with Trigonometry

Instruction Manual

By Steven P. Demme

PreCalculus

Math U See 1-888-854-MATH (6284)

www.MathUSee.com

1-888-854-MATH (6284)

www.MathUSee.com

Copyright © 2010 by Steven P. Demme

Graphic Design by Christine MinnichIllustrations by Gregory Snader

All rights reserved. No part of this book may be reproduced, stored in a retrieval system, or transmitted in any form by any means—electronic, mechanical, photocopying, recording, or otherwise.

In other words, thou shalt not steal.

Printed in the United States of America

PreCalculus

Scope & SeqUence

HonorS LeSSon TopicS

HoW To USe MATH-U-See

SUpporT AnD reSoUrceS

LeSSon 1 Introduction to TrigonometryLeSSon 2 Reciprocal Trigonometric RatiosLeSSon 3 Interpreting the Trigonometry TablesLeSSon 4 Use the Trig Table to Solve for the UnknownLeSSon 5 Using a Calculator and Arc Functions

LeSSon 6 Angles of Elevation and DepressionLeSSon 7 Angles < 0°, > 360°, and Reference AnglesLeSSon 8 Cofunctions; Negative Angle RelationshipsLeSSon 9 Proving Trigonometric IdentitiesLeSSon 10 Verifying Trig Expressions and Identities

LeSSon 11 Sum and Difference Identities LeSSon 12 The Double-Angle and Half-Angle IdentitiesLeSSon 13 Law of SinesLeSSon 14 Law of CosinesLeSSon 15 Radian Measure

LeSSon 16 Polar Coordinates; Rectangular CoordinatesLeSSon 17 Polar Equations and Polar GraphsLeSSon 18 VectorsLeSSon 19 Functions, Relations, Domain, and RangeLeSSon 20 Composite Functions

LeSSon 21 LogarithmsLeSSon 22 Natural Exponential and Logarithm FunctionsLeSSon 23 Graphing Sine and Cosine FunctionsLeSSon 24 Graphing the Cosecant and SecantLeSSon 25 Graphing the Tangent and CotangentLeSSon 26 Arithmetic Sequences and SeriesLeSSon 27 Geometric Sequences and SeriesLeSSon 28 Equations with Radicals and Absolute ValueLeSSon 29 Inequalities with Absolute Value and RadicalsLeSSon 30 Limits

AppenDix A Pythagorean TheoremAppenDix B Special Triangles (45º-45º-90º; 30º-60º-90º)

STUDenT SoLUTionS

HonorS SoLUTionS

TeST SoLUTionS

SYMBoLS & TABLeS

GLoSSArY oF TerMS

SeconDArY LeVeLS MASTer inDex

precALcULUS inDex

4

5Scope & SeqUence

Scope & SeqUence Math-U-See is a complete and comprehensive K-12 math curriculum. While each book focuses on a specific theme, Math-U-See continuously reviews and inte-grates topics and concepts presented in previous levels.

Primer

α Alpha | Focus: Single-Digit Addition and Subtraction

β Beta | Focus: Multiple-Digit Addition and Subtraction

γ Gamma | Focus: Multiplication

δ Delta | Focus: Division

ε Epsilon | Focus: Fractions

ζ Zeta | Focus: Decimals and percents

Pre-Algebra

Algebra 1

Stewardship*

Geometry

Algebra 2

Pre Calculus With Trigonometry

Calculus*Stewardship is a biblical approach to personal finance. The requisite knowledge for this cur-riculum is a mastery of the four basic operations, as well as fractions, decimals, and percents. In the Math-U-See sequence these topics are thoroughly covered in Alpha through Zeta. We also recommend Pre-Algebra and Algebra 1 since over half of the lessons require some knowledge of algebra. Stewardship may be studied as a one-year math course or in conjunction with any of the secondary math levels.

6

7

LeSSon Topic

Kinds of mathematics01 Symmetry of graphs and equations02 Interpolation03 Factoring techniques04 Using factoring techniques to simplify rational expressions05

Trig applications06 More trig applications07 Bearings and headings08 Navigation 09 More navigation 10

Proving cosine difference identity11 Unit circle 12 Using trigonometry to find the area of a triangle 13 Rational roots test and the fundamental theorem of algebra14 Synthetic division15

More on the fundamental theorem of algebra16 The remainder theorem17 Vector applications - navigation and force18 Graphs of common functions19 Transformations of common functions20

Inverse functions21 More on natural log; the “most beautiful equation”22 Continuous and discontinuous functions23 Application of sine curve24 Graphing calculators25

Three-dimensional coordinate systems26 Using series to find the natural log27 Solving expressions with two inequality signs28 Solving algebraic inequalities with quotients 29 Definition of the derivative 30

Here are the topics for the special challenge lessons included in the student text. You will find one honors lesson at the end of each regular lesson. The instructions are on the honors pages in the student text.

HonorS LeSSon TopicS

8

9

HoW To USe MATH-U-See

Welcome to PreCalculus. I believe you will have a positive experience with the unique Math-U-See approach to teaching math. These first few pages explain the essence of this methodology which has worked for thousands of students and teachers. I hope you will take five minutes and read through these steps carefully. If you are using the program properly and still need additional help, you may visit Math·U·See online at http://www.mathusee.com/support.html, or call us at 888-854-6284 (homeschools and individuals) or 800-454-6284 (schools and spe-cial education departments). —Steve Demme

The Goal of Math-U-See The underlying assumption or premise of Math-U-See is that the reason we study math is to apply math in everyday situations. Our goal is to help produce confident problem solvers who enjoy the study of math. These are students who learn their math facts, rules, and formulas and are able to use this knowledge in solving word problems and real-life applications. Therefore, the study of math is much more than simply committing to memory a list of facts. It includes memorization, but it also encompasses learning underlying concepts that are critical to problem solving.

More than Memorization Many people confuse memorization with understanding. Once while I was teaching seven junior high students, I asked how many pieces they would each receive if there were fourteen pieces. The students’ response was, “What do we do: add, subtract, multiply, or divide?” Knowing how to divide is important; understanding when to divide is equally important.

HoW To USe

10

THe SUGGeSTeD 3-STep MATH-U-See ApproAcH In order to train students to be confident problem solvers, here are the three steps that I suggest you use to get the most from the Math-U-See curriculum at this level:

Step 1. Preparation for the lesson. Step 2. Presentation of the new topic. Step 3. Progression after mastery.

Step 1. preparation for the lesson.

This course assumes a knowledge of Geometry and Algebra 2. There are two review lessons in the appendix covering the Pythagorean theorem and special triangles. Watch the DVD to learn the concepts. Study the written explanations and examples in the instruction manual. Many students watch the DVD along with their instructor. Students in the secondary level who have taken responsibility to study this course themselves will do well to watch the DVD and read through the instruction manual.

Step 2. presentation of the new topic.

Now that you have studied the new topic, choose problems from the instruction manual to present the new concept to your students.

a. Write: Record the step-by-step solutions on paper as you work them. b. Say: Explain the “why” and “what” of the math as you work the problems.

Do as many problems as you feel are necessary until the student is comfortable with the new material. One of the joys of teaching is hearing a student say, “Now I get it!” or “Now I see it!”

HoW To USe

11

Step 3. progression after mastery.

Once mastery of the new concept is demonstrated, begin doing the pages in the student text for that lesson. Mastery can be demonstrated by having each student teach the new material back to you. The goal is not to fill in worksheets, but to be able to teach back what has been learned. After the last student page in each lesson, you will find an “honors” lesson. These are optional, but highly recommended for students who will be taking advanced math or science courses. These challenging problems are a good way for all students to hone their problem-solving skills. Proceed to the lesson tests. They were designed to be an assessment tool to help determine mastery, but they may also be used as extra worksheets. Your students will be ready for the next lesson only after demonstrating mastery of the new concept. Confucius is reputed to have said, “Tell me, I forget; show me, I understand; let me do it, I will remember.” To which we add, “Let me teach it and I will have achieved mastery!”

Length of a Lesson So how long should a lesson take? This will vary from student to student and from topic to topic. You may spend a day on a new topic, or you may spend several days. There are so many factors that influence this process that it is impossible to predict the length of time from one lesson to another. I have spent three days on a lesson, and I have also invested three weeks in a lesson. This occurred in the same book with the same student. If you move from lesson to lesson too quickly without the student demonstrating mastery, he will become overwhelmed and discouraged as he is exposed to more new material without having learned the previous topics. But if you move too slowly, your student may become bored and lose interest in math. I believe that as you regularly spend time working along with your student, you will sense when is the right time to take the test and progress through the book. By following the three steps outlined above, you will have a much greater opportunity to succeed. Math must be taught sequentially, as it builds line upon line and precept upon precept on previously learned material. I hope you will try this methodology and move at your student’s pace. As you do, I think you will be helping to create a confident problem solver who enjoys the study of math.

HoW To USe

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Materials needed You will need the following items for this course:

A protractor for measuring angles. A ruler with inches and/or metric measure. A scientific calculator that does square roots, trigonometric functions,

logarithms, and natural log.

You can purchase an inexpensive calculator that does everything you need for this course. If you are planning on using it for more advanced courses, you can always get a more elaborate model.

HoW To USe

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onGoinG SUpporT AnD ADDiTionAL reSoUrceS

Welcome to the Math-U-See Family!Now that you have invested in your children’s education, I would like to tell you about the resources that are available to you. Allow me to introduce you to our staff, our ever improving website, the Math-U-See blog, our new free e-mail newsletter, and other online resources. Many of our customer service representatives have been with us for over 10 years. What makes them unique is their desire to serve and their expertise. They are able to answer your questions, place your student(s) in the appropriate level, and provide knowledgeable support throughout the school year. Come to your local curriculum fair where you can meet us face-to-face, see the latest products, attend a workshop, meet other MUS users at the booth, and be re-freshed. We are at most curriculum fairs and events. To find the fair nearest you, click on “Events” under “E-sources.”

The Website, at www.MathUSee.com, is continually being updated and im-proved.It has many excellent tools to enhance your teaching and provide more prac-tice for your student(s).

Math-U-See BlogInteresting insights and up-to-date information appear regularly on the Math-U-See Blog. The blog features updates, rep highlights, fun pictures,

and stories from other users. Visit us and get the latest scoop on what is happening.

email newsletter For the latest news and practical teaching tips, sign up online for the free Math-U-See e-mail newsletter. Each month you will receive an e-mail

with a teaching tip from Steve as well as the latest news from the website. It’s short, beneficial, and fun. Sign up today!

SUpporT AnD reSoUrceS

14

online Support You will find a variety of helpful tools on our website, including correc-tions lists, placement tests, answers to questions, and support options.

For Specific Math Help When you have watched the DVD and read the instruction manual and still have a question, we are here to help. Call us or click the support link. Our trained staff are available to answer a question or walk you through a specific lesson.

Feedback Send us an e-mail by clicking the feedback link. We are here to serve you and help you teach math. Ask a question, leave a comment, or tell us how you and your student are doing with Math-U-See. Our hope and prayer is that you and your students will be equipped to have a successful experience with math!

Blessings,

Steve Demme

15inTroDUcTion To TriGonoMeTrY - LeSSon 1

LeSSon 1 Introduction to Trigonometry

The word trigonometry comes from the Greek words trigonos (trigonos), which means triangle, and metreo (metreo), which means to measure. The Webster’s 1828 Dictionary defines it as “the measuring of triangles; the science of determining the sides and angles of triangles, by means of certain parts which are given.” From geometry, remember that a triangle is composed of three angles and three sides. Throughout this course, unless specified otherwise, we will be dealing with right triangles. We already know that one of the angles in a right triangle is 90° and that the side opposite the 90° angle is the hypotenuse. Since we are dealing primarily with right triangles, the Pythagorean theorem also applies, so leg squared plus leg squared equals the hypotenuse squared. This is reviewed in Appendix A. Special right triangles, such as the 30°– 60°– 90° and 45°– 45°– 90° triangles, pertain to our study as well. There is a brief refresher unit on this in Appendix B. Make sure the student understands the material in these two review lessons before proceeding further. They lay the foundation for our study of trigonometry. In order to measure “the sides and angles of triangles by means of certain parts which are given,” as our definition tells us, we need to name the angles and sides. For example, in the 30°– 60°– 90° right triangle, the three sides are referred to as the short leg, the long leg, and the hypotenuse. Notice that the 30° angle (the smallest angle) is opposite the smaller leg, the 60° angle is opposite the longer leg, and the 90° right angle is opposite the hypotenuse. In the 45°– 45°– 90° triangle, the two legs are congruent, so we can’t describe them in terms of length. Let’s learn a new system of identification for trigonometry. We need to describe all the angles and all the sides. Since this is a right triangle, we already know that one angle is 90° and the side opposite the right angle is the

LeSSon 1

inTroDUcTion To TriGonoMeTrY

16 LeSSon 1 - inTroDUcTion To TriGonoMeTrY precALcULUS

hypotenuse. This leaves two angles and two sides to name. I’ve decided to call the angles θ (theta) and α (alpha), both letters in the Greek alphabet. The sides will be described in reference to the angles. If I am standing in angle θ (∠θ), then the leg furthest away from me will be the opposite side. The side that touches me (where my feet are standing in the triangle) is the adjacent side. If I move to stand in angle α, then the side furthest away from me becomes the opposite side, and the side touching me is the adjacent side. The names for the sides depend on what angle you are referring to. I’ll illustrate this with our old friend, the 3–4–5 right triangle.

3

4

5

θ

α

Standing at θ, the opposite side is three units long and the adjacent side is four units long. If I move to α, then four is opposite and three is adjacent. In both instances, the hypotenuse is five units long. Now we come to the six trigonometric ratios, using the terminology of oppo-site, adjacent, and hypotenuse. The three main ratios are sine, cosine, and tangent. The sine of either angle in a right triangle (in our example θ or α) is described as the ratio of the opposite side over the hypotenuse.

oppositehypotenuse

Using the 3–4–5 right triangle, the sine of angle θ is shown below.

opposite

hypotenuse or sin θ = 35

The cosine of θ is the adjacent over the hypotenuse, or cos θ = 45

.

And the tangent of θ is the opposite over the adjacent, or tan θ = 34

.

17inTroDUcTion To TriGonoMeTrY - LeSSon 1precALcULUS

A fun way to remember these three trigonometric ratios is the result of drop-ping a brick on your big toe. What would you do? Probably get a pan of water and “soak a toe,” or SOH–CAH–TOA.

SoH stands for sin = opposite

hypotenuse ⇒ S oH

=

cAH stands for cos = adjacent

hypotenuse ⇒ c AH

=

ToA stands for tan = oppositeadjacent ⇒ T o

A=

Another relationship found in these trigonometric expressions is that sine and cosine are complementary angles. In a right triangle (with one right angle), the other two angles always add up to 90°, so they are complementary. Let’s look at the ratios again. They are constant for a 45°– 45°– 90° triangle or a 30°– 60°– 90° triangle. (See Appendix B.) In a 30°– 60°– 90° triangle, the short side is always one-half the hypotenuse. The length of the sides of all the 30°– 60°– 90° right triangles may vary, but the short side will always be one-half of the hypot-enuse. Using our new terminology, we can now show it as follows.

sin º side30 = =oppositehypotenuse

shorthypotenuse

== 12

30º

60º1

2

30º

60º105

5 3

sin º 30 510

12

= = =oppositehypotenuse

In the next triangle notice that the short side, which is opposite 30º, is 5 unites long. The hypotenuse is 10 unites long. So the sine ratio is 5/10 or 1/2.

18 LeSSon 1 - inTroDUcTion To TriGonoMeTrY precALcULUS

Here is another possibility.

30º

60º84

4 3

sin º 30 48

12

= = =oppositehypotenuse

Observe that the ratio of the small side to the hypotenuse remains constant.

Example 1

Find the sine, cosine, and tangent ratios for 30º and 60º in the triangle.

30º

60º105

5 3

sin º sin º

cos º

30 510

12

60 5 310

32

30 5 310

= = = =

= = 332

60 510

12

30 5

5 3

33

60 5 35

3

cos º

tan º n º

= =

= = = =ta

The ratios opp/hyp, adj/hyp, and opp/adj always depend on what angle is being referred to. That is why the sin 30° and the sin 60° are different, even though they are both opposite over hypotenuse.

Example 2

Find the sine, cosine, and tangent ratios for 45º. (remember the

relationships we found earlier between the lengths of the sides of a

45º–45º–90º triangle.)

4 2

45º

45º

4

4

sin º

cos º

tan º

45 4

4 2

4 28

22

45 4

4 2

4 28

22

45 44

= = =

= = =

= = 11

No matter what the lengths of the sides, the ratio for any given angle will always be the same when reduced to its simplest form.

19inTroDUcTion To TriGonoMeTrY - LeSSon 1precALcULUS

It is very important that you memorize the fractional trig ratios for 30°, 45°, and 60° since they will be used in future lessons.

Practice Problems 1

Find the sine, cosine, and tangent of θ and α, and express the ratios in

fraction form.

θ

α

12

513

1. sin θ = 4. sin α =

2. cos θ = 5. cos α =

3. tan θ = 6. tan α =

θ

α

24

725

7. sin θ = 10. sin α =

8. cos θ = 11. cos α =

9. tan θ = 12. tan α =

20 LeSSon 1 - inTroDUcTion To TriGonoMeTrY precALcULUS

θ

α 186

17

13. sin θ = 16. sin α =

14. cos θ = 17. cos α =

15. tan θ = 18. tan α =

θ

α

106

8

19. sin θ = 22. sin α =

20. cos θ = 23. cos α =

21. tan θ = 24. tan α =

21inTroDUcTion To TriGonoMeTrY - LeSSon 1precALcULUS

Solutions 1

1. sin θ = 513

4. sin α = 1213

2. cos θ = 1213

5. cos α = 513

3. tan θ = 512

6. tan α = 125

or 2 25

7. sin θ = 725

10. sin α = 2425

8. cos θ = 2425

11. cos α = 725

9. tan θ = 724

12. tan α = 247

or 3 37

13. sin θ = 618

or 13

16. sin α = 1718

14. cos θ = 1718

17. cos α = 618

or 13

15. tan θ = 617

18. tan α = 176

or 2 56

19. sin θ = 810

or 45

22. sin α = 610

or 35

20. cos θ = 610

or 35

23. cos α = 810

or 45

21. tan θ = 86

or 1 13

24. tan α = 68

or 34

22 LeSSon 1 - inTroDUcTion To TriGonoMeTrY precALcULUS

23reciprocAL TriGonoMeTric rATioS - LeSSon 2

LeSSon 2 Reciprocal Trigonometric Ratios

There are three other basic trigonometric ratios in addition to sine = opp/hyp, cosine = adj/hyp, and tangent = opp/adj. These ratios are the reciprocals of our original soh–cah–toa.

Where the sine (sin) = opphyp , the cosecant (csc) =

hypopp .

Where the cosine (cos) = adjhyp , the secant (sec) =

hypadj .

Where tangent (tan) = oppadj , the cotangent (cot) =

adjopp .

Now we have our stable full of all the possible trigonometric ratios. Study the following example to clarify the relationships between the ratios.

Example 1

Find all six trig ratios for θ.

θ

α

4

35

sin csc

cos sec

tan

θ θ

θ θ

θ

= =

= =

35

53

45

54

== =34

43

cot θ

LeSSon 2

reciprocAL TriGonoMeTric rATioS

24 LeSSon 2 - reciprocAL TriGonoMeTric rATioS precALcULUS

Since the new ratios are reciprocals of the sine, cosine, and tangent, they can also be expressed as shown below.

cscsin

θθ

= 1

sec

cos

θ

θ= 1

cot

tan

θ

θ= 1

Example 2

Show that the relationships above are true using the ratios from

example 1.

cscsin

θθ

=

=

=×

×

=

1

53

135

53

1 53

35

53

53

53

seccos

θθ

=

=

=×

×

=

1

54

145

54

1 54

45

54

54

54

cottan

θθ

=

=

=×

×

=

1

43

134

43

1 43

34

43

43

43

Practice Problems 1

Find all six trig ratios for θ.

