MTH-4103-1 Trigonometry I - SOFAD · 2014-05-15 · MTH-4103-1 C1-C4 Tige I_La 1 10-10-18 13:57...

Trigonometry I

MTH-4103-1

MTH-4103-1 C1-C4 Trigonometry I_Layout 1 10-10-18 13:57

TRIGONOMETRY I

MTH-4103-1

This course was produced in collaboration with the Service de l'éducationdes adultes de la Commission scolaire catholique de Sherbrooke and theDepartment of the Secretary of State of Canada.

Author: Monique Pagé

Content revision: Jean-Paul Groleau

Daniel Gélineau

Mireille Moisan-Sanscartier

Adult education consultant: Serge Vallières

Coordinator for the DGFD: Jean-Paul Groleau

Coordinator for the DFGA: Ronald Côté

Photocomposition and layout: Multitexte Plus

Translation: Consultation en éducation Zegray

Linguistic revision: Kay Flanagan and Leslie Macdonald

Translation of updated sections: Claudia de Fulviis

First edition: 1991

Reprint: 2004

© Société de formation à distance des commissions scolaires du Québec

All rights for translation and adaptation, in whole or in part, reserved for all countries.Any reproduction, by mechanical or electronic means, including micro-reproduction, isforbidden without the written permission of a duly authorized representative of theSociété de formation à distance des commissions scolaires du Québec (SOFAD).

Legal Deposit – 2004

Bibliothèque et Archives nationales du Québec

Bibliothèque et Archives Canada

ISBN 2-89493-281-0

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MTH-4103-1 Trigonometry I

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TABLE OF CONTENTS

Introduction to the Program Flowchart ................................................... 0.4

The Program Flowchart ............................................................................ 0.5

How to Use this Guide .............................................................................. 0.6

General Introduction................................................................................. 0.9

Intermediate and Terminal Objectives of the Module ............................ 0.11

Diagnostic Test on the Prerequisites ....................................................... 0.15

Answer Key for the Diagnostic Test on the Prerequisites ...................... 0.21

Analysis of the Diagnostic Test Results ................................................... 0.23

Information for Distance Education Students......................................... 0.25

UNITS

1. Right Triangles .......................................................................................... 1.1

2. Trigonometry and Trigonometric Ratios.................................................. 2.1

3. Using Trigonometric Ratios to Determine Angles .................................. 3.1

4. Solving Right Triangles ............................................................................ 4.1

5. Everyday Problems ................................................................................... 5.1

6. Solving Any Given Triangle...................................................................... 6.1

Final Summary.......................................................................................... 7.1

Terminal Objectives .................................................................................. 7.9

Self-evaluation Test .................................................................................. 7.11

Answer Key for the Self-evaluation Test ................................................. 7.19

Analysis of the Self-evaluation Test Results ........................................... 7.25

Final Evaluation........................................................................................ 7.26

Answer Key for the Exercises ................................................................... 7.27

Glossary ..................................................................................................... 7.93

List of Symbols .......................................................................................... 7.96

Bibliography .............................................................................................. 7.97

Review Activities ....................................................................................... 8.1

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INTRODUCTION TO THE PROGRAM FLOWCHART

Welcome to the World of Mathematics!

This mathematics program has been developed for the adult students of the

Adult Education Services of school boards and distance education. The learning

activities have been designed for individualized learning. If you encounter

difficulties, do not hesitate to consult your teacher or to telephone the resource

person assigned to you. The following flowchart shows where this module fits

into the overall program. It allows you to see how far you have progressed and

how much you still have to do to achieve your vocational goal. There are several

possible paths you can take, depending on your chosen goal.

The first path consists of modules MTH-3003-2 (MTH-314) and MTH-4104-2

(MTH-416), and leads to a Diploma of Vocational Studies (DVS).

The second path consists of modules MTH-4109-1 (MTH-426), MTH-4111-2

(MTH-436) and MTH-5104-1 (MTH-514), and leads to a Secondary School

Diploma (SSD), which allows you to enroll in certain Gegep-level programs that

do not call for a knowledge of advanced mathematics.

The third path consists of modules MTH-5109-1 (MTH-526) and MTH-5111-2

(MTH-536), and leads to Cegep programs that call for a solid knowledge of

mathematics in addition to other abiliies.

