the famous “stone circle” in England, was builtbetween 2750 B.C. and 1300 B.C. using solid stone blocksweighing over 99,000 pounds each. It required 550 people topull a single stone up a ramp inclined at a 9° angle. Describehow right triangle trigonometry can be used to determinethe distance the 550 workers had to drag a stone in order toraise it to a height of 30 feet.

ucos u,sin u

Critical Thinking ExercisesMake Sense? In Exercises 94–97, determine whether eachstatement makes sense or does not make sense, and explainyour reasoning.

94. For a given angle I found a slight increase in as thesize of the triangle increased.

95. Although I can use an isosceles right triangle to determinethe exact value of I can also use my calculator toobtain this value.

96. I can rewrite as as well as

97. Standing under this arch, I can determine its height bymeasuring the angle of elevation to the top of the arch andmy distance to a point directly under the arch.

sin ucos u

.1

cot u,tan u

sin p4 ,

sin uu,

U 0.4 0.3 0.2 0.1 0.01 0.001 0.0001 0.00001

sin U

sin UU

93. Use a calculator in the radian mode to fill in the values in

the following table.Then draw a conclusion about as approaches 0.u

cos u - 1u

U 0.4 0.3 0.2 0.1 0.01 0.001 0.0001 0.00001

cos U

cos U � 1U

Technology Exercises

92. Use a calculator in the radian mode to fill in the values in the

following table. Then draw a conclusion about as

approaches 0.

usin uu

Delicate Arch in Arches National Park, Utah

In Exercises 98–101, determine whether each statement is true orfalse. If the statement is false, make the necessary change(s) toproduce a true statement.

98. 99.

100. 101.

102. Explain why the sine or cosine of an acute angle cannot begreater than or equal to 1.

103. Describe what happens to the tangent of an acute angle asthe angle gets close to 90°. What happens at 90°?

tan2 5° = tan 25°sin 45° + cos 45° = 1

tan2 15° - sec2 15° = -1tan 45°tan 15°

= tan 3°

A-BLTZMC05_481-604-hr4 13-10-2008 15:55 Page 512

Section 5.3 Trigonometric Functions of Any Angle 513

104. From the top of a 250-foot lighthouse, a plane is sightedoverhead and a ship is observed directly below the plane.The angle of elevation of the plane is 22° and the angle ofdepression of the ship is 35°. Find a. the distance of theship from the lighthouse; b. the plane’s height above thewater. Round to the nearest foot.

Preview ExercisesExercises 105–107 will help you prepare for the material coveredin the next section. Use these figures to solve Exercises 105–106.

y

x

r

u

y

x

P � (x, y)

(a) u lies inquadrant I.

y

x

r

u

y

x

P � (x, y)

(b) u lies inquadrant II.

105. a. Write a ratio that expresses for the right triangle inFigure (a).

b. Determine the ratio that you wrote in part (a) forFigure (b) with and Is this ratio positiveor negative?

106. a. Write a ratio that expresses for the right trianglein Figure (a).

b. Determine the ratio that you wrote in part (a) forFigure (b) with and Is this ratio positiveor negative?

107. Find the positive angle formed by the terminal side of and the

a.

b. y

xu�

u � 5p6

u � 345�

u�

y

x

x-axis.uu¿

y = 5.x = -3

cos u

y = 4.x = -3

sin u

5.3 Trigonometric Functions of Any Angle

There is something comfort-ing in the repetition of

some of nature’s patterns. Theocean level at a beach variesbetween high and low tideapproximately every 12 hours.The number of hours ofdaylight oscillates from amaximum on the summersolstice, June 21, to a minimumon the winter solstice, Decem-ber 21. Then it increases to thesame maximum the followingJune 21. Some believe thatcycles, called biorhythms,represent physical, emotional,and intellectual aspects of ourlives. Throughout the remain-der of this chapter, we willsee how the trigonometricfunctions are used to model

phenomena that occur again and again. To do this, we need to move beyond righttriangles.

Objectives

� Use the definitions oftrigonometric functions ofany angle.

� Use the signs of thetrigonometric functions.

� Find reference angles.

� Use reference angles toevaluate trigonometricfunctions.

Sec t i on

A-BLTZMC05_481-604-hr4 13-10-2008 15:55 Page 513