θ

106

8

1. sin θ = 4. csc θ =

2. cos θ = 5. sec θ =

3. tan θ = 6. cot θ =

25reciprocAL TriGonoMeTric rATioS - LeSSon 2precALcULUS

θ

10

26

24

7. sin θ = 10. csc θ =

8. cos θ = 11. sec θ =

9. tan θ = 12. cot θ =

θ

48.5

7.5

13. sin θ = 16. csc θ =

14. cos θ = 17. sec θ =

15. tan θ = 18. cot θ =

The other new concept in this lesson involves changing the fractions to deci-mals rounded to the ten-thousandths place. Look at the solutions for #1–6, and then go back and complete #7–18 by changing each fraction to a decimal.

26 LeSSon 2 - reciprocAL TriGonoMeTric rATioS precALcULUS

Solutions 1

Here are #1–6 written as fractions and as decimals.

1. sin θ = 610

= .6000 4. csc θ = 106

= 1.6667

2. cos θ = 810

= .8000 5. sec θ = 108

= 1.2500

3. tan θ = 68

= .7500 6. cot θ = 86

= 1.3333

Here are #7–18 written as fractions and as decimals.

7. sin θ = 2426

= .9231 13. sin θ = 48 5.

= .4706

8. cos θ = 1026

= .3846 14. cos θ = 7 58 5..

= .8824

9. tan θ = 2410

= 2.400 15. tan θ = 47 5.

= .5333

10. csc θ = 2624

= 1.0833 16. csc θ = 8 54. = 2.1250

11. sec θ = 2610

= 2.6000 17. sec θ = 8 57 5

.

. = 1.1333

12. cot θ = 1024

= .4167 18. cot θ = 7 54. = 1.8750

27reciprocAL TriGonoMeTric rATioS - LeSSon 2precALcULUS

Practice Problems 2

Find the missing side, and then name all six trigonometric ratios of θ

for each triangle. State your answers as fractions and as decimals. if an

answer has a radical in it, leave it as a radical.

1. 2.

3. 4.

θ

X5

3

θ

12

5

H

θ

15

X

17

θ

10

8

H

28 LeSSon 2 - reciprocAL TriGonoMeTric rATioS precALcULUS

5. 6.

7. 8.

9.

θ

24

X

25

θ

36

X

39

θ

5

5

H

θ

Y8

4 3

θ

H

3 2

3 2

29reciprocAL TriGonoMeTric rATioS - LeSSon 2precALcULUS

Solutions 2

1 3 59 25

164

45

80002 2 2

2

. sin . csc+ =+ =

==

= = =xx

xx

θ θ 554

1 2500

35

6000 53

1 6667

43

=

= = = =

=

.

cos . sec .

tan

θ θ

θ == = =1 3333 34

7500. cot . θ

2 5 12

25 144

16913

1213

92312 2 2

2

2

. sin .+ =

+ =

==

= =H

H

HH

θ

csc .

cos . sec .

θ

θ θ

= =

= = = =

1312

1 0833

513

3846 135

2 60000

125

2 4000 512

4167tan . cot . θ θ= = = =

3 15 17

648

1517

8824 171

2 2 2

2

. sin . csc+ =

==

= = =x

xx

θ θ55

1 1333

817

4706 178

2 1250

1

=

= = = =

=

.

cos . sec .

tan

θ θ

θ 558

1 8750 815

5333= = =. cot . θ

4 8 10

64 100

164

2 41

10

2 41

5 414

2 2 2

2

2

. sin+ =

+ =

=

=

= =H

H

H

H

θ11

2 4110

415

8

2 41

4 4141

2 418

41

csc

cos sec

θ

θ θ

= =

= = = =44

108

1 2500 810

8000tan . cot . θ θ= = = =

5 24 25

497

2425

9600 2524

2 2 2

2

. sin . csc+ =

==

= = =x

xx

θ θ ==

= = = =

=

1 0417

725

2800 257

3 5714

24

.

cos . sec .

tan

θ θ

θ77

3 4286 724

2917= = =. cot . θ

30 LeSSon 2 - reciprocAL TriGonoMeTric rATioS precALcULUS

6 36 39

22515

3639

9231 32 2 2

2

. sin . csc+ =

==

= = =x

xx

θ θ 9936

1 0833

1539

3846 3915

2 6000

=

= = = =

.

cos . sec .

tan

θ θ

θ θ= = = =3615

2 4000 1536

4167. cot .

7 5 5

25 25

50

5 2

5

5 2

22

2 2 2

2

2

. sin csc+ =

+ =

=

=

= =H

H

H

H

θ θ == =

= = = =

= = = =

2

22

5

5 2

22

2

22

55

1 55

cos sec

tan cot

θ θ

θ θ 11

8 4 3 8

164

48

12

21

22

2 2

2

. sin csc

co

( ) + =

==

= = = =Y

YY

θ θ

ss sec

tan cot

θ θ

θ θ

= = = =

= = = =

4 38

32

2

3

2 33

4

4 3

33

3

33

9 3 2 3 2

18 18

366

3 26

22

2 22

2

2

. sin c( ) + ( ) =

+ =

==

= =H

H

HH

θ ssc

cos sec

tan cot

θ

θ θ

θ

= =

= = = =

= =

2

22

3 26

22

2

22

3 2

3 21 θθ = =3 2

3 21

31inTerpreTinG THe TriGonoMeTrY TABLeS - LeSSon 3

LeSSon 3 Interpreting the Trigonometry Tables

In this lesson, we’ll work on reading and understanding the trigonometry tables. Remember that a trigonometric ratio may be represented as a fraction, or ratio. The fraction can then be transformed to a decimal by dividing the denominator into the numerator. Trigonometry (or trig) tables are simply lists of ratios or frac-tions expressed as decimals. Turn the page to find a table of trig ratios. These ratios may also be referred to as trigonometric functions. To find the trig ratio when the measure of the angle is given, move down the first column to find the angle measure. The angle measures are listed in degrees. Then choose the appropriate column from the three columns which represent the trig functions: sine, cosine, or tangent. Now read the decimal ratio.

Example 1

Find the decimal ratio for cos 43º.

Locate 43º in the first column and move across to the cosine column.

The decimal ratio is .7314.

Practice Problems 1

Using the trig table, find the ratios for the following trigonometric

ratios, or functions.

1. sin 32º = 2. cos 46º =

3. tan 15º = 4. tan 45º =

LeSSon 3

inTerpreTinG THe TriGonoMeTrY TABLeS

32 LeSSon 3 - inTerpreTinG THe TriGonoMeTrY TABLeS precALcULUS

5. sin 60º = 6. tan 27º =

7. cos 58º = 8 sin 0º =

9. cos 73º =

Solutions 1

1. sin 32º = .5299 2. cos 46º = .6947

3. tan 15º = .2679 4. tan 45º = 1.0000

5. sin 60º = .8660 6. tan 27º = .5095

7. cos 58º = .5299 8. sin 0º = .0000

9. cos 73º =.2924

You’ve probably observed that the ratios are the same when they are computed from two complementary angles. In a 3–4–5 right triangle, the sine ratio for θ is 3/5, which is also the cosine ratio for α. Since sin θ = 3/5 and cos α = 3/5, we see that sin θ = cos α. Since this is a right triangle, θ and α are complementary angles (they add up to 90°). The trig table may also be used to find the measure of an angle when the ratio is known. Choose the appropriate column (sine, cosine, or tangent), look down the column to find the four-digit decimal which is closest to the ratio given, and then read across to find the measure of the angle.

Example 2

Given the ratio .7813 for the tangent, find the measure of the angle.

Locate .7813 in the third (tangent) column, and then move left to

the degrees in the angle column.

The measure of the angle is 38º.

33inTerpreTinG THe TriGonoMeTrY TABLeS - LeSSon 3precALcULUS

Table ofTrig Ratios

degree sine cosine tangent degree sine cosine tangent

0 .0000 1.0000 .0000 46 .7193 .6947 1.0355

1 .0175 .9998 .0175 47 .7314 .6820 1.0724

2 .0349 .9994 .0349 48 .7431 .6691 1.1106

3 .0523 .9986 .0524 49 .7547 .6561 1.1504

4 .0698 .9976 .0699 50 .7660 .6428 1.1918

5 .0872 .9962 .0875 51 .7771 .6293 1.2349

6 .1045 .9945 .1051 52 .7880 .6157 1.2799

7 .1219 .9925 .1228 53 .7986 .6018 1.3270

8 .1392 .9903 .1405 54 .8090 .5878 1.3764

9 .1564 .9877 .1584 55 .8192 .5736 1.4281

10 .1736 .9848 .1763 56 .8290 .5592 1.4826

11 .1908 .9816 .1944 57 .8387 .5446 1.5399

12 .2079 .9781 .2126 58 .8480 .5299 1.6003

13 .2250 .9744 .2309 59 .8572 .5150 1.6643

14 .2419 .9703 .2493 60 .8660 .5000 1.7321

15 .2588 .9659 .2679 61 .8746 .4848 1.8040

16 .2756 .9613 .2867 62 .8829 .5695 1.8807

17 .2924 .9563 .3057 63 .8910 .4540 1.9626

18 .3090 .9511 .3249 64 .8988 .4384 2.0503

19 .3256 .9455 .3443 65 .9063 .4226 2.1445

20 .3420 .9397 .3640 66 .9135 .4067 2.2460

21 .3584 .9336 .3839 67 .9205 .3907 2.3559

22 .3746 .9272 .4040 68 .9272 .3746 2.4751

23 .3907 .9205 .4245 69 .9336 .3584 2.6051

24 .4067 .9135 .4452 70 .9397 .3420 2.7475

25 .4226 .9063 .4663 71 .9455 .3256 2.9042

26 .4384 .8988 .4877 72 .9511 .3090 3.0777

27 .4540 .8910 .5095 73 .9563 .2924 3.2709

28 .4695 .8829 .5317 74 .9613 .2756 3.4874

29 .4848 .8746 .5543 75 .9659 .2588 3.7321

30 .5000 .8660 .5774 76 .9703 .2419 4.0108

31 .5150 .8572 .6009 77 .9744 .2250 4.3315

32 .5299 .8480 .6249 78 .9781 .2079 4.7046

33 .5446 .8387 .6494 79 .9816 .1908 5.1446

34 .5592 .8290 .6745 80 .9848 .1736 5.6713

35 .5736 .8192 .7002 81 .9877 .1564 6.3138

36 .5878 .8090 .7265 82 .9903 .1392 7.1154

37 .6018 .7986 .7536 83 .9925 .1219 8.1443

38 .6157 .7880 .7813 84 .9945 .1045 9.5144

39 .6293 .7771 .8098 85 .9962 .0872 11.4301

40 .6428 .7660 .8391 86 .9976 .0698 14.3007

41 .6561 .7547 .8693 87 .9986 .0523 19.0811

42 .6691 .7431 .9004 88 .9994 .0349 28.6363

43 .6820 .7314 .9325 89 .9998 .0175 57.2900

44 .6947 .7193 .9657 90 1.0000 .0000 undefined

45 .7071 .7071 1.0000

34 LeSSon 3 - inTerpreTinG THe TriGonoMeTrY TABLeS precALcULUS

Practice Problems 2

Using the trig table, find the angle measure that corresponds to the

ratio for each of the following trigonometric functions.

1. tan ___º = 1.0724 2. cos ___º = .9063

3. tan ___º = 2.7475 4. tan ___º = .1944

5. sin ___º = .1564 6. sin ___º = .8290

7. cos ___º = .6157 8. sin ___º = .5592

9. cos ___º = .1564

Solutions 2

1. tan 47º = 1.0724 2. cos 25º = .9063

3. tan 70º = 2.7475 4. tan 11º = .1944

5. sin 9º = .1564 6. sin 56º = .8290

7. cos 52º = .6157 8. sin 34º = .5592

9. cos 81º = .1564

Let’s take it a step further and find the decimal ratio by studying the 30° angle in a right triangle. Sin 30° = .5000, which is another way of saying that the ratio of the small side (opposite the small angle) to the hypotenuse is 1 to 2, or 1/2, or .5 or .5000. Notice that cos 60° is also 1/2 or .5000. As an example, let’s use the trig ratios found in the table to discover the measures of the angles in the 3–4–5 right triangle. These angle measures will be approximate since none of the ratios exactly match our results when we change the fractions to decimals.

35inTerpreTinG THe TriGonoMeTrY TABLeS - LeSSon 3precALcULUS

Example 3

Find the measure of angle θ to the nearest degree using at least two

trig functions to confirm your answer.

Figure 1

θ

35

4

α

sin θ = 3/5 = .6000.

Looking in the table in the sine column,

we see that this is closest to .6018 which

indicates 37º.

cos θ = 4/5 = .8000, which is closest to

.7986 in the cosine column, indicating

and confirming 37º.

Example 4

Using the same triangle, find the measure of angle θ to the nearest

degree. Use at least two trig functions to confirm your answer.

To find α, we can choose from any of these three ratios:

sin α = 4/5 = .8000

cos α = 3/5 = .6000

tan α = 4/3 ≈ 1.3333

These ratios all indicate 53º for α. Since α and θ are complementary

angles, we can also confirm this by 90º – 37º = 53º.

θ = 37º

35

4

α = 53º

36 LeSSon 3 - inTerpreTinG THe TriGonoMeTrY TABLeS precALcULUS

The tangent may be expressed as the sine over the cosine as well as opposite over the adjacent. Since the sine is the opposite over the hypotenuse and the cosine is the adjacent over the hypotenuse, if you put them over each other, the hypote-nuse is canceled and you are left with the opposite over the adjacent. We’ll work this out using the triangle in example 4.

sin º

cos ºtan º sin º

cos º

37 35

37 45

37 3737

3545

34

=

== = =

== opposite

adjacent

This relationship will be used in upcoming lessons, but I wanted you to notice it now. Conversely, the cotangent may be expressed as cosine over the sine.

tan sincos

cot= = =oppositeadjacent

eine

adjacentoppossite

inee

= cossin

Example 5

Find the measures of the angles in the following triangle, showing all

three ratios for each angle.

θ

206

19

α

sin θ = 6/20 = .3000 sin α = 19/20 = .9500

cos θ = 19/20 = .9500 cos α = 6/20 = .3000

tan θ = 6/19 ≈ .3158 tan α = 19/6 ≈ 3.1667

Using the table, θ = 18º Using the table, α = 72º

37inTerpreTinG THe TriGonoMeTrY TABLeS - LeSSon 3precALcULUS

Practice Problems 3

There are three ways to find the measure of an angle: sin, cos, and tan.

in each of these problems, use all three ratios to confirm the measures

of angle θ and angle α.

1.

θ5.8

20

19.1

α

sin . sin .

cos .

θ α

θ

= = = =

= =

cos .

tan . tan

α

θ α

= =

= = = ==

= =

.

º º

θ α

2.

sin . sin .

cos .

θ α

θ

= = = =

= =

cos .

tan . tan

α

θ α

= =

= = = ==

= =

.

º º

θ α

θ

38

23.4

30

α

38 LeSSon 3 - inTerpreTinG THe TriGonoMeTrY TABLeS precALcULUS

3.

sin . sin .

cos .

θ α

θ

= = = =

= =

cos .

tan . tan

α

θ α

= =

= = = ==

= =

.

º º

θ α

4.

sin . sin .

cos .

θ α

θ

= = = =

= =

cos .

tan . tan

α

θ α

= =

= = = ==

= =

.

º º

θ α

θ

2730

13

α

θ

9.219

16.6

α

39inTerpreTinG THe TriGonoMeTrY TABLeS - LeSSon 3precALcULUS

Solutions 3

1 5 820

2900 19 120

9550. sin . . sin . .

cos

θ α

θ

= = = =

= 119 120

9550 5 820

2900

5 819 1

. . cos . .

tan ..

= = =

= =

α

θ .. tan ..

.

º º

3037 19 15 8

3 2931

17 73

α

θ α

= =

= =

2 23 438

6158 3038

7895

3038

. sin . . sin .

cos

θ α

θ

= = = =

= == = =

= =

. cos . .

tan . .

7895 23 438

6158

23 430

7800

α

θ tan.

.

º º

α

θ α

= =

= =

3023 4

1 2821

38 52

3 2730

9000 1330

4333

133

. sin . sin .

cos

θ α

θ

= = = =

=00

4333 2730

9000

2713

2 0769

= = =

= =

. cos .

tan .

α

θ tan .

º º

α

θ α

= =

= =

1327

4815

64 26

4 9 219

4842 16 619

8737

1

. sin . . sin . .

cos

θ α

θ

= = = =

= 66 619

8737 9 219

4842

9 216 6

55

. . cos . .

tan ..

.

= = =

= =

α

θ 442 16 69 2

1 8043

29 61

tan ..

.

º º

α

θ α

= =

= =

40 LeSSon 3 - inTerpreTinG THe TriGonoMeTrY TABLeS precALcULUS

41USe THe TriG TABLe To SoLVe For THe UnKnoWn - LeSSon 4

In lesson 3, we learned how to find the measure of an angle when given the length of the two sides. In this lesson, we are going to learn how to find the length of a side when given the angle and the length of another side. Let’s start with an easier problem and then move to more difficult problems. In a 30°– 60°– 90° triangle, we know the ratio of the short side to the hypotenuse is 1 to 2 or 1/2 or .5000. When the short side is 8, we know the hypotenuse is 16.

Example 1

Find H.

8H 60º

30º

The equation looks like this:

sin 30º = 8H

Multiply both sides by H.

H sin 30º = 8 Divide both sides by sin 30º.

H = 830sin º

replace sin 30º with the ratio, which is .5000.

H = 85000.

Solve for H.

H = 16

LeSSon 4 Use the Trig Table to Solve for the Unknown

LeSSon 4

USe THe TriG TABLe To SoLVe For THe UnKnoWn

42 LeSSon 4 - USe THe TriG TABLe To SoLVe For THe UnKnoWn precALcULUS

We knew two pieces of information, so we were able to solve for the missing information, in this case the length of the hypotenuse.

Example 2

Find A.

A6

40º

sin 40º = A6 Multiply both sides by 6.

6 sin 40º = A replace sin 40º with .6428.

(6)(.6428) = A Multiply.

3.8568 = A We choose to round to hundredths.

3.86 = A

It was natural to use the sine to solve for the missing side since we were given an angle, the side opposite the angle, and the hypotenuse. We could also have solved this problem by using the cosine. The missing angle must be 50°, since the angles are complementary. So cos 50° = A/6, and we can continue the process.

A6 50º

40º

cos 50º = A6

Multiply both sides by 6.

6 cos 50º = A replace cos 50º with .6428.

(6)(.6428) = A Multiply and round to hundreths.

3.86 = A in real life, the context will determine how

many places to use for your answer.

43USe THe TriG TABLe To SoLVe For THe UnKnoWn - LeSSon 4precALcULUS

Example 3

Find B.

4

B29º

in this example, we choose to use the tangent to solve the problem

efficiently, since we are interested in the sides opposite and adjacent to

the angle

tan 29º = B4 Multiply both sides by B.

B tan 29º = 4 Divide both sides by tan 29º.

B = 429tan º

replace tan 29º with the ratio, which is .5543.

B = 4

5543. Divide and round to hundredths.

B = 7.22

We could have found the other angle by subtracting (90° – 29° = 61°), and used the equation tan 61° = B/4 to find B. The equation would have been easier than the one we used in example 3, since no division is involved.

Example 4

tan 61º = B4

Multiply both sides by 4.

4 tan 61º = B replace tan 61º with 1.804.

(4)(1.804) = B Multiply and round to hundredths.