If this is your first contact with this mathematics program, consult the flowchart

on the next page and then read the section “How to Use This Guide.” Otherwise,

go directly to the section entitled “General Introduction.” Enjoy your work!

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CEGEP

MTH-5110-1 Introduction to Vectors

MTH-5109-1 Geometry IV

MTH-5108-1 Trigonometric Functions and Equations

MTH-5107-1 Exponential and Logarithmic Functions and Equations

MTH-5106-1 Real Functions and Equations

MTH-5105-1 Conics

MTH-5104-1 Optimization II

MTH-5103-1 Probability II

MTH-5102-1 Statistics III

MTH-5101-1 Optimization I

MTH-4110-1 The Four Operations on Algebraic Fractions

MTH-4109-1 Sets, Relations and Functions

MTH-4108-1 Quadratic Functions

MTH-4107-1 Straight Lines II

MTH-4106-1 Factoring and Algebraic Functions

MTH-4105-1 Exponents and Radicals


MTH-4102-1 Geometry III

MTH-536

MTH-526

MTH-514

MTH-436

MTH-426

MTH-416

MTH-314

MTH-216

MTH-116

MTH-3002-2 Geometry II

MTH-3001-2 The Four Operations on Polynomials

MAT-2008-2 Statistics and Probabilities I

MTH-2007-2 Geometry I

MTH-2006-2 Equations and Inequalities I

MTH-1007-2 Decimals and Percent

MTH-1006-2 The Four Operations on Fractions

MTH-1005-2 The Four Operations on Integers

MTH-5111-2 Complement and Synthesis II

MTH-4111-2 Complement and Synthesis I

MTH-4101-2 Equations and Inequalities II

MTH-3003-2 Straight Lines I

TradesDVS

MTH-5112-1 Logic

25 hours = 1 credit

50 hours = 2 credits

MTH-4104-2 Statistics II

THE PROGRAM FLOWCHART

You ar e here

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Hi! My name is Monica and I have beenasked to tell you about this math module.What’s your name?

I’m Andy.

Whether you areregistered at anadult education

center or atFormation àdistance, ...

You’ll see that with this method, math isa real breeze!

My results on the testindicate that I should beginwith this module.

Now, the module you have in yourhand is divided into threesections. The first section is...

... the entry activity, whichcontains the test on theprerequisites.

By carefully correcting this test using thecorresponding answer key, and record-ing your results on the analysis sheet ...

HOW TO USE THIS GUIDE

... you have probably taken aplacement test which tells youexactly which module youshould start with.

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?

The memo pad signals a brief reminder ofconcepts which you have already studied.

The calculator symbol reminds you thatyou will need to use your calculator.

The sheaf of wheat indicates a review designed toreinforce what you have just learned. A row ofsheaves near the end of the module indicates thefinal review, which helps you to interrelate all thelearning activities in the module.

The starting lineshows where thelearning activitiesbegin.

The little white question mark indicates the questionsfor which answers are given in the text.?

... you can tell if you’re well enoughprepared to do all the activities in themodule.

The boldface question markindicates practice exerciceswhich allow you to try out whatyou have just learned.

And if I’m not, if I need a littlereview before moving on, whathappens then?

In that case, before you start theactivities in the module, the resultsanalysis chart refers you to a reviewactivity near the end of the module.

In this way, I can be sure Ihave all the prerequisitesfor starting.

Exactly! The second sectioncontains the learning activities. It’sthe main part of the module.

Look closely at the box tothe right. It explains thesymbols used to identify thevarious activities.

The target precedes theobjective to be met.

I see!

?

START

Lastly, the finish line indicatesthat it is time to go on to the self-evaluationtest to verify how well you have understoodthe learning activities.

FINISH

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A “Did you know that...”?

Later ...

For example. words in bold-face italics appear in theglossary at the end of themodule...

G r e a t !

... statements in boxes are importantpoints to remember, like definitions, for-mulas and rules. I’m telling you, the for-mat makes everything much easier.

The third section contains the final re-view, which interrelates the differentparts of the module.

Yes, for example, short tidbitson the history of mathematicsand fun puzzles. They are in-teresting and relieve tension atthe same time.

No, it’s not part of the learn-ing activity. It’s just there togive you a breather.

There are also many fun thingsin this module. For example,when you see the drawing of asage, it introduces a “Did youknow that...”

Must I memorize what the sage says?

It’s the same for the “math whiz”pages, which are designed espe-cially for those who love math.

They are so stimulating thateven if you don’t have to dothem, you’ll still want to.

And the whole module hasbeen arranged to makelearning easier.

There is also a self-evaluationtest and answer key. They tellyou if you’re ready for the finalevaluation.

Thanks, Monica, you’ve been a bighelp.

I’m glad! Now,I’ve got to run.

See you!This is great! I never thought that I wouldlike mathematics as much as this!

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GENERAL INTRODUCTION

DISCOVERING THE PROPERTIES OF TRIANGLES

As everyone knows, surveyors measure the dimensions of lots. Most complex

geometric figures are formed by the juxtaposition of simple figures. Surveyors

therefore split up the area to be surveyed into triangles and then measure the

sides or angles in the triangles in order to determine the exact boundaries of the

lot in question.

Problems in subdividing land into lots or of measuring area led people in ancient

times to develop techniques which allowed them to state the relationships

between the lengths of the sides in a triangle and the size of its angles.

Astronomy was also developed in order to allow navigators to determine their

bearings and their location by taking the position of the stars as reference points:

a star, its projection on the earth's surface, and the position of an observer form

a triangle.

Our predecessors thus developed a great deal of knowledge based on the ratios

between the measures of the sides in a right triangle. In so doing, they gave

birth to trigonometry, often abbreviated to "trig" (triangle measurement),

which, when developed further, allows you to discover the relationships between

the sides or angles in any triangle.

To reach the objective of this module, you should be able to solve triangulation

problems, that is, problems that involve measuring angles and sides in

triangles. You should also be able to evaluate trigonometric ratios and solve

problems involving carpentry or navigation by using trigonometric ratios.

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Lastly, you will extend this knowledge to surveying by trying to solve problems

related to any triangle. To do so, you will apply the sine and cosine laws which

are the key tools for solving these problems. As you will see, this knowledge is

very useful in many sectors of human activity. In addition to its role in surveying,

carpentry and astronomy, it is useful in sewing, metalwork, architecture,

engineering and so on.

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INTERMEDIATE AND TERMINAL OBJECTIVES OFTHE MODULE

Module MTH-4103-1 consists of nine units and requires fifty hours of study

distributed as shown below. Each unit covers either an intermediate or a

terminal objective. The terminal objectives appear in boldface.

Objectives Number of Hours* % (evaluation)

1 to 4 7 30%

5 9 40%

6 8 30%

* One hour is allotted for the final evaluation.

1. Right Triangles

To solve a right triangle, that is, to determine the measure of the three angles

and the three sides in this triangle, using the Pythagorean Theorem, and to

establish the relationships between these angles. Two cases are possible:

• a right triangle, given the measures of one acute angle and two sides

• a right triangle, given an angle which measures either 30° or 45°, and the

length of one side

The situations are presented in the form of word problems borrowed from

everyday life. The steps in the solution must be described.

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2. Trigonometry and Trigonometric Ratios

To calculate the numerical value of any one of the three trigonometric ratios

of an angle A:

• sine A (sin A)

• cosine A (cos A)

• tangent A (tan A)

given the lengths of the three sides in a right triangle ABC, which is right-

angled in C.

3. Using Trigonometric Ratios to Determine Angles

To calculate the measures of the acute angles in right triangles, given the

lengths of two of the sides, by applying the properties of trigonometric ratios

and by using a trigonometric table or a scientific calculator.

4. Solving Right Triangles

To determine the missing dimensions in right triangles by applying

the properties of trigonometric ratios, the Pythagorean Theorem

and the relationships between the angles in this triangle, given:

• the length of two sides, or

• the length of one side and the measure of one acute angle.

The problems are solved either by using a trigonometric table or a

scientific calculator. The steps in the solution must be described.

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5. Everyday Problems

To solve word problems dealing with right triangles which require

the application of the properties of the six trigonometric ratios and

the use of a trigonometric table or a scientific calculator. The

measures to be determined may be those of angles, sides, or angles

and sides in a right triangle that is given or is to be drawn. The

situations are borrowed from everyday life and the steps in the

solution must be described.