7.22 = B

44 LeSSon 4 - USe THe TriG TABLe To SoLVe For THe UnKnoWn precALcULUS

Practice Problems 1

To solve for the unknown, carefully examine what information is

given, and choose the trig function that will solve the problem most

efficiently. remember that there is more than one way to solve each

problem. estimate your answer.

1. 2.

3. 4.

5. 6.

5A

47º

B

8.2

33º

3C

25º

D

1464º

E

.99º

F

2338º

45USe THe TriG TABLe To SoLVe For THe UnKnoWn - LeSSon 4precALcULUS

Solutions 1

5 9 9

99

91

. cos º .

.cos º

.

=

=

=

e

e

e

6 382323 3817 97

. tan º

tan º.

=

==

F

FF

1 475

5 473 66

. sin º

sin º.

=

==

A

AA

2 338 28 2 336 88

. cos º.. cos º.

=

==

B

BB

3 25 3

25 3325

7 10

. sin º

sin º

sin º.

=

=

=

=

cc

c

c

4 641414 6428 70

. tan º

tan º.

=

==

D

DD

46 LeSSon 4 - USe THe TriG TABLe To SoLVe For THe UnKnoWn precALcULUS

47USinG A cALcULATor AnD Arc FUncTionS - LeSSon 5

I wanted you to use the trig table so you would understand that it is a compilation of rational numbers, or ratios, written as decimals. But it is easier, quicker, and more accurate to use a scientific calculator that has the trigonometric functions. Here is how to select and use one. (This is how they usually operate. Consult the operating instructions that came with your calculator to be sure.) If you have a calculator with keys for sinsin , coscos , tantan , then you are all set. Usually you will also have a button that says INV for inverse or buttons that enable you to get to functions labeled sinsin–1–1 , coscos–1–1 , or tantan–1–1 . Hopefully yours also has a 1/x1/x key. If not, when you go shopping, pick one up with these characteristics.

Generally, if it has these capabilities, it will also be able to solve for squares and square roots and find logarithms. Later, we will be studying graphing. While a graphing calculator isn’t neces-sary for our study of trigonometry, one is nice if you will be doing more advanced math in the future. There is also a key that switches the value of the numbers on the screen from “degree” to “grad” to “radian” (or rad). At this point in our study you'll want to have “degree” showing. Now here’s how to use your calculator. Compare your results with the trig table in lesson 3 to make sure you are on the right track. To find a trig ratio with a calculator instead of the table, punch in the number of degrees, hit the trig func-tion, and your decimal ratio will be shown.

LeSSon 5 Using a Calculator and Arc Functions

LeSSon 5

USinG A cALcULATor AnD Arc FUncTionS

48 LeSSon 5 - USinG A cALcULATor AnD Arc FUncTionS precALcULUS

Example 1

Find the decimal ratio for sin 30º.

enter 30 into your calculator and then hit sinsin .

The screen should read .5.

Example 2

Find the decimal ratio for tan 81º.

enter 81 into your calculator and then hit tantan .

The screen should read 6.313751 , or 6.3138

when rounded to four places or ten-thousandths as in the table.

Practice Problems 1

Find the decimal ratio for each angle. Use your calculator and round to

ten-thousandths.

1. sin 14º 2. tan 88º

3. tan 67º 4. cos 36º

5. sin 52º 6. cos 45º

Solutions 1

1. sin 14º = .2419 2. tan 88º = 28.6363

3. tan 67º = 2.3559 4. cos 36º = .8090

5. sin 52º = .7880 6. cos 45º = .7071

49USinG A cALcULATor AnD Arc FUncTionS - LeSSon 5precALcULUS

Arc FUncTionS The third part of this lesson tells us how to use the calculator to find degrees when we are given the decimal ratio. We will also learn a new term to indicate this operation. We read sin 30° = 1/2 as “the sin of 30° is one-half.” To go in the oppo-site direction, we say, “The angle whose sine is one-half is 30°.” This expression is written as: arcsin 1/2 = 30°, so arc stands for “the angle whose.” Using a decimal, the expression is arcsin (.5000) = 30°.

To summarize: Given the angle, find the ratio.

For sine 30º = .5000, say, “The sine of 30° is .5000.”

Given the ratio, find the angle.

For arcsin (.5000) = 30°, say, “The angle whose sine

is .5000 is 30°.”

For arccos (.5000) = 60°, say, “The angle whose cosine

is .5000 is 60°.”

For arctan (1.000) = 45º, say, “The angle whose tangent

is 1.000 is 45º.”

The arcsin is referred to as the inverse function. Sin–1 is the inverse of sin. When given the ratio, we use the arc, or inverse, in order to find the angle measure. In lesson 3 we divided two numbers to find the ratio, and then looked at the table to find the nearest angle. Now, instead of looking at the table, we can use a calcu-lator to find the angle. To use the calculator, first find out whether there is an arcarc button and an INVINV button, or if you need to push 2ND2ND sinsin to get sin–1, which is above the sinsin button.

If you are given sin .5000 and want to find the measure of the angle, enter .5 and do one of the following:

2ND2ND sinsin or INVINV sinsin or arcarc sinsin to get sin–1.

Work with your calculator until you understand how to find the ratio when given the angle and how to find the angle when given the ratio.

50 LeSSon 5 - USinG A cALcULATor AnD Arc FUncTionS precALcULUS

Practice Problems 2

Use a calculator to find the degrees rounded to the hundredths place.

1. arcsin (.8387) = 2. arccos (.3855) =

3. arccos (.0954) = 4. arccos (.6378) =

5. arctan (3.5971) = 6. arcsin (.9874) =

Solutions 2

1. arcsin (.8387) = 57.00º 2. arccos (.3855) = 67.33º

3. arccos (.0954) = 84.53º 4. arccos (.6378) = 50.37º

5. arctan (3.5971) = 74.46º 6. arcsin (.9874) = 80.89º

DeGreeS - MinUTeS - SeconDS Degrees are often given in a form other than decimals. For more accuracy, they are broken down into minutes and seconds. The expression 34° 20' 45" is read as “thirty-four degrees, twenty minutes, and forty-five seconds.” To change this to a decimal so that we can use the calculator, remember that there are 60 seconds in a minute and 60 minutes in a degree.

Example 3

change 34º 20' 45" to a decimal, using unit multipliers.

change the seconds to minutes first:

45

11

604560

seconds minute seconds

× = =minutes .75 minutes

51USinG A cALcULATor AnD Arc FUncTionS - LeSSon 5precALcULUS

change the minutes to degrees:

20 751

160

20 7560

. . deg min deg min

× = = .3458 degrees

(rounds to .35º)

So 34º 20' 45" = 34.35º.

now we can find the trig ratio by entering 34.35º and using sine,

cosine, or tangent to find the ratio.

Practice Problems 3

change the degrees–minutes–seconds to a decimal number using

unit multipliers.

1. 28º 30' 25" 2. 53º 42' 10"

3. 36º 24' 4. 41º 15' 48"

5. 18º 50' 08" 6. 62º 00' 06"

Solutions 3

1. 28.51º 2. 53.70º

3. 36.4º 4. 41.26º

5. 18.84º 6. 62.00º

52 LeSSon 5 - USinG A cALcULATor AnD Arc FUncTionS precALcULUS

Now we will learn how to change from degrees–minutes–seconds to a decimal number by using a calculator. This procedure may vary from one calculator to another. DMS to DD is the key button. DD stands for decimal degrees, and DMS is degrees-minutes-seconds. If your calculator does not have these buttons, refer to your owner’s manual.

Example 4

enter 28º 30' 25" as 28.3025.

push the button DMS→DDDMS→DD .

it shows 28.50694 or 28.51º (rounded).

Do #2–5 from practice problems 3 again to check your work.

Now change from a decimal to degrees-minutes-seconds. DD to DMS is the key button.

Example 5

enter 28.50694 (DD).

push the button DD→DMSDD→DMS .

it shows 28.30249, representing 28º 30' 25" (DMS).

Change your answers to #2–5 from practice problems 3 back to DMS in order to check your work.

Now that we can use a calculator to change back and forth from DMS to DD, let’s take it another step and find a trig ratio.

53USinG A cALcULATor AnD Arc FUncTionS - LeSSon 5precALcULUS

Example 6

Find the decimal ratio for sin 15º 24' 50".

enter 15.2450 (DMS).

push the button DMS→DDDMS→DD . it shows 15.413888 (DD).

push the button sinsin .

it reads .2657898, which may be rounded to .2658

Practice Problems 4

Use the calculator to find the decimal ratio for each angle.

1. cos 7º 34' 2. sin 68º 27' 43"

3. sin 36º 48' 4. tan 24º 12' 12"

5. tan 50º 08' 22" 6. cos 47º 00' 05"

Solutions 4

1. cos 7º 34' = .9913 2. sin 68º 27' 43" = .9302

3. sin 36º 48' = .5990 4. tan 24º 12' 12" = .4495

5. tan 50º 08' 22" = 1.1977 6. cos 47º 00' 05" = .6820

54 LeSSon 5 - USinG A cALcULATor AnD Arc FUncTionS precALcULUS

Example 7

What is arcsin (.5222)?

arcsin (.5222) = 31.48º or 31º 28' 48"

When estimating this kind of problem, think of .48° as a little less than half of a degree. A little less than one-half of 60 minutes would be 28 or 29. So 28' 48" (28 minutes 48 seconds) is what we should expect.

Practice Problems 5

1. arctan (.6565) = ____º

2. arcsin (.0911) = ____º

3. arcsin (.8874) = ____º

4. arccos (.9600) = ____º

5. arccos (.3131) = ____º

6. arctan (1.5988) = ____º

Solutions 5

1. arctan (.6565) = 33.28º or 33º 17' 06"

2. arcsin (.0911) = 5.23º or 5º 13' 37"

3. arcsin (.8874) = 62.55º or 62º 32' 54"

4. arccos (.9600) = 16.26º or 16º 15' 37"

5. arccos (.3131) = 71.75º or 71º 45' 14"

6. arctan (1.5988) = 57.98º or 57º 58' 31"

55USinG A cALcULATor AnD Arc FUncTionS - LeSSon 5precALcULUS

Example 8

B

817.3º

cos . º

cos . º .

.

17 3 8

817 3

89548

8 38

=

= =

=

B

B

B

Practice Problems 6

Use your calculator to solve for the missing side or angle.

1. 2.

3. 4.

5. 6.

A

1227º

θ29.4

17.5

B

5.739.2º

θ

9.7

12

C

10.438º

D

44.1º

.87

56 LeSSon 5 - USinG A cALcULATor AnD Arc FUncTionS precALcULUS

Solutions 6

5 38 10 4

10 438

10 47880

13 20

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.cos º

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c

c

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B

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4 129 7

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==

261pYTHAGoreAn THeoreM - AppenDix A

AppenDix A Pythagorean Theorem

AppenDix A

pYTHAGoreAn THeoreM

This is a right triangle. There are three sides. The two sides that join to form the right angle are called the legs. You can remember this by the letter L for Leg, because L makes a right angle. Besides the two legs, there is the longest side, which is called the hypot-enuse. (Hypotenuse is the longest word and the longest side!)

The most familiar right triangle is the 3–4–5 right triangle. Here is a picture of this triangle, showing the Pythagorean theorem, which is “leg squared plus leg squared equals hypotenuse squared.”

25 = 25

24 = 16

23 = 9 Leg Leg Hypotenuse2 2 2

2 2 23 4 59 16 25

25 25

+ =

+ =+ =

=

Leg

Leg

Hypotenuse

262 AppenDix A - pYTHAGoreAn THeoreM precALcULUS

If you have a right triangle, then the Pythagorean theorem works. The converse is also true. If leg squared plus leg squared equals hypotenuse squared, then the triangle is is a right triangle. Look at these examples to make sure you understand this important theorem.

Example 1

7

H6

Leg Leg Hyp

H

H

H

H

2 2 2

2 2 2

2

2

6 7

36 49

85

85

+ =

+ =

+ =

=

=

Example 2

A triangle has sides of 6, 8, and 10. is it a right triangle?

62 + 82 = 102

36 + 64 = 100 Yes, it is a right triangle.

Example 3

8

L

3

Leg Leg Hyp

L

L

L

L

2 2 2

2 2 2

2

2

3 8

9 64

55

55

+ =

+ =

+ =

=

=

Example 4

A triangle has sides of 4, 5, and 6. is it a right triangle?

42 + 52 = 62

16 + 25 ≠ 36 No, it is not a right triangle.

263SpeciAL TriAnGLeS (45º-45º-90º; 30º-60º-90º) - AppenDix B

AppenDix B Special Triangles (45º-45º-90º; 30º-60º-90º)

AppenDix B

SpeciAL TriAnGLeS (45º-45º-90º; 30º-60º-90º)

Have you noticed the relationship between the legs and the hypotenuse of a 45°-45°-90° right triangle? Since the angles are congruent, the opposite sides are also congruent. Think of the triangle shown as half of a square.

45º

45º

45º

45º45º

45º

Find the length of the hypotenuse of the following triangles, and observe the common thread.

6

6

6 6

36 36

72

72

36 2

6 2

2 2 2

2

2

+ =

+ =

=

=

=

=

H

H

H

H

H

H

10

10

10 10

100 100

200

200

100 2

10 2

2 2 2

2

2

+ =

+ =

=

=

=

=

H

H

H

H

H

H

264 AppenDix B - SpeciAL TriAnGLeS (45º-45º-90º; 30º-60º-90º) precALcULUS

Now let's try a tricky one with A as the length of the legs.

A

A

A A H

A H

A H

A H

A H

2 2 2

2 2

2 2

2

2

2

2

2

+ =

=

=

=

=

The hypotenuse is 2 times the leg. If the leg is 5, then the hypotenuse will be 5 2 . If the leg is x, then the hypotenuse is x 2 . We employed the Pythagorean theorem to discover this unique relation-ship between the legs and the hypotenuse in a 45°-45°-90° right triangle. Notice that this is true regardless of the lengths of the legs or of the hypotenuse. For any 45°-45°-90° triangle, whatever the length of one leg is, the other leg will be the same length, and the hypotenuse will be 2 times that length! Not only does the Pythagorean theorem reveal this, but it also proves it to be so for any 45°-45°-90° right triangle. The ratio between the legs, and between a leg and the hypotenuse, is constant, i.e. always true. Another special right triangle is the 30°-60°-90° right triangle. Let’s do a couple of examples to observe the relationship and discover the pattern.

H60º

30º

4

4 3

4 4 3

16 16 3

16 48

64

8

2 2 2

2

2

2

+ ( ) =

+ ( ) =

+ =

=

=

H

H

H

H

H

⋅

H60º

30º

7

7 3

7 7 3

49 49 3

49 147

19614

2 2 2

2

2

2

+ ( ) =

+ ( ) =

+ =

==

H

H

H

HH

⋅

precALcULUS 265SpeciAL TriAnGLeS (45º-45º-90º; 30º-60º-90º) - AppenDix B

Now predict the hypotenuse in this triangle (based on your observations); and then work it out to confirm your hypothesis.

H60º

30º

S

S 3

S S H

S S H

S S H

S HS H

2 2 2

2 2 2

2 2 2

2 2

3

3

3

42

+ ( ) =

+ ( ) =

+ =

==

⋅

The hypotenuse is twice the length of the short side (the side opposite the smallest angle). If you are given the short leg, you double it to find the hypotenuse. Now let’s find the relationship between the short leg and the long leg. We double the short leg to find the hypotenuse, which is 10. Our equation to find the long leg is:

1060º

30º

5

L L

5 10

25 100

75

25 3

5 3

2 2 2

2

2

+ ( ) =

+ ( ) =

( ) =

=

=

LL

LL

LL

LL

LL

Now let’s use variables to confirm our guess. If the hypotenuse is 2B, then the short leg is B.

2B60º

30º

B

L L

B LL B

B LL B

LL B

LL B

LL B

( ) + ( ) = ( )+ ( ) =

( ) =

=

=

2 2 2

2 2 2

2 2

2

4

3

3

3

The long leg is 3 times the short leg.

266 AppenDix B - SpeciAL TriAnGLeS (45º-45º-90º; 30º-60º-90º) precALcULUS

In the 30°-60°-90° right triangle, the relationship between the sides is not determined by the lengths of the sides, but by the size or measure of the angles. The side opposite the 30° angle is always one-half the length of the hypotenuse (the side opposite the right angle). The side opposite the 60° angle is always 3 times the side opposite the 30° angle. These ratios are “set in concrete” so to speak—that is, they are always true.

ALGeBrA 2 267LeSSon 1A - LeSSon 1BprecALcULUS 267

STUDenT SoLUTionS

cosStudent SolutionsSolutions are shown in detail. The student may use canceling and other shortcuts as long as the answers match. If you see an error, see online solutions mentioned on page 14.Lesson 1ALLeessssoonn11AA

1.

2.

sin

cos

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θ

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4.

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7.

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32

8.

9. θθ

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1

3

33

13 326

32

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11.

sin

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α

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1326

12

13 313

31

312.

13..

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sin

cos

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θ

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130

7 130130

9

130

9 1130130

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130

9 13

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16.

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20.

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cos

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θ

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22

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4.

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sin

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7.

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tan

α

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cos

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θ

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1326

12

13 326

32

8.

9. θθ

α

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13

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1

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33

13 326

32

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11.

sin

ccos

tan

α

α

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1326

12

13 313

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312.

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sin

cos

θ

θ

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130

7 130130

9

130

9 1130130

79

9

130

9 13

15.

16.

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sin

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130

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sin

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θ

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5 2

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2

22

5

55 2

1

2

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1

5

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1

2

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4.

tan

sin

θ

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7.

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12

13 326

32

8.

9. θθ

α

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1

3

33

13 326

32

10.

11.

sin

ccos

tan

α

α

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adjhyp

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1326

12

13 313

31

312.

13..

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sin

cos

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θ

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130

7 130130

9

130

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79

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130

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16.

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sin

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sin

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sin

cos

θ

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sin

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5

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4.

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sin

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22

5

5 2

1

2

22

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1

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7.

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13 326

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8.

9. θθ

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13 313

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312.

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sin

cos

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130

7 130130

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130

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130

9 13

15.

16.

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sin

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130

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17.

18.

cos

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21.

sin

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203

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Lesson 1BLesson1B1.

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sin

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8.

9.

10..

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12.

sin

cos

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α

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8

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cos

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=

precALcULUS

LeSSon 1B - LeSSon 2A

SoLUTionS268

Lesson1B1.

2.

sin

cos

θ

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8.

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= =

135

2 6000

1312

1 0833

125

2 4000

≈

precALcULUS

LeSSon 2B - LeSSon 2c

SoLUTionS270

Lesson 2B1.

4 5 6 6

20 25 43 56

63 81

63

2 2 2

2

2

. .

. .

.

+ =

+ =

=

H

H

H

..

.

sin ..

.

cos ..

.

81

7 99

4 57 99

5632

6 67 99

82

=

=

=

H

H ≈

≈

≈

θ

θ 660

4 56 6

6818

7 994 5

1 7756

7 9

tan ..

.

csc ..

.

sec .

θ

θ

θ

=

=

=

≈

≈

996 6

1 2106

6 64 5

1 4667

6 8 51

4

2 2 2

2

..

cot ..

.

.

≈

≈θ =

+ =

+

2.

L

L 66 24 51

51 46 24

4 76

4 76

2 18

51 7 14

2

2

.

.

.

.

.

.

=

= −

=

=

L

L

L

L ≈

≈

ssin ..

.

cos ..

.

tan .

θ

θ

θ

=

=

=

2 187 14

3053

6 87 14

9524

2 18

≈

≈

66 83206

7 142 18

3 2752

7 146 8

1 05

..

csc ..

.

sec ..

.