6. Solving any Given Triangle

To solve word problems dealing with any given triangle, by applying:

• either the sine law: asin A

= bsin B

= csin C

,

• or the cosine law: a2 = b2 + c2 – 2bc cos A.

The measures to be determined may be those of angles, sides, or

angles and sides in any type of triangle that is given or is to be drawn.

Use of trigonometric tables or a scientific calculator is required. The

situations are borrowed from everyday life and the steps in the

solution must be described.

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DIAGNOSTIC TEST ON THE PREREQUISITES

Instructions

1. Answer as many questions as you can.

2. To answer these questions, you should have a protractor and acalculator.

3. Write your answers on the test paper.

4. Do not waste any time. If you cannot answer a question, go on

to the next one immediately.

5. When you have answered as many questions as you can, correct

your answers, using the answer key which follows the diagnos-

tic test.

6. To be considered correct, your answers must be identical to

those in the key. For example, if you are asked to describe the

steps involved in solving a problem, your answer must contain

all the steps.

7. Transcribe your results onto the chart which follows the answer

key. It gives an analysis of the diagnostic test results.

8. Do only the review activities listed for each of your incorrect

answers.

9. If all your answers are correct, you have the prerequisites to

begin working on this module.

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1. Match the names of the different triangles with the corresponding geometric

figures below.

a) An isosceles right triangle: ....................

b) An isosceles triangle: ....................

c) A scalene triangle: ....................

d) An equilateral triangle: ....................

2. a) Measure the three angles

in the adjacent triangle,

using a protractor.

N.B. Your measurements

must be accurate to the

nearest 2°.

m∠A = ......................, m∠B = ......................, m∠C = ......................

b) Which angle in the triangle is obtuse? ......................................................

A

B C

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3. In the adjacent figure, identify the following:

a) the hypotenuse: .................................

b) an acute angle: ..................................

c) the sides adjacent to angle A:

...........................................................

d) the side opposite angle A: .................

e) the right angle: ..................................

4. Given the adjacent triangle ABC,

determine the measure of angle A

without using a protractor. A com-

plete solution is required.

5. a) Referring to the adjacent figure,

calculate the length of the hypot-

enuse in this right triangle by

applying the Pythagorean Theo-

rem. Round your answers to the

nearest centimetre. A complete

solution is required.

A

B

C

a

b

c

A

B C

20°

c4 cm

B

C A10 cm

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b) Determine the length of side u in

adjacent triangle TUV, using the

Pythagorean Theorem. A comp-

lete solution is required.

6. Convert each of the following fractions to a decimal. Your answers must

contain 4 decimal places.

a) 716 = .............. b) 12

17 =............... c) 49 = ................

7. Calculate the value of x in the following expressions by applying the funda-

mental property of proportions. A complete solution is required. If necessary,

round your answers to the nearest hundredth.

a) 725x = 125

150 b) x210 = 240

715

U

V

T

u = ?

v = 4 cm

t = 5 cm

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8. a) Garfield deposits $75.00 each week in a savings account. The remainder

of his salary, namely $250.00, is used for various expenses such as rent,

food, travel and so on. Garfield wants to save enough money to buy a new

television and a video cassette recorder worth $925.00. If he already has

$150.00 in his account, how many more weeks will it take him to

accumulate this amount? A complete solution is required.

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b) Julianna is having fun flying a kite. This kite is shaped like a quadrilat-

eral with two 40 cm sides and two 55 cm sides. Julianna is standing

50 m from her house and she has let out all the string. If she sights the

kite at a 30° angle from where she stands and if the string attached to the

kite is 45 m long, calculate how high above Julianna's head the kite is

flying as well as the horizontal distance which separates Julianna and the

kite. The steps in the solution as well as the answers are required.

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ANSWER KEY FOR THE DIAGNOSTICTEST ON THE PREREQUISITES

1. a) ➃ b) ➁ or ➃ or ➅ c) ➄ d) ➅

2. a) m∠A = 32°; m∠B = 130°; m∠C = 18°

N.B. The sum of the measures of the three angles of a triangle must equal

180°.

b) ∠B

3. a) c or AB b) ∠A or ∠B c) b and c or AC and AB

d) a or BC e) ∠C

4. m∠A = 90° – 20° = 70°

5. a) c2 = a2 + b2 b) u2 = t2 – v2

c2 = 42 + 102 u2 = 52 – 42

c2 = 16 + 100 u2 = 25 – 16

c2 = 116 u2 = 9

c = 116 u = 9

c = 10.77 cm u = 3 cm

c = 11 cm

6. a) 0.437 5 b) 0.705 9 c) 0.444 4

7. a) 725x = 125

150 b) x210 = 240

715125 × x = 725 × 150 x × 715 = 240 × 210

x = 725 × 150125 x = 240 × 210

715x = 870 x = 70.49

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8. a) • We want to determine how many more weeks it will take Garfield to

save this amount.

• ($925 – $150) ÷ $75/week = 10.3 weeks

• Garfield must save for 11 weeks.

b)

C

A

30°

22.5 m

39m

45 m

B

• We want to determine how high above Julianna’s head the kite is

flying.

• In a right triangle, the length of the side opposite the 30° angle is equal

to half that of the hypotenuse.

b = 45 m2 = 22.5 m.

The kite is flying 22.5 m above Julianna’s head.

• We want to determine the horizontal distance which separates

Julianna from the kite.

• The horizontal distance required is calculated by applying the

Pythagorean Theorem.

c2 = a2 + b2

a2 = c2 – b2

a2 = 452 – 22.52

a2 = 2 025 – 506.25

a2 = 1 518.75

a = 1 518.75

a = 38.97

The horizontal distance between Julianna and the kite is 39 m.

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ANALYSIS OF THE DIAGNOSTICTEST RESULTS

QuestionAnswer Review Before Going

Correct Incorrect Section Page to Unit(s)

1. a) 8.2 8.20 1b) 8.2 8.20 1c) 8.2 8.20 1d) 8.2 8.20 1

2. a) 8.2 8.22 1b) 8.2 8.22 1

3. a) 8.2 8.26 1b) 8.2 8.26 1c) 8.2 8.26 1d) 8.2 8.26 1e) 8.2 8.26 1

4. 8.3 8.29 15. a) 8.4 8.32 1

b) 8.4 8.32 16. a) 8.5 8.39 3

b) 8.5 8.39 37. a) 8.6 8.41 4

b) 8.6 8.41 48. a) 8.1 8.4 1, 5 and 6

b) 8.1 8.4 1, 5 and 6

• If all your answers are correct, you may begin working on this module.

• For each incorrect answer, find the related section listed in the Review

column. Do the review activities for that section before beginning the units

listed in the right-hand column under the heading Before Going to Unit(s).

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INFORMATION FOR DISTANCEEDUCATION STUDENTS

You now have the learning material for MTH-4103-1 together with the home-

work assignments. Enclosed with this material is a letter of introduction from

your tutor indicating the various ways in which you can communicate with him

or her (e.g. by letter, telephone) as well as the times when he or she is available.

Your tutor will correct your work and help you with your studies. Do not hesitate

to make use of his or her services if you have any questions.

DEVELOPING EFFECTIVE STUDY HABITS

Distance education is a process which offers considerable flexibility, but which

also requires active involvement on your part. It demands regular study and

sustained effort. Efficient study habits will simplify your task. To ensure

effective and continuous progress in your studies, it is strongly recommended

that you:

• draw up a study timetable that takes your working habits into account and

is compatible with your leisure time and other activities;

• develop a habit of regular and concentrated study.

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The following guidelines concerning the theory, examples, exercises and assign-

ments are designed to help you succeed in this mathematics course.

Theory

To make sure you thoroughly grasp the theoretical concepts:

1. Read the lesson carefully and underline the important points.

2. Memorize the definitions, formulas and procedures used to solve a given

problem, since this will make the lesson much easier to understand.

3. At the end of an assignment, make a note of any points that you do not

understand. Your tutor will then be able to give you pertinent explanations.

4. Try to continue studying even if you run into a particular problem. However,

if a major difficulty hinders your learning, ask for explanations before

sending in your assignment. Contact your tutor, using the procedure

outlined in his or her letter of introduction.

Examples

The examples given throughout the course are an application of the theory you

are studying. They illustrate the steps involved in doing the exercises. Carefully

study the solutions given in the examples and redo them yourself before starting

the exercises.

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Exercises

The exercises in each unit are generally modelled on the examples provided.

Here are a few suggestions to help you complete these exercises.