≈

≈θ

θ

=

= = 000

6 82 18

3 1193

10 3

100 3

103

2 2 2

2

2

cot ..

.θ =

+ =

+ =

=

≈

3. H

H

H

1103

10 15

3 1 73

1 7310 15

1704

10=

=

=H

H ≈

≈

≈

.

.

sin ..

.

cos

θ

θ110 15

9852

1 7310

1730

10 151 73

5 8

..

tan . .

csc ..

.

≈

≈

θ

θ

= =

= 6671

10 1510

1 0150

101 73

5 7803

7 2

sec . .

cot.

.

θ

θ

= =

=

( )

≈

4.

22 22

2

7 3

7 7 2 2 7 7 3 3

49 2 49 3

+ ( ) =

( )( ) + ( )( ) =

( )( ) + ( )( ) =

H

H

HH

H

H

H

2

2

2

98 147

245

15 65

7 2 9 9

7 3 12 12

9

+ =

=

=

≈

≈

≈

.

.

.

sin .θ 9915 65

6326

12 1215 65

7744

9 912 12

..

cos ..

.

tan ..

≈

≈θ

θ

=

= ≈≈

≈

≈

.

csc ..

.

sec ..

.

8168

15 659 9

1 5808

15 6512 12

1 29

θ

θ

=

= 113

12 129 9

1 2242

2 3 11

2 2 3 3 1

22 2

cot ..

.θ =

( ) + =

( )( ) +

≈

5. H

221

4 3 121

133

11 53

2 3 3 46

3 462

2

2

=

( ) + =

=

=H

H

H

H ≈

≈

.

.

sin .θ111 53

3001

1111 53

9540

3 4611

3145

..

cos.

.

tan . .

≈

≈

≈

θ

θ

=

=

ccsc ..

.

sec . .

cot

θ

θ

θ

=

=

=

11 533 46

3 3324

11 5311

1 0482

1

≈

≈

113 46

3 1792

5 12

25 144

169

13

1

2 2 2

2

2

2

..≈

6. + =

+ =

=

=

=

H

H

H

H

H 33

513

3846

1213

9231

512

4167

sin .

cos .

tan .

c

θ

θ

θ

=

=

=

≈

≈

≈

ssc .

sec .

cot .

θ

θ

θ

= =

=

= =

135

2 6000

1312

1 0833

125

2 4000

≈

Lesson 2B1.

4 5 6 6

20 25 43 56

63 81

63

2 2 2

2

2

. .

. .

.

+ =

+ =

=

H

H

H

..

.

sin ..

.

cos ..

.

81

7 99

4 57 99

5632

6 67 99

82

=

=

=

H

H ≈

≈

≈

θ

θ 660

4 56 6

6818

7 994 5

1 7756

7 9

tan ..

.

csc ..

.

sec .

θ

θ

θ

=

=

=

≈

≈

996 6

1 2106

6 64 5

1 4667

6 8 51

4

2 2 2

2

..

cot ..

.

.

≈

≈θ =

+ =

+

2.

L

L 66 24 51

51 46 24

4 76

4 76

2 18

51 7 14

2

2

.

.

.

.

.

.

=

= −

=

=

L

L

L

L ≈

≈

ssin ..

.

cos ..

.

tan .

θ

θ

θ

=

=

=

2 187 14

3053

6 87 14

9524

2 18

≈

≈

66 83206

7 142 18

3 2752

7 146 8

1 05

..

csc ..

.

sec ..

.

≈

≈θ

θ

=

= = 000

6 82 18

3 1193

10 3

100 3

103

2 2 2

2

2

cot ..

.θ =

+ =

+ =

=

≈

3. H

H

H

1103

10 15

3 1 73

1 7310 15

1704

10=

=

=H

H ≈

≈

≈

.

.

sin ..

.

cos

θ

θ110 15

9852

1 7310

1730

10 151 73

5 8

..

tan . .

csc ..

.

≈

≈

θ

θ

= =

= 6671

10 1510

1 0150

101 73

5 7803

7 2

sec . .

cot.

.

θ

θ

= =

=

( )

≈

4.

22 22

2

7 3

7 7 2 2 7 7 3 3

49 2 49 3

+ ( ) =

( )( ) + ( )( ) =

( )( ) + ( )( ) =

H

H

HH

H

H

H

2

2

2

98 147

245

15 65

7 2 9 9

7 3 12 12

9

+ =

=

=

≈

≈

≈

.

.

.

sin .θ 9915 65

6326

12 1215 65

7744

9 912 12

..

cos ..

.

tan ..

≈

≈θ

θ

=

= ≈≈

≈

≈

.

csc ..

.

sec ..

.

8168

15 659 9

1 5808

15 6512 12

1 29

θ

θ

=

= 113

12 129 9

1 2242

2 3 11

2 2 3 3 1

22 2

cot ..

.θ =

( ) + =

( )( ) +

≈

5. H

221

4 3 121

133

11 53

2 3 3 46

3 462

2

2

=

( ) + =

=

=H

H

H

H ≈

≈

.

.

sin .θ111 53

3001

1111 53

9540

3 4611

3145

..

cos.

.

tan . .

≈

≈

≈

θ

θ

=

=

ccsc ..

.

sec . .

cot

θ

θ

θ

=

=

=

11 533 46

3 3324

11 5311

1 0482

1

≈

≈

113 46

3 1792

5 12

25 144

169

13

1

2 2 2

2

2

2

..≈

6. + =

+ =

=

=

=

H

H

H

H

H 33

513

3846

1213

9231

512

4167

sin .

cos .

tan .

c

θ

θ

θ

=

=

=

≈

≈

≈

ssc .

sec .

cot .

θ

θ

θ

= =

=

= =

135

2 6000

1312

1 0833

125

2 4000

≈

Lesson 2CLesson 2C1. 9 8

81 64

145

145

12 04

2 2 2

2

2

2

+ =

+ =

=

=

H

H

H

H

H ≈ .

ssin.

.

cos.

.

tan .

θ

θ

θ

=

=

=

812 04

6645

912 04

7475

89

888

≈

≈

≈ 99

12 048

1 5050

12 049

1 3378

98

1

csc . .

sec . .

cot

θ

θ

θ

=

=

= =

≈

≈

..

.

sin

1250

5 5

25 25

50

50

7 07

57

2 2 2

2

2

2. + =

+ =

=

=

=H

H

H

H

H ≈

θ..

.

cos.

.

tan .

csc .

077072

57 07

7072

55

1 0000

7

≈

≈θ

θ

θ

=

= =

= 0075

1 4140

7 075

1 4140

55

1 0000

32

=

= =

= =

+

.

sec . .

cot .

θ

θ

3. 33

9 9

18

18

4 24

34 24

70752 2

2

2

=

+ =

=

=

=H

H

H

H

H ≈

≈

.

sin.

.

cos

θ

θ ==

= =

=

34 24

7075

33

1 0000

4 243

1 4133

..

tan .

csc . .

sec

≈

≈

θ

θ

θθ

θ

=

= =

+ =

+

4 243

1 4133

33

1 0000

12 13

144

2 2 2

2

. .

cot .

≈

4. L

L ==

= −

=

=

=

=

=169

169 144

25

25

5

513

3846

12

2

L

L

L

L

sin .

cos

θ

θ

≈

2213

9231

512

4167

135

2 6000

131

≈

≈

.

tan .

csc .

sec

θ

θ

θ

=

= =

=22

1 0833

125

2 4000

4 6

16 36

52

2 2 2

2

2

≈ .

cot .θ = =

+ =

+ =

=

5. H

H

H

552

7 21

67 21

8322

47 21

5548

=

=

=

H

H ≈

≈

≈

.

sin.

.

cos.

.

tan

θ

θ

θθ

θ

θ

= =

=

= =

64

1 5000

7 216

1 2017

7 214

1 8025

.

csc . .

sec . .

c

≈

oot .

.

θ =

+ =

+ =

=

=

46

6667

7 7

49 49

98

98

9 90

2 2 2

2

2

≈

≈

6. H

H

H

H

H

ssin.

.

cos.

.

tan .

cs

θ

θ

θ

=

=

= =

79 9

7071

79 9

7071

77

1 0000

≈

≈

cc . .

sec . .

cot .

θ

θ

θ

=

=

= =

9 97

1 4143

9 97

1 4143

77

1 0000

≈

≈

precALcULUS

LeSSon 2c - LeSSon 3A

SoLUTionS 271

Lesson 2C1. 9 8

81 64

145

145

12 04

2 2 2

2

2

2

+ =

+ =

=

=

H

H

H

H

H ≈ .

ssin.

.

cos.

.

tan .

θ

θ

θ

=

=

=

812 04

6645

912 04

7475

89

888

≈

≈

≈ 99

12 048

1 5050

12 049

1 3378

98

1

csc . .

sec . .

cot

θ

θ

θ

=

=

= =

≈

≈

..

.

sin

1250

5 5

25 25

50

50

7 07

57

2 2 2

2

2

2. + =

+ =

=

=

=H

H

H

H

H ≈

θ..

.

cos.

.

tan .

csc .

077072

57 07

7072

55

1 0000

7

≈

≈θ

θ

θ

=

= =

= 0075

1 4140

7 075

1 4140

55

1 0000

32

=

= =

= =

+

.

sec . .

cot .

θ

θ

3. 33

9 9

18

18

4 24

34 24

70752 2

2

2

=

+ =

=

=

=H

H

H

H

H ≈

≈

.

sin.

.

cos

θ

θ ==

= =

=

34 24

7075

33

1 0000

4 243

1 4133

..

tan .

csc . .

sec

≈

≈

θ

θ

θθ

θ

=

= =

+ =

+

4 243

1 4133

33

1 0000

12 13

144

2 2 2

2

. .

cot .

≈

4. L

L ==

= −

=

=

=

=

=169

169 144

25

25

5

513

3846

12

2

L

L

L

L

sin .

cos

θ

θ

≈

2213

9231

512

4167

135

2 6000

131

≈

≈

.

tan .

csc .

sec

θ

θ

θ

=

= =

=22

1 0833

125

2 4000

4 6

16 36

52

2 2 2

2

2

≈ .

cot .θ = =

+ =

+ =

=

5. H

H

H

552

7 21

67 21

8322

47 21

5548

=

=

=

H

H ≈

≈

≈

.

sin.

.

cos.

.

tan

θ

θ

θθ

θ

θ

= =

=

= =

64

1 5000

7 216

1 2017

7 214

1 8025

.

csc . .

sec . .

c

≈

oot .

.

θ =

+ =

+ =

=

=

46

6667

7 7

49 49

98

98

9 90

2 2 2

2

2

≈

≈

6. H

H

H

H

H

ssin.

.

cos.

.

tan .

cs

θ

θ

θ

=

=

= =

79 9

7071

79 9

7071

77

1 0000

≈

≈

cc . .

sec . .

cot .

θ

θ

θ

=

=

= =

9 97

1 4143

9 97

1 4143

77

1 0000

≈

≈

Lesson 2C1. 9 8

81 64

145

145

12 04

2 2 2

2

2

2

+ =

+ =

=

=

H

H

H

H

H ≈ .

ssin.

.

cos.

.

tan .

θ

θ

θ

=

=

=

812 04

6645

912 04

7475

89

888

≈

≈

≈ 99

12 048

1 5050

12 049

1 3378

98

1

csc . .

sec . .

cot

θ

θ

θ

=

=

= =

≈

≈

..

.

sin

1250

5 5

25 25

50

50

7 07

57

2 2 2

2

2

2. + =

+ =

=

=

=H

H

H

H

H ≈

θ..

.

cos.

.

tan .

csc .

077072

57 07

7072

55

1 0000

7

≈

≈θ

θ

θ

=

= =

= 0075

1 4140

7 075

1 4140

55

1 0000

32

=

= =

= =

+

.

sec . .

cot .

θ

θ

3. 33

9 9

18

18

4 24

34 24

70752 2

2

2

=

+ =

=

=

=H

H

H

H

H ≈

≈

.

sin.

.

cos

θ

θ ==

= =

=

34 24

7075

33

1 0000

4 243

1 4133

..

tan .

csc . .

sec

≈

≈

θ

θ

θθ

θ

=

= =

+ =

+

4 243

1 4133

33

1 0000

12 13

144

2 2 2

2

. .

cot .

≈

4. L

L ==

= −

=

=

=

=

=169

169 144

25

25

5

513

3846

12

2

L

L

L

L

sin .

cos

θ

θ

≈

2213

9231

512

4167

135

2 6000

131

≈

≈

.

tan .

csc .

sec

θ

θ

θ

=

= =

=22

1 0833

125

2 4000

4 6

16 36

52

2 2 2

2

2

≈ .

cot .θ = =

+ =

+ =

=

5. H

H

H

552

7 21

67 21

8322

47 21

5548

=

=

=

H

H ≈

≈

≈

.

sin.

.

cos.

.

tan

θ

θ

θθ

θ

θ

= =

=

= =

64

1 5000

7 216

1 2017

7 214

1 8025

.

csc . .

sec . .

c

≈

oot .

.

θ =

+ =

+ =

=

=

46

6667

7 7

49 49

98

98

9 90

2 2 2

2

2

≈

≈

6. H

H

H

H

H

ssin.

.

cos.

.

tan .

cs

θ

θ

θ

=

=

= =

79 9

7071

79 9

7071

77

1 0000

≈

≈

cc . .

sec . .

cot .

θ

θ

θ

=

=

= =

9 97

1 4143

9 97

1 4143

77

1 0000

≈

≈

Lesson 3ALesson 3A1.

2.

3.

cos º .

tan º .

sin

37 7986

51 1 2349

2

=

=

00 3420

49 7547

65 4226

6249

º .

sin º .

cos º .

. t

=

=

=

=

4.

5.

6. aan º

. cos º

. tan º

. sin

32

4540 63

0875 5

9781

7.

8.

9.

=

=

= º

. tan º

sin.

.

cos

78

14 3007 86

2026 9

7435

10.

11.

=

=θ

θ

≈

==

=

=

=

1826 9

6691

2018

1 1111

48

1826 9

..

tan .

º

sin.

≈

≈

≈

θ

θ

α ..

cos.

.

tan .

º

s

6691

2026 9

7435

1820

9000

42

α

α

α

=

= =

=

≈

12. iin.

.

cos.

.

tan .

θ

θ

θ

=

=

= =

712 2

5738

1012 2

8197

710

7000

≈

≈

θθ

α

α

α

=

=

=

=

35

1012 2

8197

712 2

5738

107

º

sin.

.

cos.

.

tan

≈

≈

≈≈

≈

≈

1 4286

55

100119

8403

64 5119

.

º

sin .

cos . .

α

θ

θ

=

=

=

13.

55420

10064 5

1 5504

57

64 5119

5420

tan.

.

º

sin . .

θ

θ

α

=

=

=

≈

≈

ccos .

tan . .

º

α

α

α

=

= =

=

100119

8403

64 5100

6450

33

≈

precALcULUS

LeSSon 3A - LeSSon 3B

SoLUTionS272

Lesson 3A1.

2.

3.

cos º .

tan º .

sin

37 7986

51 1 2349

2

=

=

00 3420

49 7547

65 4226

6249

º .

sin º .

cos º .

. t

=

=

=

=

4.

5.

6. aan º

. cos º

. tan º

. sin

32

4540 63

0875 5

9781

7.

8.

9.

=

=

= º

. tan º

sin.

.

cos

78

14 3007 86

2026 9

7435

10.

11.

=

=θ

θ

≈

==

=

=

=

1826 9

6691

2018

1 1111

48

1826 9

..

tan .

º

sin.

≈

≈

≈

θ

θ

α ..

cos.

.

tan .

º

s

6691

2026 9

7435

1820

9000

42

α

α

α

=

= =

=

≈

12. iin.

.

cos.

.

tan .

θ

θ

θ

=

=

= =

712 2

5738

1012 2

8197

710

7000

≈

≈

θθ

α

α

α

=

=

=

=

35

1012 2

8197

712 2

5738

107

º

sin.

.

cos.

.

tan

≈

≈

≈≈

≈

≈

1 4286

55

100119

8403

64 5119

.

º

sin .

cos . .

α

θ

θ

=

=

=

13.

55420

10064 5

1 5504

57

64 5119

5420

tan.

.

º

sin . .

θ

θ

α

=

=

=

≈

≈

ccos .

tan . .

º

α

α

α

=

= =

=

100119

8403

64 5100

6450

33

≈

Lesson 3BLesson 3B1.

2.

3.

sin º .

cos º .

sin

12 2079

86 0698

40

=

=

ºº .

tan º .

cos º .

. s

=

=

=

=

6428

22 4040

18 9511

9336

4.

5.

6. iin º

. tan º

. sin º

.

69

4 0108 76

9986 87

1 4826

7.

8.

9.

=

=

= ttan º

. cos º

sin . .

cos

56

9925 7

4 855

9700

10.

11.

=

= =

=

θ

θ 11 25

2400

4 851 2

4 0417

76

1 25

2

. .

tan ..

.

º

sin . .

=

=

=

= =

θ

θ

α

≈

4400

4 855

9700

1 24 85

2474

14

cos . .

tan ..

.

º

s

α

α

α

= =

=

=

≈

12. iin.

.

cos.

.

tan .

θ

θ

θ

=

=

= =

1419 8

7071

1419 8

7071

1414

1 0

≈

≈

0000

45

1419 8

7071

1419 8

7071

θ

α

α

α

=

=

=

º

sin.

.

cos.

.

tan

≈

≈

== =

=

= =

=

1414

1 0000

45

34100

3400

94100

.

º

sin .

cos

α

θ

θ

13.

==

=

=

= =

.

tan .

º

sin .

cos

9400

3494

3617

20

94100

9400

θ

θ

α

α

≈

== =

=

=

34100

3400

9434

2 7647

70

.

tan .

º

α

α

≈

Lesson 3B1.

2.

3.

sin º .

cos º .

sin

12 2079

86 0698

40

=

=

ºº .

tan º .

cos º .

. s

=

=

=

=

6428

22 4040

18 9511

9336

4.

5.

6. iin º

. tan º

. sin º

.

69

4 0108 76

9986 87

1 4826

7.

8.

9.

=

=

= ttan º

. cos º

sin . .

cos

56

9925 7

4 855

9700

10.

11.

=

= =

=

θ

θ 11 25

2400

4 851 2

4 0417

76

1 25

2

. .

tan ..

.

º

sin . .

=

=

=

= =

θ

θ

α

≈

4400

4 855

9700

1 24 85

2474

14

cos . .

tan ..

.

º

s

α

α

α

= =

=

=

≈

12. iin.

.

cos.

.

tan .

θ

θ

θ

=

=

= =

1419 8

7071

1419 8

7071

1414

1 0

≈

≈

0000

45

1419 8

7071

1419 8

7071

θ

α

α

α

=

=

=

º

sin.

.

cos.

.

tan

≈

≈

== =

=

= =

=

1414

1 0000

45

34100

3400

94100

.

º

sin .

cos

α

θ

θ

13.

==

=

=

= =

.

tan .

º

sin .

cos

9400

3494

3617

20

94100

9400

θ

θ

α

α

≈

== =

=

=

34100

3400

9434

2 7647

70

.

tan .

º

α

α

≈

precALcULUS

LeSSon 3c - LeSSon 3c

SoLUTionS 273

Lesson 3CLesson 3C1. sin .

cos . .

tan

θ

θ

θ

= =

= =

1020

5000

17 320

8650

==

=

= =

= =

1017 3

5780

30

17 320

8650

1020

..

º

sin . .

cos .

≈

θ

α

α 55000

17 310

1 7300

60

78

8750

tan . .

º

sin .

cos

α

α

θ

θ

= =

=

= =2.

== =

=

=

= =

3 98

4875

73 9

1 7949

61

3 98

487

. .

tan.

.

º

sin . .

θ

θ

α

≈

55

78

8750

3 97

5571

29

17 172

cos .

tan . .