1. Write up your solutions, using the examples in the unit as models. It is

important not to refer to the answer key found on the coloured pages at the

end of the module until you have completed the exercises.

2. Compare your solutions with those in the answer key only after having done

all the exercises. Careful! Examine the steps in your solution carefully even

if your answers are correct.

3. If you find a mistake in your answer or your solution, review the concepts that

you did not understand, as well as the pertinent examples. Then, redo the

exercise.

4. Make sure you have successfully completed all the exercises in a unit before

moving on to the next one.

Homework Assignments

Module MTH-4103-1 contains three assignments. The first page of each

assignment indicates the units to which the questions refer. The assignments

are designed to evaluate how well you have understood the material studied.

They also provide a means of communicating with your tutor.

When you have understood the material and have successfully done the perti-

nent exercises, do the corresponding assignment immediately. Here are a few

suggestions.

1. Do a rough draft first and then, if necessary, revise your solutions before

submitting a clean copy of your answer.

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2. Copy out your final answers or solutions in the blank spaces of the document

to be sent to your tutor. It is preferable to use a pencil.

3. Include a clear and detailed solution with the answer if the problem involves

several steps.

4. Mail only one homework assignment at a time. After correcting the assign-

ment, your tutor will return it to you.

In the section “Student’s Questions,” write any questions which you may wish

to have answered by your tutor. He or she will give you advice and guide you in

your studies, if necessary.

In this course

Homework Assignment 1 is based on units 1 to 4.

Homework Assignment 2 is based on units 5 and 6.

Homework Assignment 3 is based on units 1 to 6.

CERTIFICATION

When you have completed all the work, and provided you have maintained an

average of at least 60%, you will be eligible to write the examination for this

course.

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UNIT 1

RIGHT TRIANGLES

1.1 SETTING THE CONTEXT

Triangles Everywhere

Young Trigon really likes kites. His father, Mr Trigon, capitalizes on this one day

when they are out flying a kite in order to teach his son some physics concepts.

"There are two forces," he says, "which work together to prevent your kite from

being carried off into the sky by the wind; one comes from the tension which you

apply with your arm on the string which is holding the kite and the other is

caused by the earth's gravitational attraction.* These two forces form a right

triangle with a line which is parallel to the ground (Fig. 1.1)."

* All objects are attracted towards the centre of the earth. This gravitational force is thereforeexerted downwards, perpendicular to the earth's surface.

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AT

37°

Ground

The forces actingon the kite

37°

Fig. 1.1 The physics lesson

Thus we can see the shape of a right triangle each time we project a perpendicu-

lar and an oblique (slanted) line to the ground from the same point above the

ground. Knowing how to measure the sides and the angles in this type of

triangle is very useful in everyday situations for measuring distances or

determining the measures of various angles.

Have you noticed to what extent right angles, which measure 90°, are found

everywhere? A pole carrying electrical wires forms a right angle with the ground;

a staircase which connects one floor to another meets the floor at a right angle,

and the stairs-wall-floor together define a right triangle. Look around you

carefully: you will discover a great many right angles and right triangles.

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A triangle is a polygon with 3 sides and with 3 angles, the sumof whose measures is always equal to 180°. The point ofintersection of two sides is called a vertex.There are 5 types of triangles:1. equilateral triangles, which have 3 congruent angles of 60°

and 3 congruent sides;2. isosceles triangles, which have 2 congruent sides and 2

congruent angles;3. right triangles, which have 1 right angle;4. isosceles right triangles, which have 1 right angle, 2 con-

gruent angles of 45° and 2 congruent sides;5. scalene triangles, which have sides and angles that are not

congruent.

You know that a right triangle is a triangle which has an angle of 90°. It isimportant to remember that the side opposite the right angle is called thehypotenuse.

Since a triangle is the simplest polygon possible, any other polygon can bedecomposed into triangles by drawing appropriate diagonals inside the figure(Fig. 1.2). Also, by dropping a perpendicular from a vertex in each of these 3 types

of triangles (equilateral, isosceles or scalene) we obtain right triangles (Fig. 1.2).

A

C

G

F

E

D

B

A

G

F

A

G G

FH H

H

Fig. 1.2 Decomposition of polygons into right triangles

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The study of various geometric shapes, for example in surveying or in landscap-

ing, usually involves the study of right triangles.