º

α

α

α

= =

=

=

( ) + ( )

≈

3.22

2

2

2

17 17

34

34

5 83

17 4 124 125

=

+ =

=

=

=

H

H

H

H

H ≈

≈

.

.

sin ..

θ883

7067

4 125 83

7067

4 124 12

1 000

≈

≈

.

cos ..

.

tan ..

.

θ

θ

=

= = 00

5 834 12

1 4150

5 834 12

1 4150

csc ..

.

sec ..

.

cot

θ

θ

θ

=

=

=

≈

≈

44 124 12

1 0000

6 8 4

36 70 56

106 56

2 2 2

2

.

..

.

.

.

=

+ =

+ =

=

4. H

H

H22

106 56

10 32

610 32

5814

8 410 3

.

.

sin.

.

cos ..

=

=

=

H

H ≈

≈θ

θ22

8140

68 4

7143

10 326

1 7200

1

≈

≈

.

tan.

.

csc . .

sec

θ

θ

θ

=

= =

= 00 328 4

1 2286

8 46

1 4000

9 18

81

2 2 2

2

.

..

cot . .

≈

θ = =

+ =

+

5. L

L ==

= −

=

=

= =

324

324 81

243

243

15 59

918

5000

2

2

L

L

L

L ≈ .

sin .θ

ccos . .

tan.

.

csc .

θ

θ

θ

=

=

= =

15 5918

8661

915 59

5773

189

2

≈

≈

00000

1815 59

1 1546

15 599

1 7322

52

sec.

.

cot . .

θ

θ

=

=

+

≈

≈

6. 66

25 36

61

61

7 81

57 81

64022 2

2

2

=

+ =

=

=

=H

H

H

H

H ≈

≈

.

sin.

.θ

ccos.

.

tan .

csc . .

θ

θ

θ

=

=

= =

67 81

7682

56

8333

7 815

1 5620

≈

≈

ssec . .

cot .

θ

θ

=

= =

7 816

1 3017

65

1 2000

≈

Lesson 3C1. sin .

cos . .

tan

θ

θ

θ

= =

= =

1020

5000

17 320

8650

==

=

= =

= =

1017 3

5780

30

17 320

8650

1020

..

º

sin . .

cos .

≈

θ

α

α 55000

17 310

1 7300

60

78

8750

tan . .

º

sin .

cos

α

α

θ

θ

= =

=

= =2.

== =

=

=

= =

3 98

4875

73 9

1 7949

61

3 98

487

. .

tan.

.

º

sin . .

θ

θ

α

≈

55

78

8750

3 97

5571

29

17 172

cos .

tan . .

º

α

α

α

= =

=

=

( ) + ( )

≈

3.22

2

2

2

17 17

34

34

5 83

17 4 124 125

=

+ =

=

=

=

H

H

H

H

H ≈

≈

.

.

sin ..

θ883

7067

4 125 83

7067

4 124 12

1 000

≈

≈

.

cos ..

.

tan ..

.

θ

θ

=

= = 00

5 834 12

1 4150

5 834 12

1 4150

csc ..

.

sec ..

.

cot

θ

θ

θ

=

=

=

≈

≈

44 124 12

1 0000

6 8 4

36 70 56

106 56

2 2 2

2

.

..

.

.

.

=

+ =

+ =

=

4. H

H

H22

106 56

10 32

610 32

5814

8 410 3

.

.

sin.

.

cos ..

=

=

=

H

H ≈

≈θ

θ22

8140

68 4

7143

10 326

1 7200

1

≈

≈

.

tan.

.

csc . .

sec

θ

θ

θ

=

= =

= 00 328 4

1 2286

8 46

1 4000

9 18

81

2 2 2

2

.

..

cot . .

≈

θ = =

+ =

+

5. L

L ==

= −

=

=

= =

324

324 81

243

243

15 59

918

5000

2

2

L

L

L

L ≈ .

sin .θ

ccos . .

tan.

.

csc .

θ

θ

θ

=

=

= =

15 5918

8661

915 59

5773

189

2

≈

≈

00000

1815 59

1 1546

15 599

1 7322

52

sec.

.

cot . .

θ

θ

=

=

+

≈

≈

6. 66

25 36

61

61

7 81

57 81

64022 2

2

2

=

+ =

=

=

=H

H

H

H

H ≈

≈

.

sin.

.θ

ccos.

.

tan .

csc . .

θ

θ

θ

=

=

= =

67 81

7682

56

8333

7 815

1 5620

≈

≈

ssec . .

cot .

θ

θ

=

= =

7 816

1 3017

65

1 2000

≈

precALcULUS

LeSSon 3D - LeSSon 4A

SoLUTionS274

Lesson 3DLesson 3D1. sin .

..

cos ..

.

θ

θ

=

=

12 312 5

9840

2 1812 5

17

≈

≈ 444

12 32 18

5 6422

802 1812 5

1744

tan ..

.

º

sin ..

.

θ

θ

α

=

=

=

≈

≈

ccos ..

.

tan ..

.

º

s

α

α

α

=

=

=

12 312 5

9840

2 1812 3

1772

10

≈

≈

2. iin .

cos . .

tan.

.

θ

θ

θ

θ

=

=

=

811

7273

7 511

6818

87 5

1 0667

≈

≈

≈

==

=

=

= =

477 511

6818

811

7273

7 58

9

º

sin . .

cos .

tan . .

α

α

α

≈

≈

3375

43

7 12

49 144

193

193

13 89

2 2 2

2

2

α =

+ =

+ =

=

=

º

.

3.

H

H

H

H

H ≈

sin.

.

cos.

.

tan

θ

θ

θ

=

=

=

1213 89

8639

713 89

5040

127

1

≈

≈

≈ ..

csc . .

sec . .

cot

7143

13 8912

1 1575

13 897

1 9843

θ

θ

θ

=

=

≈

≈

==

+ =

+ =

=

=

712

5833

12 13

144 169

313

313

1

2 2 2

2

2

≈

≈

.

4. H

H

H

H

H 77 69

1317 69

7349

1217 69

6783

.

sin.

.

cos.

.

tan

θ

θ

θ

=

=

=

≈

≈

11312

1 0833

17 6913

1 3608

17 6912

1 4

≈

≈

≈

.

csc . .

sec . .

θ

θ

=

= 7742

1213

9231

9 12

81 144

225

2 2 2

2

2

cot .θ =

+ =

+ =

==

≈

5. H

H

HH 115

915

6000

1215

8000

912

7500

sin .

cos .

tan .

θ

θ

θ

= =

= =

= =

ccsc .

sec .

cot .

θ

θ

θ

=

= =

=

159

1 667

1512

1 2500

129

1 3333

≈

≈

6..

5 2 14

5 2 196

25 2 196

50 196

22 2

22

2

2

( ) + =

( ) + =

( ) + =

+ =

H

H

H

H22

2246

246

15 68

7 0715 68

4509

1

=

=

=

=

H

H

H ≈

≈

.

sin ..

.

cos

θ

θ 4415 68

8929

7 0714

5050

15 687 07

2

..

tan . .

csc ..

.

≈

≈

θ

θ

= =

= 22178

15 6814

1 1200

147 07

1 9802

sec . .

cot.

.

θ

θ

= =

= ≈

Lesson 3D1. sin .

..

cos ..

.

θ

θ

=

=

12 312 5

9840

2 1812 5

17

≈

≈ 444

12 32 18

5 6422

802 1812 5

1744

tan ..

.

º

sin ..

.

θ

θ

α

=

=

=

≈

≈

ccos ..

.

tan ..

.

º

s

α

α

α

=

=

=

12 312 5

9840

2 1812 3

1772

10

≈

≈

2. iin .

cos . .

tan.

.

θ

θ

θ

θ

=

=

=

811

7273

7 511

6818

87 5

1 0667

≈

≈

≈

==

=

=

= =

477 511

6818

811

7273

7 58

9

º

sin . .

cos .

tan . .

α

α

α

≈

≈

3375

43

7 12

49 144

193

193

13 89

2 2 2

2

2

α =

+ =

+ =

=

=

º

.

3.

H

H

H

H

H ≈

sin.

.

cos.

.

tan

θ

θ

θ

=

=

=

1213 89

8639

713 89

5040

127

1

≈

≈

≈ ..

csc . .

sec . .

cot

7143

13 8912

1 1575

13 897

1 9843

θ

θ

θ

=

=

≈

≈

==

+ =

+ =

=

=

712

5833

12 13

144 169

313

313

1

2 2 2

2

2

≈

≈

.

4. H

H

H

H

H 77 69

1317 69

7349

1217 69

6783

.

sin.

.

cos.

.

tan

θ

θ

θ

=

=

=

≈

≈

11312

1 0833

17 6913

1 3608

17 6912

1 4

≈

≈

≈

.

csc . .

sec . .

θ

θ

=

= 7742

1213

9231

9 12

81 144

225

2 2 2

2

2

cot .θ =

+ =

+ =

==

≈

5. H

H

HH 115

915

6000

1215

8000

912

7500

sin .

cos .

tan .

θ

θ

θ

= =

= =

= =

ccsc .

sec .

cot .

θ

θ

θ

=

= =

=

159

1 667

1512

1 2500

129

1 3333

≈

≈

6..

5 2 14

5 2 196

25 2 196

50 196

22 2

22

2

2

( ) + =

( ) + =

( ) + =

+ =

H

H

H

H22

2246

246

15 68

7 0715 68

4509

1

=

=

=

=

H

H

H ≈

≈

.

sin ..

.

cos

θ

θ 4415 68

8929

7 0714

5050

15 687 07

2

..

tan . .

csc ..

.

≈

≈

θ

θ

= =

= 22178

15 6814

1 1200

147 07

1 9802

sec . .

cot.

.

θ

θ

= =

= ≈

Lesson 3D1. sin .

..

cos ..

.

θ

θ

=

=

12 312 5

9840

2 1812 5

17

≈

≈ 444

12 32 18

5 6422

802 1812 5

1744

tan ..

.

º

sin ..

.

θ

θ

α

=

=

=

≈

≈

ccos ..

.

tan ..

.

º

s

α

α

α

=

=

=

12 312 5

9840

2 1812 3

1772

10

≈

≈

2. iin .

cos . .

tan.

.

θ

θ

θ

θ

=

=

=

811

7273

7 511

6818

87 5

1 0667

≈

≈

≈

==

=

=

= =

477 511

6818

811

7273

7 58

9

º

sin . .

cos .

tan . .

α

α

α

≈

≈

3375

43

7 12

49 144

193

193

13 89

2 2 2

2

2

α =

+ =

+ =

=

=

º

.

3.

H

H

H

H

H ≈

sin.

.

cos.

.

tan

θ

θ

θ

=

=

=

1213 89

8639

713 89

5040

127

1

≈

≈

≈ ..

csc . .

sec . .

cot

7143

13 8912

1 1575

13 897

1 9843

θ

θ

θ

=

=

≈

≈

==

+ =

+ =

=

=

712

5833

12 13

144 169

313

313

1

2 2 2

2

2

≈

≈

.

4. H

H

H

H

H 77 69

1317 69

7349

1217 69

6783

.

sin.

.

cos.

.

tan

θ

θ

θ

=

=

=

≈

≈

11312

1 0833

17 6913

1 3608

17 6912

1 4

≈

≈

≈

.

csc . .

sec . .

θ

θ

=

= 7742

1213

9231

9 12

81 144

225

2 2 2

2

2

cot .θ =

+ =

+ =

==

≈

5. H

H

HH 115

915

6000

1215

8000

912

7500

sin .

cos .

tan .

θ

θ

θ

= =

= =

= =

ccsc .

sec .

cot .

θ

θ

θ

=

= =

=

159

1 667

1512

1 2500

129

1 3333

≈

≈

6..

5 2 14

5 2 196

25 2 196

50 196

22 2

22

2

2

( ) + =

( ) + =

( ) + =

+ =

H

H

H

H22

2246

246

15 68

7 0715 68

4509

1

=

=

=

=

H

H

H ≈

≈

.

sin ..

.

cos

θ

θ 4415 68

8929

7 0714

5050

15 687 07

2

..

tan . .

csc ..

.

≈

≈

θ

θ

= =

= 22178

15 6814

1 1200

147 07

1 9802

sec . .

cot.

.

θ

θ

= =

= ≈

Lesson 4ALesson 4A1. sin º .

sin º ..

sin º.

15 7 5

15 7 57 515

7

=

=

=

AA

A ≈ 552588

28 98

37 20

37 202037

2

..

tan º

tan º

tan º

≈

≈

2. =

=

=

BB

B 007536

26 54

69 12

69 121269

1

..

sin º

sin º

sin º

≈

≈

3. =

=

=

cc

c 229336

12 85

571414 57

14 1 5399

..

tan º

tan º

.

≈

≈

4. =

=

( )

D

D

D ≈≈

≈

21 56

19 10

19 101019

103443

.

tan º

tan º

tan º .

5. =

=

=

ee

e ≈≈

≈ ≈

29 04

8011 3

11 3 80

11 3 9848

.

sin º.

. sin º

. .

6. =

=

( )

F

F

F 111 13.

precALcULUS

LeSSon 4A - LeSSon 4c

SoLUTionS 275

Lesson 4A1. sin º .

sin º ..

sin º.

15 7 5

15 7 57 515

7

=

=

=

AA

A ≈ 552588

28 98

37 20

37 202037

2

..

tan º

tan º

tan º

≈

≈

2. =

=

=

BB

B 007536

26 54

69 12

69 121269

1

..

sin º

sin º

sin º

≈

≈

3. =

=

=

cc

c 229336

12 85

571414 57

14 1 5399

..

tan º

tan º

.

≈

≈

4. =

=

( )

D

D

D ≈≈

≈

21 56

19 10

19 101019

103443

.

tan º

tan º

tan º .

5. =

=

=

ee

e ≈≈

≈ ≈

29 04

8011 3

11 3 80

11 3 9848

.

sin º.

. sin º

. .

6. =

=

( )

F

F

F 111 13.

Lesson 4A1. sin º .

sin º ..

sin º.

15 7 5

15 7 57 515

7

=

=

=

AA

A ≈ 552588

28 98

37 20

37 202037

2

..

tan º

tan º

tan º

≈

≈

2. =

=

=

BB

B 007536

26 54

69 12

69 121269

1

..

sin º

sin º

sin º

≈

≈

3. =

=

=

cc

c 229336

12 85

571414 57

14 1 5399

..

tan º

tan º

.

≈

≈

4. =

=

( )

D

D

D ≈≈

≈

21 56

19 10

19 101019

103443

.

tan º

tan º

tan º .

5. =

=

=

ee

e ≈≈

≈ ≈

29 04

8011 3

11 3 80

11 3 9848

.

sin º.

. sin º

. .

6. =

=

( )

F

F

F 111 13.

Lesson 4A1. sin º .

sin º ..

sin º.

15 7 5

15 7 57 515

7

=

=

=

AA

A ≈ 552588

28 98

37 20

37 202037

2

..

tan º

tan º

tan º

≈

≈

2. =

=

=

BB

B 007536

26 54

69 12

69 121269

1

..

sin º

sin º

sin º

≈

≈

3. =

=

=

cc

c 229336

12 85

571414 57

14 1 5399

..

tan º

tan º

.

≈

≈

4. =

=

( )

D

D

D ≈≈

≈

21 56

19 10

19 101019

103443

.

tan º

tan º

tan º .

5. =

=

=

ee

e ≈≈

≈ ≈

29 04

8011 3

11 3 80

11 3 9848

.

sin º.

. sin º

. .

6. =

=

( )

F

F

F 111 13.

Lesson 4A1. sin º .

sin º ..

sin º.

15 7 5

15 7 57 515

7

=

=

=

AA

A ≈ 552588

28 98

37 20

37 202037

2

..

tan º

tan º

tan º

≈

≈

2. =

=

=

BB

B 007536

26 54

69 12

69 121269

1

..

sin º

sin º

sin º

≈

≈

3. =

=

=

cc

c 229336

12 85

571414 57

14 1 5399

..

tan º

tan º

.

≈

≈

4. =

=

( )

D

D

D ≈≈

≈

21 56

19 10

19 101019

103443

.

tan º

tan º

tan º .

5. =

=

=

ee

e ≈≈

≈ ≈

29 04

8011 3

11 3 80

11 3 9848

.

sin º.

. sin º

. .

6. =

=

( )

F

F

F 111 13.

Lesson 4A1. sin º .

sin º ..

sin º.

15 7 5

15 7 57 515

7

=

=

=

AA

A ≈ 552588

28 98

37 20

37 202037

2

..

tan º

tan º

tan º

≈

≈

2. =

=

=

BB

B 007536

26 54

69 12

69 121269

1

..

sin º

sin º

sin º

≈

≈

3. =

=

=

cc

c 229336

12 85

571414 57

14 1 5399

..

tan º

tan º

.

≈

≈

4. =

=

( )

D

D

D ≈≈

≈

21 56

19 10

19 101019

103443

.

tan º

tan º

tan º .

5. =

=

=

ee

e ≈≈

≈ ≈

29 04

8011 3

11 3 80

11 3 9848

.

sin º.

. sin º

. .

6. =

=

( )

F

F

F 111 13.

Lesson 4BLesson 4B1. tan º

tan º

. .

2515

15 25

15 4663 6 99

=

=

( )

G

G

G ≈ ≈

22. cos º

cos º

cos º ..

32 22

32 222232

228480

25 94

=

=

=

HH

H ≈ ≈

33. cos º

cos º

.

. .

4117

17 41

17 7547

4 12 7547

=

=

( )( )

J

J

J

J

≈

≈ (( )

=

=

( )( )

≈

≈ ≈

3 11

6213

13 62

13 1 8807 24

.

tan º

tan º

.

4. K

K

K ..

cos º.

. cos º

. . .

45

739 8

9 8 73

9 8 2924 2 87

5. =

=

( )( )

L

L

L ≈ ≈

66. tan º

tan º

4516

16 45

16 1 16

=

=

= ( )( ) =

M

M

M

Lesson 4B1. tan º

tan º

. .

2515

15 25

15 4663 6 99

=

=

( )

G

G

G ≈ ≈

22. cos º

cos º

cos º ..

32 22

32 222232

228480

25 94

=

=

=

HH

H ≈ ≈

33. cos º

cos º

.

. .

4117

17 41

17 7547

4 12 7547

=

=

( )( )

J

J

J

J

≈

≈ (( )

=

=

( )( )

≈

≈ ≈

3 11

6213

13 62

13 1 8807 24

.

tan º

tan º

.

4. K

K

K ..

cos º.

. cos º

. . .

45

739 8

9 8 73

9 8 2924 2 87

5. =

=

( )( )

L

L

L ≈ ≈

66. tan º

tan º

4516

16 45

16 1 16

=

=

= ( )( ) =

M

M

M

Lesson 4B1. tan º

tan º

. .

2515

15 25

15 4663 6 99

=

=

( )

G

G

G ≈ ≈

22. cos º

cos º

cos º ..

32 22

32 222232

228480

25 94

=

=

=

HH

H ≈ ≈

33. cos º

cos º

.

. .

4117

17 41

17 7547

4 12 7547

=

=

( )( )

J

J

J

J

≈

≈ (( )

=

=

( )( )

≈

≈ ≈

3 11

6213

13 62

13 1 8807 24

.

tan º

tan º

.

4. K

K

K ..

cos º.

. cos º

. . .

45

739 8

9 8 73

9 8 2924 2 87

5. =

=

( )( )

L

L

L ≈ ≈

66. tan º

tan º

4516

16 45

16 1 16

=

=

= ( )( ) =

M

M

M

Lesson 4B1. tan º

tan º

. .

2515

15 25

15 4663 6 99

=

=

( )

G

G

G ≈ ≈

22. cos º

cos º

cos º ..

32 22

32 222232

228480

25 94

=

=

=

HH

H ≈ ≈

33. cos º

cos º

.

. .

4117

17 41

17 7547

4 12 7547

=

=

( )( )

J

J

J

J

≈

≈ (( )

=

=

( )( )

≈

≈ ≈

3 11

6213

13 62

13 1 8807 24

.

tan º

tan º

.

4. K

K

K ..

cos º.

. cos º

. . .

45

739 8

9 8 73

9 8 2924 2 87

5. =

=

( )( )

L

L

L ≈ ≈

66. tan º

tan º

4516

16 45

16 1 16

=

=

= ( )( ) =

M

M

M

Lesson 4B1. tan º

tan º

. .

2515

15 25

15 4663 6 99

=

=

( )

G

G

G ≈ ≈

22. cos º

cos º

cos º ..

32 22

32 222232

228480

25 94

=

=

=

HH

H ≈ ≈

33. cos º

cos º

.

. .

4117

17 41

17 7547

4 12 7547

=

=

( )( )

J

J

J

J

≈

≈ (( )

=

=

( )( )

≈

≈ ≈

3 11

6213

13 62

13 1 8807 24

.

tan º

tan º

.

4. K

K

K ..

cos º.

. cos º

. . .

45

739 8

9 8 73

9 8 2924 2 87

5. =

=

( )( )

L

L

L ≈ ≈

66. tan º

tan º

4516

16 45

16 1 16

=

=

= ( )( ) =

M

M

M

Lesson 4B1. tan º

tan º

. .

2515

15 25

15 4663 6 99

=

=

( )

G

G

G ≈ ≈

22. cos º

cos º

cos º ..

32 22

32 222232

228480

25 94

=

=

=

HH

H ≈ ≈

33. cos º

cos º

.

. .

4117

17 41

17 7547

4 12 7547

=

=

( )( )

J

J

J

J

≈

≈ (( )

=

=

( )( )

≈

≈ ≈

3 11

6213

13 62

13 1 8807 24

.

tan º

tan º

.

4. K

K

K ..

cos º.

. cos º

. . .

45

739 8

9 8 73

9 8 2924 2 87

5. =

=

( )( )

L

L

L ≈ ≈

66. tan º

tan º

4516

16 45

16 1 16

=

=

= ( )( ) =

M

M

M

Lesson 4CLesson 4C1. sin º

sin º

. .

2125

25 21

25 3584 8 96

=

=

( )

n

n

N ≈ ≈

22. tan º .

tan º ..

tan º.

.

36 1 8

36 1 81 8

361 87265

=

=

=

pp

p

P ≈ ≈ 22 48

6819

19 68

19 2 4751 47 03

.

tan º

tan º

. .

3.

4.

=

=

( )

q

q

Q ≈ ≈

ssin º

sin º

.

. .

4232

32 42

32 6691

5 66 6691

=

=

( )( )( )

r

r

r

r

≈

≈ (( )

=

=

( )

≈

≈ ≈

3 79

10100

100 10

100 1736 17

.

sin º

sin º

.

5. S

S

S ..

tan º.

. tan º

. .

36

5416 1

16 1 54

16 11 3764 22

6. =

=

( )

T

T

T ≈ ≈ ..16

Lesson 4C1. sin º

sin º

. .

2125

25 21

25 3584 8 96

=

=

( )

n

n

N ≈ ≈

22. tan º .

tan º ..

tan º.

.

36 1 8

36 1 81 8

361 87265

=

=

=

pp

p

P ≈ ≈ 22 48

6819

19 68

19 2 4751 47 03

.

tan º

tan º

. .

3.

4.

=

=

( )

q

q

Q ≈ ≈

ssin º

sin º

.

. .

4232

32 42

32 6691

5 66 6691

=

=

( )( )( )

r

r

r

r

≈

≈ (( )

=

=

( )

≈

≈ ≈

3 79

10100

100 10

100 1736 17

.

sin º

sin º

.

5. S

S

S ..

tan º.

. tan º

. .

36

5416 1

16 1 54

16 11 3764 22

6. =

=

( )

T

T

T ≈ ≈ ..16

Lesson 4C1. sin º

sin º

. .

2125

25 21

25 3584 8 96

=

=

( )

n

n

N ≈ ≈

22. tan º .

tan º ..

tan º.

.

36 1 8

36 1 81 8

361 87265

=

=

=

pp

p

P ≈ ≈ 22 48

6819

19 68

19 2 4751 47 03

.

tan º

tan º

. .

3.

4.

=

=

( )

q

q

Q ≈ ≈

ssin º

sin º

.

. .

4232

32 42

32 6691

5 66 6691

=

=

( )( )( )

r

r

r

r

≈

≈ (( )

=

=

( )

≈

≈ ≈

3 79

10100

100 10

100 1736 17

.

sin º

sin º

.

5. S

S

S ..

tan º.

. tan º

. .

36

5416 1

16 1 54

16 11 3764 22

6. =

=

( )

T

T

T ≈ ≈ ..16

Lesson 4C1. sin º

sin º

. .

2125

25 21

25 3584 8 96

=

=

( )

n

n

N ≈ ≈

22. tan º .

tan º ..

tan º.

.

36 1 8

36 1 81 8

361 87265

=

=

=

pp

p

P ≈ ≈ 22 48

6819

19 68

19 2 4751 47 03

.

tan º

tan º

. .

3.

4.

=

=

( )

q

q

Q ≈ ≈

ssin º

sin º

.

. .

4232

32 42

32 6691

5 66 6691

=

=

( )( )( )

r

r

r

r

≈

≈ (( )

=

=

( )

≈

≈ ≈

3 79

10100

100 10

100 1736 17

.

sin º

sin º

.

5. S

S

S ..

tan º.

. tan º

. .

36

5416 1

16 1 54

16 11 3764 22

6. =

=

( )

T

T

T ≈ ≈ ..16

Lesson 4C1. sin º

sin º

. .

2125

25 21

25 3584 8 96

=

=

( )

n

n

N ≈ ≈

22. tan º .

tan º ..

tan º.

.

36 1 8

36 1 81 8

361 87265

=

=

=

pp

p

P ≈ ≈ 22 48

6819

19 68

19 2 4751 47 03

.

tan º

tan º

. .

3.

4.

=

=

( )

q

q

Q ≈ ≈

ssin º

sin º

.

. .

4232

32 42

32 6691

5 66 6691

=

=

( )( )( )

r

r

r

r

≈

≈ (( )

=

=

( )

≈

≈ ≈

3 79

10100

100 10

100 1736 17

.

sin º

sin º

.

5. S

S

S ..

tan º.

. tan º

. .

36

5416 1

16 1 54

16 11 3764 22

6. =

=

( )

T

T

T ≈ ≈ ..16

Lesson 4C1. sin º

sin º

. .

2125

25 21

25 3584 8 96

=

=

( )

n

n

N ≈ ≈

22. tan º .

tan º ..

tan º.

.

36 1 8

36 1 81 8

361 87265

=

=

=

pp

p

P ≈ ≈ 22 48

6819

19 68

19 2 4751 47 03

.

tan º

tan º

. .

3.

4.

=

=

( )

q

q

Q ≈ ≈

ssin º

sin º

.

. .

4232

32 42

32 6691

5 66 6691

=

=

( )( )( )

r

r

r

r

≈

≈ (( )

=

=

( )

≈

≈ ≈

3 79

10100

100 10

100 1736 17

.

sin º

sin º

.

5. S

S

S ..

tan º.

. tan º

. .

36

5416 1

16 1 54

16 11 3764 22

6. =

=

( )

T

T

T ≈ ≈ ..16

precALcULUS

LeSSon 4D - LeSSon 5A

SoLUTionS276

Lesson 4DLesson 4D1. cos º

cos º

cos º

.

75 30

75 303075

302

=

=

=

UU

U

U ≈5588

115 92

28 9

28 9

28 3328

≈ .

sin º

sin º

sin

sin

2. =

=

=

=

V

V

V

Vºº

..

tan º

tan º

tan º

V

WW

W

W

≈ ≈

≈

34695

6 39

18 7

18 7718

7

3. =

=

=

...

cos º .

cos º ..

cos

324921 55

64 4 75

64 4 754 75

≈

4. =

=

=

xx

x664

4 754384

10 83

4222

22 42

22

º.

..

sin º

sin º

x

Y

Y

Y

≈ ≈

≈

5. =

=

.. .

tan º

tan º

.

6691 14 72

5118

18 51

18 1 2349

( )

=

=

(

≈

≈

6. Z

Z

Z )) ≈ 22 23.

Lesson 4D1. cos º

cos º

cos º

.

75 30

75 303075

302

=

=

=

UU

U

U ≈5588

115 92

28 9

28 9

28 3328

≈ .

sin º

sin º

sin

sin

2. =

=

=

=

V

V

V

Vºº

..

tan º

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V

WW

W

W

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34695

6 39

18 7

18 7718

7

3. =

=

=

...

cos º .

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cos

324921 55

64 4 75

64 4 754 75

≈

4. =

=

=

xx

x664

4 754384

10 83

4222

22 42

22

º.

..

sin º

sin º

x

Y

Y

Y

≈ ≈

≈

5. =

=

.. .

tan º

tan º

.

6691 14 72

5118

18 51

18 1 2349

( )

=

=

(

≈

≈

6. Z

Z

Z )) ≈ 22 23.

Lesson 4D1. cos º

cos º

cos º

.

75 30

75 303075

302

=

=

=

UU

U

U ≈5588

115 92

28 9

28 9

28 3328

≈ .

sin º

sin º

sin

sin

2. =

=

=

=

V

V

V

Vºº

..

tan º

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V

WW

W

W

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34695

6 39

18 7

18 7718

7

3. =

=

=

...

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cos º ..

cos

324921 55

64 4 75

64 4 754 75

≈

4. =

=

=

xx

x664

4 754384

10 83

4222

22 42

22

º.

..

sin º

sin º

x

Y

Y

Y

≈ ≈

≈

5. =

=

.. .

tan º

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.

6691 14 72

5118

18 51

18 1 2349

( )

=

=

(

≈

≈

6. Z

Z

Z )) ≈ 22 23.

Lesson 4D1. cos º

cos º

cos º

.

75 30

75 303075

302

=

=

=

UU

U

U ≈5588

115 92

28 9

28 9

28 3328

≈ .

sin º

sin º

sin

sin

2. =

=

=

=

V

V

V

Vºº

..

tan º

tan º

tan º

V

WW

W

W

≈ ≈

≈

34695

6 39

18 7

18 7718

7

3. =

=

=

...

cos º .

cos º ..

cos

324921 55

64 4 75

64 4 754 75

≈

4. =

=

=

xx

x664

4 754384

10 83

4222

22 42

22

º.

..

sin º

sin º

x

Y

Y

Y

≈ ≈

≈

5. =

=

.. .

tan º

tan º

.

6691 14 72

5118

18 51

18 1 2349

( )

=

=

(

≈

≈

6. Z

Z

Z )) ≈ 22 23.

Lesson 4D1. cos º

cos º

cos º

.

75 30

75 303075

302

=

=

=

UU

U

U ≈5588

115 92

28 9

28 9

28 3328

≈ .

sin º

sin º

sin

sin

2. =

=

=

=

V

V

V

Vºº

..

tan º

tan º

tan º

V

WW

W

W

≈ ≈

≈

34695

6 39

18 7

18 7718

7

3. =

=

=

...

cos º .

cos º ..

cos

324921 55

64 4 75

64 4 754 75

≈

4. =

=

=

xx

x664

4 754384

10 83

4222

22 42

22

º.

..

sin º

sin º

x

Y

Y

Y

≈ ≈

≈

5. =

=

.. .

tan º

tan º

.

6691 14 72

5118

18 51

18 1 2349

( )

=

=

(

≈

≈

6. Z

Z

Z )) ≈ 22 23.

Lesson 4D1. cos º

cos º

cos º

.

75 30

75 303075

302

=

=

=

UU

U

U ≈5588

115 92

28 9

28 9

28 3328

≈ .

sin º

sin º

sin

sin

2. =

=

=

=

V

V

V

Vºº

..

tan º

tan º

tan º

V

WW

W

W

≈ ≈

≈

34695

6 39

18 7

18 7718

7

3. =

=

=

...

cos º .

cos º ..

cos

324921 55

64 4 75

64 4 754 75

≈

4. =

=

=

xx

x664

4 754384

10 83

4222

22 42

22

º.

..

sin º

sin º

x

Y

Y

Y

≈ ≈

≈

5. =

=

.. .

tan º

tan º

.

6691 14 72

5118

18 51

18 1 2349

( )

=

=

(

≈

≈

6. Z

Z

Z )) ≈ 22 23.

Lesson 5ALesson 5A1.

2.

3.

tan º .

sin º .

cos º .

28 5317

33 5446

40

≈

≈

≈ 77660

5 0875

72 3090

69 9336

4.

5.

6.

7

tan º .

cos º .

sin º .

≈

≈

≈

..

8.

9.

arccos . º

arctan . º

arcsin .

9848 10

5317 28

9

≈

≈

5511 72

4 7046 78

4067 24

≈

≈

≈

º

arctan . º

arcsin .

10.

11. ºº

arccos . º

º ' "

" '"

(

12.

13.

4067 66

25 15 10

101

160

1

≈

× = 0060

17

15 171

160

15 1760

× =

) ' . '

. ' º'

( . )º

≈

≈≈

≈

. º

º ' " . º

º ' "

" '"

253

25 15 10 25 253

74 22 30

301

160

14.

× = (( ) ' . '

. ' º'

( . )º

3060

5

22 51

160

22 560

=

× = = .. º

º ' " . º

º ' "

" '"

(

375

74 22 30 74 375

32 48 12

121

160

=

× =

15.

11260

20

48 21

160

48 260

=

× =

) ' . '

. ' º'

( . )º≈ ..

º ' " . º

º '

' º'

(

803

32 48 12 32 803

81 50

501

160

5060

≈

16.

× = )º .

º ' .

º ' " . º

≈

≈

≈

833

81 50 81 833

25 15 10 25 25317.

118.

19.

20.

74 22 30 74 375

32 48 12 32 803

81

º ' " . º

º ' " . º

º

=

≈

550 1 833

5825 54 37

2 4

' . º

arccos . . º

arctan .

≈ 8

≈21.

22. ≈≈

≈

≈

67 38

9918 82 66

9299 2

. º

arcsin . . º

arccos .

23.

24. 11 58

5825 54 22 24

2 4 67

. º

arccos . º ' "

arctan .

25.

26.

≈

≈ ºº ' "

arcsin . º ' "

arccos .

22 48

9918 82 39 27

929

27.

28.

≈

99 21 34 51≈ º ' "

Lesson 5A1.

2.

3.

tan º .

sin º .

cos º .

28 5317

33 5446

40

≈

≈

≈ 77660

5 0875

72 3090

69 9336

4.

5.

6.

7

tan º .

cos º .

sin º .

≈

≈

≈

..

8.

9.

arccos . º

arctan . º

arcsin .

9848 10

5317 28

9

≈

≈

5511 72

4 7046 78

4067 24

≈

≈

≈

º

arctan . º

arcsin .

10.

11. ºº

arccos . º

º ' "

" '"

(

12.

13.

4067 66

25 15 10

101

160

1

≈

× = 0060

17

15 171

160

15 1760

× =

) ' . '

. ' º'

( . )º

≈

≈≈

≈

. º

º ' " . º

º ' "

" '"

253

25 15 10 25 253

74 22 30

301

160

14.

× = (( ) ' . '

. ' º'

( . )º

3060

5

22 51

160

22 560

=

× = = .. º

º ' " . º

º ' "

" '"

(

375

74 22 30 74 375

32 48 12

121

160

=

× =

15.

11260

20

48 21

160

48 260

=

× =

) ' . '

. ' º'

( . )º≈ ..

º ' " . º

º '

' º'

(

803

32 48 12 32 803

81 50

501

160

5060

≈

16.

× = )º .

º ' .

º ' " . º

≈

≈

≈

833

81 50 81 833

25 15 10 25 25317.

118.

19.

20.

74 22 30 74 375

32 48 12 32 803

81

º ' " . º

º ' " . º

º

=

≈

550 1 833

5825 54 37

2 4

' . º

arccos . . º

arctan .

≈ 8

≈21.

22. ≈≈

≈

≈

67 38

9918 82 66

9299 2

. º

arcsin . . º

arccos .

23.

24. 11 58

5825 54 22 24

2 4 67

. º

arccos . º ' "

arctan .

25.

26.

≈

≈ ºº ' "

arcsin . º ' "

arccos .

22 48

9918 82 39 27

929

27.

28.

≈

99 21 34 51≈ º ' "

Lesson 5A1.

2.

3.

tan º .

sin º .

cos º .

28 5317

33 5446

40

≈

≈

≈ 77660

5 0875

72 3090

69 9336

4.

5.

6.

7

tan º .

cos º .

sin º .

≈

≈

≈

..

8.

9.

arccos . º

arctan . º

arcsin .

9848 10

5317 28

9

≈

≈

5511 72

4 7046 78

4067 24

≈

≈

≈

º

arctan . º

arcsin .

10.

11. ºº

arccos . º

º ' "

" '"

(

12.

13.

4067 66

25 15 10

101

160

1

≈

× = 0060

17

15 171

160

15 1760

× =

) ' . '

. ' º'

( . )º

≈

≈≈

≈

. º

º ' " . º

º ' "

" '"

253

25 15 10 25 253

74 22 30

301

160

14.

× = (( ) ' . '

. ' º'

( . )º

3060

5

22 51

160

22 560

=

× = = .. º

º ' " . º

º ' "

" '"

(

375

74 22 30 74 375

32 48 12

121

160

=

× =

15.

11260

20

48 21

160

48 260

=

× =

) ' . '

. ' º'

( . )º≈ ..

º ' " . º

º '

' º'

(

803

32 48 12 32 803

81 50

501

160

5060

≈

16.

× = )º .

º ' .

º ' " . º

≈

≈

≈

833

81 50 81 833

25 15 10 25 25317.

118.

19.

20.

74 22 30 74 375

32 48 12 32 803

81

º ' " . º

º ' " . º

º

=

≈

550 1 833

5825 54 37

2 4

' . º

arccos . . º

arctan .

≈ 8

≈21.

22. ≈≈

≈

≈

67 38

9918 82 66

9299 2

. º

arcsin . . º

arccos .

23.

24. 11 58

5825 54 22 24

2 4 67

. º

arccos . º ' "

arctan .

25.

26.

≈

≈ ºº ' "

arcsin . º ' "

arccos .

22 48

9918 82 39 27

929

27.

28.

≈

99 21 34 51≈ º ' "

Lesson 5A1.

2.

3.

tan º .

sin º .

cos º .

28 5317

33 5446

40

≈

≈

≈ 77660

5 0875

72 3090

69 9336

4.

5.

6.

7

tan º .

cos º .

sin º .

≈

≈

≈

..

8.

9.

arccos . º

arctan . º

arcsin .

9848 10

5317 28

9

≈

≈

5511 72

4 7046 78

4067 24

≈

≈

≈

º

arctan . º

arcsin .

10.

11. ºº

arccos . º

º ' "

" '"

(

12.

13.

4067 66

25 15 10

101

160

1

≈

× = 0060

17

15 171

160

15 1760

× =

) ' . '

. ' º'

( . )º

≈

≈≈

≈

. º

º ' " . º

º ' "

" '"

253

25 15 10 25 253

74 22 30

301

160

14.

× = (( ) ' . '

. ' º'

( . )º

3060

5

22 51

160

22 560

=

× = = .. º

º ' " . º

º ' "

" '"

(

375

74 22 30 74 375

32 48 12

121

160

=

× =

15.

11260

20

48 21

160

48 260

=

× =

) ' . '

. ' º'

( . )º≈ ..

º ' " . º

º '

' º'

(

803

32 48 12 32 803

81 50

501

160

5060

≈

16.

× = )º .

º ' .

º ' " . º

≈

≈

≈

833

81 50 81 833

25 15 10 25 25317.

118.

19.

20.

74 22 30 74 375

32 48 12 32 803

81

º ' " . º

º ' " . º

º

=

≈

550 1 833

5825 54 37

2 4

' . º

arccos . . º

arctan .

≈ 8

≈21.

22. ≈≈

≈

≈

67 38

9918 82 66

9299 2

. º

arcsin . . º

arccos .

23.

24. 11 58

5825 54 22 24

2 4 67

. º

arccos . º ' "

arctan .

25.

26.

≈

≈ ºº ' "

arcsin . º ' "

arccos .

22 48

9918 82 39 27

929

27.

28.

≈

99 21 34 51≈ º ' "

precALcULUS

LeSSon 5B - LeSSon 5c

SoLUTionS 277

Lesson 5BLesson 5B1.

2.

3.

sin º .

tan º .

sin º .

82 9903

14 2493

38

≈

≈

≈ 66157

49 6561

58 5299

77 4 331

4.

5.

6.

cos º .

cos º .

tan º .

≈

≈

≈ 55

0175 1

4848 61

7.

8.

9.

arcsin . º

arccos . º

arcsin .

≈

≈

88746 61

4245 23

1 1918 50

≈

≈

≈

º

arctan . º

arctan .

10.

11. ºº

arccos . º

º ' "

" '"

(

12.

13.

9945 6

47 36 25

251

160

256

≈

× =00

42

36 421

160

36 4260

× =

) " . '

. ' º'

( . )º .

≈

≈ 6607

47 36 25 47 607

53 28 10

101

160

1

º

º ' " . º

º ' "

" '"

(

≈

14.

× = 0060

17

28 171

160

28 1760

× =

) ' . '

. ' º'

( . )º

≈

≈≈

≈

. º

º ' " . º

º '

' º'

(

470

53 28 10 53 470

8 55

551

160

556

15.

× =00

917

8 55 8 917

88 0 40

401

16

×

)º . º

º ' . º

º ' "

" '

≈

≈

16.

004060

67

671

160

6760

º( )º . '

. ºº

( . )º

=

× =

≈

≈≈

≈

≈

. º

º ' " . º

º ' " . º

011

88 0 40 88 011

47 36 25 47 607

5

17.

18. 33 28 10 53 469

8 55 8 917

88 0 40 88

º ' " . º

º ' . º

º ' " .

≈

≈

≈

19.

20. 0011

4700 28 03

9722 13 5

º

arcsin . . º

arccos . .

21.

22.

≈

≈ 44

3 3333 73 30

9250 67 67

º

arctan . . º

arcsin . .

23.

24.

≈

≈ ºº

arcsin . º ' "

arccos . º '

25.

26.

4700 28 2 3

9722 13 32

≈

≈ 330

3 3333 73 18 00

9250 6

"

arctan . º ' "

arcsin .

27.

28.

≈

≈ 77 40 6º ' "

Lesson 5B1.

2.

3.

sin º .

tan º .

sin º .

82 9903

14 2493

38

≈

≈

≈ 66157

49 6561

58 5299

77 4 331

4.

5.

6.

cos º .

cos º .

tan º .

≈

≈

≈ 55

0175 1

4848 61

7.

8.

9.

arcsin . º

arccos . º

arcsin .

≈

≈

88746 61

4245 23

1 1918 50

≈

≈

≈

º

arctan . º

arctan .

10.

11. ºº

arccos . º

º ' "

" '"

(

12.

13.

9945 6

47 36 25

251

160

256

≈

× =00

42

36 421

160

36 4260

× =

) " . '

. ' º'

( . )º .

≈

≈ 6607

47 36 25 47 607

53 28 10

101

160

1

º

º ' " . º

º ' "

" '"

(

≈

14.

× = 0060

17

28 171

160

28 1760

× =

) ' . '

. ' º'

( . )º

≈

≈≈

≈

. º

º ' " . º

º '

' º'

(

470

53 28 10 53 470

8 55

551

160

556

15.

× =00

917

8 55 8 917

88 0 40

401

16

×

)º . º

º ' . º

º ' "

" '

≈

≈

16.

004060

67

671

160

6760

º( )º . '

. ºº

( . )º

=

× =

≈

≈≈

≈

≈

. º

º ' " . º

º ' " . º

011

88 0 40 88 011

47 36 25 47 607

5

17.

18. 33 28 10 53 469

8 55 8 917

88 0 40 88

º ' " . º

º ' . º

º ' " .

≈

≈

≈

19.

20. 0011

4700 28 03

9722 13 5

º

arcsin . . º

arccos . .

21.

22.

≈

≈ 44

3 3333 73 30

9250 67 67

º

arctan . . º

arcsin . .

23.

24.

≈

≈ ºº

arcsin . º ' "

arccos . º '

25.

26.

4700 28 2 3

9722 13 32

≈

≈ 330

3 3333 73 18 00

9250 6

"

arctan . º ' "

arcsin .

27.

28.

≈

≈ 77 40 6º ' "

Lesson 5B1.

2.

3.

sin º .

tan º .

sin º .

82 9903

14 2493

38

≈

≈

≈ 66157

49 6561

58 5299

77 4 331

4.

5.

6.

cos º .

cos º .

tan º .

≈

≈

≈ 55

0175 1

4848 61

7.

8.

9.

arcsin . º

arccos . º

arcsin .

≈

≈

88746 61

4245 23

1 1918 50

≈

≈

≈

º

arctan . º

arctan .

10.

11. ºº

arccos . º

º ' "

" '"

(

12.

13.

9945 6

47 36 25

251

160

256

≈

× =00

42

36 421

160

36 4260

× =

) " . '

. ' º'

( . )º .

≈

≈ 6607

47 36 25 47 607

53 28 10

101

160

1

º

º ' " . º

º ' "

" '"

(

≈

14.

× = 0060

17

28 171

160

28 1760

× =

) ' . '

. ' º'

( . )º

≈

≈≈

≈

. º

º ' " . º

º '

' º'

(

470

53 28 10 53 470

8 55

551

160

556

15.

× =00

917

8 55 8 917

88 0 40

401

16

×

)º . º

º ' . º

º ' "

" '

≈

≈

16.

004060

67

671

160

6760

º( )º . '

. ºº

( . )º

=

× =

≈

≈≈

≈

≈

. º

º ' " . º

º ' " . º

011

88 0 40 88 011

47 36 25 47 607

5

17.

18. 33 28 10 53 469

8 55 8 917

88 0 40 88

º ' " . º

º ' . º

º ' " .

≈

≈

≈

19.

20. 0011

4700 28 03

9722 13 5

º

arcsin . . º

arccos . .

21.

22.

≈

≈ 44

3 3333 73 30

9250 67 67

º

arctan . . º

arcsin . .

23.

24.

≈

≈ ºº

arcsin . º ' "

arccos . º '

25.

26.

4700 28 2 3

9722 13 32

≈

≈ 330

3 3333 73 18 00

9250 6

"

arctan . º ' "

arcsin .

27.

28.

≈

≈ 77 40 6º ' "

Lesson 5B1.

2.

3.

sin º .

tan º .

sin º .

82 9903

14 2493

38

≈

≈

≈ 66157

49 6561

58 5299

77 4 331

4.

5.

6.

cos º .

cos º .

tan º .

≈

≈

≈ 55

0175 1

4848 61

7.

8.

9.

arcsin . º

arccos . º

arcsin .

≈

≈

88746 61

4245 23

1 1918 50

≈

≈

≈

º

arctan . º

arctan .

10.

11. ºº

arccos . º

º ' "

" '"

(

12.

13.

9945 6

47 36 25

251

160

256

≈

× =00

42

36 421

160

36 4260

× =

) " . '

. ' º'

( . )º .

≈

≈ 6607

47 36 25 47 607

53 28 10

101

160

1

º

º ' " . º

º ' "

" '"

(

≈

14.

× = 0060

17

28 171

160

28 1760

× =

) ' . '

. ' º'

( . )º

≈

≈≈

≈

. º

º ' " . º

º '

' º'

(

470

53 28 10 53 470

8 55

551

160

556

15.

× =00

917

8 55 8 917

88 0 40

401

16

×

)º . º

º ' . º

º ' "

" '

≈

≈

16.

004060

67

671

160

6760

º( )º . '

. ºº

( . )º

=

× =

≈

≈≈

≈

≈

. º

º ' " . º

º ' " . º

011

88 0 40 88 011

47 36 25 47 607

5

17.

18. 33 28 10 53 469

8 55 8 917

88 0 40 88

º ' " . º

º ' . º

º ' " .

≈

≈

≈

19.

20. 0011

4700 28 03

9722 13 5

º

arcsin . . º

arccos . .

21.

22.

≈

≈ 44

3 3333 73 30

9250 67 67

º

arctan . . º

arcsin . .

23.

24.

≈

≈ ºº

arcsin . º ' "

arccos . º '

25.

26.

4700 28 2 3

9722 13 32

≈

≈ 330

3 3333 73 18 00

9250 6

"

arctan . º ' "

arcsin .

27.

28.

≈

≈ 77 40 6º ' "

Lesson 5CLesson 5CThere are several ways to solve each off

these problems. one method will be

shown for eeach.

1. sin . º

sin . º

sin . º

22 5 18

22 5 181822 5

47

=

=

=

BB

B ≈

ttan . º

tan . º .

º . º

67 518

18 67 5 43 5

90 22 5 6

=

= ( )( )= − =

A

A ≈

α 77 5

47 311

11 47 3 8 1

42 7

. º

sin . º

sin . º .

sin . º

2. =

= ( )( )

D

D ≈

==

= ( )= − =

c

c1111 42 7 7 5

90 47 3 42 7

35

sin . º .

º . º . º

tan

≈

α

3. .. º

tan . º .

cos . º

cos

629

29 35 6 20 8

35 6 29

35

=

= ( )( )

=

F

F

ee

≈

.. º

cos . º.

º . º . º

º

6 292935 6

35 7

90 35 6 54 4

50 4

=

=

= − =

E ≈

α

4. 22 30 50 71

50 71250

250 50 71 1

' " . º

sin . º

sin . º

≈

≈

=

= ( )( )

G

G 993 5

50 71250

250 50 71 158 3

90

.

cos .

cos . º .

º

=

= ( )( )=

H

H ≈

α −− =

=

50 42 30 39 17 30

1213

9231

9

º ' " º ' "

tan .

arctan .

5. θ ≈

2231 42 71 42 42 36

1312

1 0833

90 47

≈ ≈

≈

. º º ' "

tan .

.

α

α θ

=

= − = 229 47 17 24

3427

1 2593

1 2593 5

º º ' "

tan .

arctan .

≈

≈

≈

6. θ =

11 55 51 32 50

90 51 55 38 45 38 27

. º º ' "

º . º . º º

or

orα = − = ≈ ''

sin .

arcsin . . º º '

7. θ = 1629

5517

5517 33 49 33 29 24

≈

≈ ≈ ""

º . º . º º ' "

cos .

α

θ

= − =

= =

90 33 49 56 51 56 30 36

76100

76

≈

8. 000

7600 40 54 40 32 9

90 40 54 49

arccos . . º º ' "

º . º .

≈ ≈

α = − = 446 49 27 36º º ' "≈

Lesson 5CThere are several ways to solve each off

these problems. one method will be

shown for eeach.

1. sin . º

sin . º

sin . º

22 5 18

22 5 181822 5

47

=

=

=

BB

B ≈

ttan . º

tan . º .

º . º

67 518

18 67 5 43 5

90 22 5 6

=

= ( )( )= − =

A

A ≈

α 77 5

47 311

11 47 3 8 1

42 7

. º

sin . º

sin . º .

sin . º

2. =

= ( )( )

D

D ≈

==

= ( )= − =

c

c1111 42 7 7 5

90 47 3 42 7

35

sin . º .

º . º . º

tan

≈

α

3. .. º

tan . º .

cos . º

cos

629

29 35 6 20 8

35 6 29

35

=

= ( )( )

=

F

F

ee

≈

.. º

cos . º.

º . º . º

º

6 292935 6

35 7

90 35 6 54 4

50 4

=

=

= − =

E ≈

α

4. 22 30 50 71

50 71250

250 50 71 1

' " . º

sin . º

sin . º

≈

≈

=

= ( )( )

G

G 993 5

50 71250

250 50 71 158 3

90

.

cos .

cos . º .

º

=

= ( )( )=

H

H ≈

α −− =

=

50 42 30 39 17 30

1213

9231

9

º ' " º ' "

tan .

arctan .

5. θ ≈

2231 42 71 42 42 36

1312

1 0833

90 47

≈ ≈

≈

. º º ' "

tan .

.

α

α θ

=

= − = 229 47 17 24

3427

1 2593

1 2593 5

º º ' "

tan .

arctan .

≈

≈

≈

6. θ =

11 55 51 32 50

90 51 55 38 45 38 27

. º º ' "

º . º . º º

or

orα = − = ≈ ''

sin .

arcsin . . º º '

7. θ = 1629

5517

5517 33 49 33 29 24

≈

≈ ≈ ""

º . º . º º ' "

cos .

α

θ

= − =

= =

90 33 49 56 51 56 30 36

76100

76

≈

8. 000

7600 40 54 40 32 9

90 40 54 49

arccos . . º º ' "

º . º .

≈ ≈

α = − = 446 49 27 36º º ' "≈

Lesson 5CThere are several ways to solve each off

these problems. one method will be

shown for eeach.

1. sin . º

sin . º

sin . º

22 5 18

22 5 181822 5

47

=

=

=

BB

B ≈

ttan . º

tan . º .

º . º

67 518

18 67 5 43 5

90 22 5 6

=

= ( )( )= − =

A

A ≈

α 77 5

47 311

11 47 3 8 1

42 7

. º

sin . º

sin . º .

sin . º

2. =

= ( )( )

D

D ≈

==

= ( )= − =

c

c1111 42 7 7 5

90 47 3 42 7

35

sin . º .

º . º . º

tan

≈

α

3. .. º

tan . º .

cos . º

cos

629

29 35 6 20 8

35 6 29

35

=

= ( )( )

=

F

F

ee

≈

.. º

cos . º.

º . º . º

º

6 292935 6

35 7

90 35 6 54 4

50 4

=

=

= − =

E ≈

α

4. 22 30 50 71

50 71250

250 50 71 1

' " . º

sin . º

sin . º

≈

≈

=

= ( )( )

G

G 993 5

50 71250

250 50 71 158 3

90

.

cos .

cos . º .

º

=

= ( )( )=

H

H ≈

α −− =

=

50 42 30 39 17 30

1213

9231

9

º ' " º ' "

tan .

arctan .

5. θ ≈

2231 42 71 42 42 36

1312

1 0833

90 47

≈ ≈

≈

. º º ' "

tan .

.

α

α θ

=

= − = 229 47 17 24

3427

1 2593

1 2593 5

º º ' "

tan .

arctan .

≈

≈

≈

6. θ =

11 55 51 32 50

90 51 55 38 45 38 27

. º º ' "

º . º . º º

or

orα = − = ≈ ''

sin .

arcsin . . º º '

7. θ = 1629

5517

5517 33 49 33 29 24

≈

≈ ≈ ""

º . º . º º ' "

cos .

α

θ

= − =

= =

90 33 49 56 51 56 30 36

76100

76

≈

8. 000

7600 40 54 40 32 9

90 40 54 49

arccos . . º º ' "

º . º .

≈ ≈

α = − = 446 49 27 36º º ' "≈

Lesson 5CThere are several ways to solve each off

these problems. one method will be

shown for eeach.

1. sin . º

sin . º

sin . º

22 5 18

22 5 181822 5

47

=

=

=

BB

B ≈

ttan . º

tan . º .

º . º

67 518

18 67 5 43 5

90 22 5 6

=

= ( )( )= − =

A

A ≈

α 77 5

47 311

11 47 3 8 1

42 7

. º

sin . º

sin . º .

sin . º

2. =

= ( )( )

D

D ≈

==

= ( )= − =

c

c1111 42 7 7 5

90 47 3 42 7

35

sin . º .

º . º . º

tan

≈

α

3. .. º

tan . º .

cos . º

cos

629

29 35 6 20 8

35 6 29

35

=

= ( )( )

=

F

F

ee

≈

.. º

cos . º.

º . º . º

º

6 292935 6

35 7

90 35 6 54 4

50 4

=

=

= − =

E ≈

α

4. 22 30 50 71

50 71250

250 50 71 1

' " . º

sin . º

sin . º

≈

≈

=

= ( )( )

G

G 993 5

50 71250

250 50 71 158 3

90

.

cos .

cos . º .

º

=

= ( )( )=

H

H ≈

α −− =

=

50 42 30 39 17 30

1213

9231

9

º ' " º ' "

tan .

arctan .

5. θ ≈

2231 42 71 42 42 36

1312

1 0833

90 47

≈ ≈

≈

. º º ' "

tan .

.

α

α θ

=

= − = 229 47 17 24

3427

1 2593

1 2593 5

º º ' "

tan .

arctan .

≈

≈

≈

6. θ =

11 55 51 32 50

90 51 55 38 45 38 27

. º º ' "

º . º . º º

or

orα = − = ≈ ''

sin .

arcsin . . º º '

7. θ = 1629

5517

5517 33 49 33 29 24

≈

≈ ≈ ""

º . º . º º ' "

cos .

α

θ

= − =

= =

90 33 49 56 51 56 30 36

76100

76

≈

8. 000

7600 40 54 40 32 9

90 40 54 49

arccos . . º º ' "

º . º .

≈ ≈

α = − = 446 49 27 36º º ' "≈

precALcULUS

LeSSon 5c - LeSSon 5D

SoLUTionS278

Lesson 5CThere are several ways to solve each off

these problems. one method will be

shown for eeach.

1. sin . º

sin . º

sin . º

22 5 18

22 5 181822 5

47

=

=

=

BB

B ≈

ttan . º

tan . º .

º . º

67 518

18 67 5 43 5

90 22 5 6

=

= ( )( )= − =

A

A ≈

α 77 5

47 311

11 47 3 8 1

42 7

. º

sin . º

sin . º .

sin . º

2. =

= ( )( )

D

D ≈

==

= ( )= − =

c

c1111 42 7 7 5

90 47 3 42 7

35

sin . º .

º . º . º

tan

≈

α

3. .. º

tan . º .

cos . º

cos

629

29 35 6 20 8

35 6 29

35

=

= ( )( )

=

F

F

ee

≈

.. º

cos . º.

º . º . º

º

6 292935 6

35 7

90 35 6 54 4

50 4

=

=

= − =

E ≈

α

4. 22 30 50 71

50 71250

250 50 71 1

' " . º

sin . º

sin . º

≈

≈

=

= ( )( )

G

G 993 5

50 71250

250 50 71 158 3

90

.

cos .

cos . º .

º

=

= ( )( )=

H

H ≈

α −− =

=

50 42 30 39 17 30

1213

9231

9

º ' " º ' "

tan .

arctan .

5. θ ≈

2231 42 71 42 42 36

1312

1 0833

90 47

≈ ≈

≈

. º º ' "

tan .

.

α

α θ

=

= − = 229 47 17 24

3427

1 2593

1 2593 5

º º ' "

tan .

arctan .

≈

≈

≈

6. θ =

11 55 51 32 50

90 51 55 38 45 38 27

. º º ' "

º . º . º º

or

orα = − = ≈ ''

sin .

arcsin . . º º '

7. θ = 1629

5517

5517 33 49 33 29 24

≈

≈ ≈ ""

º . º . º º ' "

cos .

α

θ

= − =

= =

90 33 49 56 51 56 30 36

76100

76

≈

8. 000

7600 40 54 40 32 9

90 40 54 49

arccos . . º º ' "

º . º .

≈ ≈

α = − = 446 49 27 36º º ' "≈

Lesson 5CThere are several ways to solve each off

these problems. one method will be

shown for eeach.

1. sin . º

sin . º

sin . º

22 5 18

22 5 181822 5

47

=

=

=

BB

B ≈

ttan . º

tan . º .

º . º

67 518

18 67 5 43 5

90 22 5 6

=

= ( )( )= − =

A

A ≈

α 77 5

47 311

11 47 3 8 1

42 7

. º

sin . º

sin . º .

sin . º

2. =

= ( )( )

D

D ≈

==

= ( )= − =

c

c1111 42 7 7 5

90 47 3 42 7

35

sin . º .

º . º . º

tan

≈

α

3. .. º

tan . º .

cos . º

cos

629

29 35 6 20 8

35 6 29

35

=

= ( )( )

=

F

F

ee

≈

.. º

cos . º.

º . º . º

º

6 292935 6

35 7

90 35 6 54 4

50 4

=

=

= − =

E ≈

α

4. 22 30 50 71

50 71250

250 50 71 1

' " . º

sin . º

sin . º

≈

≈

=

= ( )( )

G

G 993 5

50 71250

250 50 71 158 3

90

.

cos .

cos . º .

º

=

= ( )( )=

H

H ≈

α −− =

=

50 42 30 39 17 30

1213

9231

9

º ' " º ' "

tan .

arctan .

5. θ ≈

2231 42 71 42 42 36

1312

1 0833

90 47

≈ ≈

≈

. º º ' "

tan .

.

α

α θ

=

= − = 229 47 17 24

3427

1 2593

1 2593 5

º º ' "

tan .

arctan .

≈

≈

≈

6. θ =

11 55 51 32 50

90 51 55 38 45 38 27

. º º ' "

º . º . º º

or

orα = − = ≈ ''

sin .

arcsin . . º º '

7. θ = 1629

5517

5517 33 49 33 29 24

≈

≈ ≈ ""

º . º . º º ' "

cos .

α

θ

= − =

= =

90 33 49 56 51 56 30 36

76100

76

≈

8. 000

7600 40 54 40 32 9

90 40 54 49

arccos . . º º ' "

º . º .

≈ ≈

α = − = 446 49 27 36º º ' "≈

Lesson 5CThere are several ways to solve each off

these problems. one method will be

shown for eeach.

1. sin . º

sin . º

sin . º

22 5 18

22 5 181822 5

47

=

=

=

BB

B ≈

ttan . º

tan . º .

º . º

67 518

18 67 5 43 5

90 22 5 6

=

= ( )( )= − =

A

A ≈

α 77 5

47 311

11 47 3 8 1

42 7

. º

sin . º

sin . º .

sin . º

2. =

= ( )( )

D

D ≈

==

= ( )= − =

c

c1111 42 7 7 5

90 47 3 42 7

35

sin . º .

º . º . º

tan

≈

α

3. .. º

tan . º .

cos . º

cos

629

29 35 6 20 8

35 6 29

35

=

= ( )( )

=

F

F

ee

≈

.. º

cos . º.

º . º . º

º

6 292935 6

35 7

90 35 6 54 4

50 4

=

=

= − =

E ≈

α

4. 22 30 50 71

50 71250

250 50 71 1

' " . º

sin . º

sin . º

≈

≈

=

= ( )( )

G

G 993 5

50 71250

250 50 71 158 3

90

.

cos .

cos . º .

º

=

= ( )( )=

H

H ≈

α −− =

=

50 42 30 39 17 30

1213

9231

9

º ' " º ' "

tan .

arctan .

5. θ ≈

2231 42 71 42 42 36

1312

1 0833

90 47

≈ ≈

≈

. º º ' "

tan .

.

α

α θ

=

= − = 229 47 17 24

3427

1 2593

1 2593 5

º º ' "

tan .

arctan .

≈

≈

≈

6. θ =

11 55 51 32 50

90 51 55 38 45 38 27

. º º ' "

º . º . º º

or

orα = − = ≈ ''

sin .

arcsin . . º º '

7. θ = 1629

5517

5517 33 49 33 29 24

≈

≈ ≈ ""

º . º . º º ' "

cos .

α

θ

= − =

= =

90 33 49 56 51 56 30 36

76100

76

≈

8. 000

7600 40 54 40 32 9

90 40 54 49

arccos . . º º ' "

º . º .

≈ ≈

α = − = 446 49 27 36º º ' "≈

Lesson 5CThere are several ways to solve each off

these problems. one method will be

shown for eeach.

1. sin . º

sin . º

sin . º

22 5 18

22 5 181822 5

47

=

=

=

BB

B ≈

ttan . º

tan . º .

º . º

67 518

18 67 5 43 5

90 22 5 6

=

= ( )( )= − =

A

A ≈

α 77 5

47 311

11 47 3 8 1

42 7

. º

sin . º

sin . º .

sin . º

2. =

= ( )( )

D

D ≈

==

= ( )= − =

c

c1111 42 7 7 5

90 47 3 42 7

35

sin . º .

º . º . º

tan

≈

α

3. .. º

tan . º .

cos . º

cos

629

29 35 6 20 8

35 6 29

35

=

= ( )( )

=

F

F

ee

≈

.. º

cos . º.

º . º . º

º

6 292935 6

35 7

90 35 6 54 4

50 4

=

=

= − =

E ≈

α

4. 22 30 50 71

50 71250

250 50 71 1

' " . º

sin . º

sin . º

≈

≈

=

= ( )( )

G

G 993 5

50 71250

250 50 71 158 3

90

.

cos .

cos . º .

º

=

= ( )( )=

H

H ≈

α −− =

=

50 42 30 39 17 30

1213

9231

9

º ' " º ' "

tan .

arctan .

5. θ ≈

2231 42 71 42 42 36

1312

1 0833

90 47

≈ ≈

≈

. º º ' "

tan .

.

α

α θ

=

= − = 229 47 17 24

3427

1 2593

1 2593 5

º º ' "

tan .

arctan .

≈

≈

≈

6. θ =

11 55 51 32 50

90 51 55 38 45 38 27

. º º ' "

º . º . º º

or

orα = − = ≈ ''

sin .

arcsin . . º º '

7. θ = 1629

5517

5517 33 49 33 29 24

≈

≈ ≈ ""

º . º . º º ' "

cos .

α

θ

= − =

= =

90 33 49 56 51 56 30 36

76100

76

≈

8. 000

7600 40 54 40 32 9

90 40 54 49

arccos . . º º ' "

º . º .

≈ ≈

α = − = 446 49 27 36º º ' "≈

Lesson 5CThere are several ways to solve each off

these problems. one method will be

shown for eeach.

1. sin . º

sin . º

sin . º

22 5 18

22 5 181822 5

47

=

=

=

BB

B ≈

ttan . º

tan . º .

º . º

67 518

18 67 5 43 5

90 22 5 6

=

= ( )( )= − =

A

A ≈

α 77 5

47 311

11 47 3 8 1

42 7

. º

sin . º

sin . º .

sin . º

2. =

= ( )( )

D

D ≈

==

= ( )= − =

c

c1111 42 7 7 5

90 47 3 42 7

35

sin . º .

º . º . º

tan

≈

α

3. .. º

tan . º .

cos . º

cos

629

29 35 6 20 8

35 6 29

35

=

= ( )( )

=

F

F

ee

≈

.. º

cos . º.

º . º . º

º

6 292935 6

35 7

90 35 6 54 4

50 4

=

=

= − =

E ≈

α

4. 22 30 50 71

50 71250

250 50 71 1

' " . º

sin . º

sin . º

≈

≈

=

= ( )( )

G

G 993 5

50 71250

250 50 71 158 3

90

.

cos .

cos . º .

º

=

= ( )( )=

H

H ≈

α −− =

=

50 42 30 39 17 30

1213

9231

9

º ' " º ' "

tan .

arctan .

5. θ ≈

2231 42 71 42 42 36

1312

1 0833

90 47

≈ ≈

≈

. º º ' "

tan .

.

α

α θ

=

= − = 229 47 17 24

3427

1 2593

1 2593 5

º º ' "

tan .

arctan .

≈

≈

≈

6. θ =

11 55 51 32 50

90 51 55 38 45 38 27

. º º ' "

º . º . º º

or

orα = − = ≈ ''

sin .

arcsin . . º º '

7. θ = 1629

5517

5517 33 49 33 29 24

≈

≈ ≈ ""

º . º . º º ' "

cos .

α

θ

= − =

= =

90 33 49 56 51 56 30 36

76100

76

≈

8. 000

7600 40 54 40 32 9

90 40 54 49

arccos . . º º ' "

º . º .

≈ ≈

α = − = 446 49 27 36º º ' "≈

Lesson 5DLesson 5D1. sin . º

sin . º .

cos .

39 16

6 39 1 3 8

39

=

= ( )( )

B

B ≈

116

6 39 1 4 7

90 39 1 50 9

º

cos . º .

º . º . º

cos

=

= ( )( )= − =

A

A ≈

α

2. 442 25 20

2042 25

27 0

42 2527 0

2

. º

cos . º.

sin . º.

=

=

=

=

D

D

c

c

≈

77 42 25 18 2

42 15 42 25

90 42 25

( )( )= =

= −

sin . º .

º ' . º

º . º

≈

θ

α ==

=

= ( )( )

=

47 75

2857

57 28 30 3

62 5

. º

tan º

tan º .

sin º

3. e

E ≈

77

5762

64 6

90 62 28

90 25 73 64

F

F =

= − =

− =

sin º.

º º º

º . º .

≈

α

4. 227

64 27 18 3

64 27 18 3

18 364 2

º

cos . º .

cos . º .

.cos .

=

=

=

HH

H77

42 2

64 2742 2

42 2 64 27 38 0

º.

sin . º.

. sin . º .

≈

≈

=

= ( )( )

G

G ºº

º ' . º

º . º . º

tan .

θ

α

θ

=

= − =

=

25 44 25 73

90 25 73 64 27

6 25

≈

5...

.

arctan . . º

. º º '

41 1481

1 1481 48 95

48 95 48 56 38

≈

≈ ≈

≈

θ ""

º . º . º º '

tan ..

.

α

θ

= − =

=

90 48 95 41 05 41 3

40 528 7

1 41

≈

≈6. 111

1 4111 54 68

54 68 54 40 34

90

arctan . . º

. º º ' "

º

≈

≈ ≈θ

α = −554 68 35 32 35 19 12

3963

6190

. º . º ' "

cos .

arccos

=

=

≈

≈7. θ

.. . º º ' "

º . º . º º

6190 51 76 51 45 25

90 51 76 38 24 38 1

≈ ≈

≈α = − = 44 24

3 87 9

4810

4810 28 75

2

' "

sin ..

.

arcsin . . º

8. θ

θ

= ≈

≈

≈ 88 75 28 45 03

90 28 75 61 25 61 15

. º º ' "

º . º . º º '

≈

α = − = ≈

Lesson 5D1. sin . º

sin . º .

cos .

39 16

6 39 1 3 8

39

=

= ( )( )

B

B ≈

116

6 39 1 4 7

90 39 1 50 9

º

cos . º .

º . º . º

cos

=

= ( )( )= − =

A

A ≈

α

2. 442 25 20

2042 25

27 0

42 2527 0

2

. º

cos . º.

sin . º.

=

=

=

=

D

D

c

c

≈

77 42 25 18 2

42 15 42 25

90 42 25

( )( )= =

= −

sin . º .

º ' . º

º . º

≈

θ

α ==

=

= ( )( )

=

47 75

2857

57 28 30 3

62 5

. º

tan º

tan º .

sin º

3. e

E ≈

77

5762

64 6

90 62 28

90 25 73 64

F

F =

= − =

− =

sin º.

º º º

º . º .

≈

α

4. 227

64 27 18 3

64 27 18 3

18 364 2

º

cos . º .

cos . º .

.cos .

=

=

=

HH

H77

42 2

64 2742 2

42 2 64 27 38 0

º.

sin . º.

. sin . º .

≈

≈

=

= ( )( )

G

G ºº

º ' . º

º . º . º

tan .

θ

α

θ

=

= − =

=

25 44 25 73

90 25 73 64 27

6 25

≈

5...

.

arctan . . º

. º º '

41 1481

1 1481 48 95

48 95 48 56 38

≈

≈ ≈

≈

θ ""

º . º . º º '

tan ..

.

α

θ

= − =

=

90 48 95 41 05 41 3

40 528 7

1 41

≈

≈6. 111

1 4111 54 68

54 68 54 40 34

90

arctan . . º

. º º ' "

º

≈

≈ ≈θ

α = −554 68 35 32 35 19 12

3963

6190

. º . º ' "

cos .

arccos

=

=

≈

≈7. θ

.. . º º ' "

º . º . º º

6190 51 76 51 45 25

90 51 76 38 24 38 1

≈ ≈

≈α = − = 44 24

3 87 9

4810

4810 28 75

2

' "

sin ..

.

arcsin . . º

8. θ

θ

= ≈

≈

≈ 88 75 28 45 03

90 28 75 61 25 61 15

. º º ' "

º . º . º º '

≈

α = − = ≈

Lesson 5D1. sin . º

sin . º .

cos .

39 16

6 39 1 3 8

39

=

= ( )( )

B

B ≈

116

6 39 1 4 7

90 39 1 50 9

º

cos . º .

º . º . º

cos

=

= ( )( )= − =

A

A ≈

α

2. 442 25 20

2042 25

27 0

42 2527 0

2

. º

cos . º.

sin . º.

=

=

=

=

D

D

c

c

≈

77 42 25 18 2

42 15 42 25

90 42 25

( )( )= =

= −

sin . º .

º ' . º

º . º

≈

θ

α ==

=

= ( )( )

=

47 75

2857

57 28 30 3

62 5

. º

tan º

tan º .

sin º

3. e

E ≈

77

5762

64 6

90 62 28

90 25 73 64

F

F =

= − =

− =

sin º.

º º º

º . º .

≈

α

4. 227

64 27 18 3

64 27 18 3

18 364 2

º

cos . º .

cos . º .

.cos .

=

=

=

HH

H77

42 2

64 2742 2

42 2 64 27 38 0

º.

sin . º.

. sin . º .

≈

≈

=

= ( )( )

G

G ºº

º ' . º

º . º . º

tan .

θ

α

θ

=

= − =

=

25 44 25 73

90 25 73 64 27

6 25

≈

5...

.

arctan . . º

. º º '

41 1481

1 1481 48 95

48 95 48 56 38

≈

≈ ≈

≈

θ ""

º . º . º º '

tan ..

.

α

θ

= − =

=

90 48 95 41 05 41 3

40 528 7

1 41

≈

≈6. 111

1 4111 54 68

54 68 54 40 34

90

arctan . . º

. º º ' "

º

≈

≈ ≈θ

α = −554 68 35 32 35 19 12

3963

6190

. º . º ' "

cos .

arccos

=

=

≈

≈7. θ

.. . º º ' "

º . º . º º

6190 51 76 51 45 25

90 51 76 38 24 38 1

≈ ≈

≈α = − = 44 24

3 87 9

4810

4810 28 75

2

' "

sin ..

.

arcsin . . º

8. θ

θ

= ≈

≈

≈ 88 75 28 45 03

90 28 75 61 25 61 15

. º º ' "

º . º . º º '

≈

α = − = ≈

Lesson 5D1. sin . º

sin . º .

cos .

39 16

6 39 1 3 8

39

=

= ( )( )

B

B ≈

116

6 39 1 4 7

90 39 1 50 9

º

cos . º .

º . º . º

cos

=

= ( )( )= − =

A

A ≈

α

2. 442 25 20

2042 25

27 0

42 2527 0

2

. º

cos . º.

sin . º.

=

=

=

=

D

D

c

c

≈

77 42 25 18 2

42 15 42 25

90 42 25

( )( )= =

= −

sin . º .

º ' . º

º . º

≈

θ

α ==

=

= ( )( )

=

47 75

2857

57 28 30 3

62 5

. º

tan º

tan º .

sin º

3. e

E ≈

77

5762

64 6

90 62 28

90 25 73 64

F

F =

= − =

− =

sin º.

º º º

º . º .

≈

α

4. 227

64 27 18 3

64 27 18 3

18 364 2

º

cos . º .

cos . º .

.cos .

=

=

=

HH

H77

42 2

64 2742 2

42 2 64 27 38 0

º.

sin . º.

. sin . º .

≈

≈

=

= ( )( )

G

G ºº

º ' . º

º . º . º

tan .

θ

α

θ

=

= − =

=

25 44 25 73

90 25 73 64 27

6 25

≈

5...

.

arctan . . º

. º º '

41 1481

1 1481 48 95

48 95 48 56 38

≈

≈ ≈

≈

θ ""

º . º . º º '

tan ..

.

α

θ

= − =

=

90 48 95 41 05 41 3

40 528 7

1 41

≈

≈6. 111

1 4111 54 68

54 68 54 40 34

90

arctan . . º

. º º ' "

º

≈

≈ ≈θ

α = −554 68 35 32 35 19 12

3963

6190

. º . º ' "

cos .

arccos

=

=

≈

≈7. θ

.. . º º ' "

º . º . º º

6190 51 76 51 45 25

90 51 76 38 24 38 1

≈ ≈

≈α = − = 44 24

3 87 9

4810

4810 28 75

2

' "

sin ..

.

arcsin . . º

8. θ

θ

= ≈

≈

≈ 88 75 28 45 03

90 28 75 61 25 61 15

. º º ' "

º . º . º º '

≈

α = − = ≈

Lesson 5D1. sin . º

sin . º .

cos .

39 16

6 39 1 3 8

39

=

= ( )( )

B

B ≈

116

6 39 1 4 7

90 39 1 50 9

º

cos . º .

º . º . º

cos

=

= ( )( )= − =

A

A ≈

α

2. 442 25 20

2042 25

27 0

42 2527 0

2

. º

cos . º.

sin . º.

=

=

=

=

D

D

c

c

≈

77 42 25 18 2

42 15 42 25

90 42 25

( )( )= =

= −

sin . º .

º ' . º

º . º

≈

θ

α ==

=

= ( )( )

=

47 75

2857

57 28 30 3

62 5

. º

tan º

tan º .

sin º

3. e

E ≈

77

5762

64 6

90 62 28

90 25 73 64

F

F =

= − =

− =

sin º.

º º º

º . º .

≈

α

4. 227

64 27 18 3

64 27 18 3

18 364 2

º

cos . º .

cos . º .

.cos .

=

=

=

HH

H77

42 2

64 2742 2

42 2 64 27 38 0

º.

sin . º.

. sin . º .

≈

≈

=

= ( )( )

G

G ºº

º ' . º

º . º . º

tan .

θ

α

θ

=

= − =

=

25 44 25 73

90 25 73 64 27

6 25

≈

5...

.

arctan . . º

. º º '

41 1481

1 1481 48 95

48 95 48 56 38

≈

≈ ≈

≈

θ ""

º . º . º º '

tan ..

.

α

θ

= − =

=

90 48 95 41 05 41 3

40 528 7

1 41

≈

≈6. 111

1 4111 54 68

54 68 54 40 34

90

arctan . . º

. º º ' "

º

≈

≈ ≈θ

α = −554 68 35 32 35 19 12

3963

6190

. º . º ' "

cos .

arccos

=

=

≈

≈7. θ

.. . º º ' "

º . º . º º

6190 51 76 51 45 25

90 51 76 38 24 38 1

≈ ≈

≈α = − = 44 24

3 87 9

4810

4810 28 75

2

' "

sin ..

.

arcsin . . º

8. θ

θ

= ≈

≈

≈ 88 75 28 45 03

90 28 75 61 25 61 15

. º º ' "

º . º . º º '

≈

α = − = ≈