To reach the objective of this unit, you should be able to solve problems

dealing with right triangles where the measures of one acute angle and

of two sides are given. In the case of a right triangle where one of the

acute angles measures 30° or 45°, you should be able to solve these

problems given only the length of one leg (side adjacent the right angle)

in the right triangle.

To solve problems concerning triangles, you must determine the measures of its

6 dimensions, that is, the measures of the 3 sides and the 3 angles.

The hypotenuse is the longest side in a right triangle. It is the side

opposite the triangle's right angle.

An acute angle is an angle whose measure is less than 90°.

The side opposite an acute angle is the side of the triangle located

across from this angle.

The side of the right angle adjacent to an angle is the side in the

triangle which is common to the right angle and to this angle.

c: hypotenuse

a: side opposite angle Aor

side adjacent to angle B

b: side opposite angle Bor

side adjacent to angle A

Fig. 1.3 Triangle, right-angled in C

B

A

C

b

a

c

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Angles are measured in degrees. Hipparchus of Alexandria (2nd century BC) is

credited with having introduced this measurement unit to the Western world.

He divided a circle into 360 parts called degrees. Each of these degrees was

divided into 60 minutes of 60 seconds each. Up to that time, this method of

dividing by 60, called the sexagesimal system, had been used only by the

Babylonians. This measurement unit can be compared to our method of

measuring time; a day is divided into 24 hours, which are in turn divided into

60 minutes, each of which contains 60 seconds. Of course, it is much more precise

to say 2 hours 4 minutes 30 seconds than it is to say about 2 hours. The

sexagesimal system is a very precise system of measurement.

1 degree = 1360 of a circle and is written 1°.

1 minute = 160 of a degree and is written 1'.

1 second = 160 of a minute and is written 1".

Example 1

Angle A measures 22° or 22360 of a circle.

Fig. 1.4 Measuring a portion of a circle

Angle measures can also be expressed in decimal form. An angle measure of

20° 15' can be converted to decimal form as follows:

15' ÷ 60' = 0.25; therefore 20° 15' = 20.25°.

You can also perform the inverse operation. For example, to convert an angle

measure of 65.32° to degrees, minutes and seconds:

A

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• Calculate how many minutes 0.32° represents. 32

100 × 60' = 19.2'

• Keep the 19' and calculate how many seconds 0.2' represents. 2

10 × 60" = 12"

• Therefore, 65.32° = 65° 19' 12".

Now see if you can measure dimensions, using the sexagesimal system, and if you

have learned the vocabulary relevant to right triangles.

Exercise 1.1

1. How many degrees are there in 14 of a circle? ...............................................

2. Convert the following angles, which are expressed in minutes or seconds, to

angles expressed in degrees and minutes.

a) 150' = .................... b) 162 000" = ............. c) 1 830' = ......................

3. Convert the following angle measures to decimal form.

a) 32° 15' = ....................................... b) 53° 32' 24" =...................................

4. Convert the following angle measures to degrees, minutes and seconds.

a) 72.2° = .......................................... b) 18.32° = ..........................................

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5. Using a ruler, measure:

a) the hypotenuse in right triangle ABC:

...............................................................

b) the side opposite the 30° angle: ...................

People have long been interested in measuring the dimensions in a right

triangle. The Babylonians already understood the concept that later made

Pythagoras famous: "If c is the length of the hypotenuse in a right triangle and

if a and b are the lengths of its two legs, then c2 = a2 + b2." This is the famous

Pythagorean Theorem.

In a right triangle, the square of the

length of the hypotenuse is equal to the

sum of the squares of the lengths of the

other two sides. If a = 3, b = 4, c = 5, then:

c2 = a2 + b2

52 = 32 + 42

25 = 9 + 16

25 = 25Fig. 1.5 Illustration of the

Pythagorean Theorem

In any triangle ABC, the length of the segment joining vertex A to vertex B is

symbolized either by mAB or by the lower-case letter of the opposite angle,

namely c.

mAB = mBA = c

mBC = mCB = a

mAC = mCA = b

Fig. 1.6 Lengths of the sides in a right triangle

B

AC

3 5

4

B

C Ab

ac

B

C A

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The measure of an angle, for example, angle A, is symbolized by m∠A, which is

read "the measure of angle A." Protractors are essential in trigonometry; proceed

now with a few short exercises in order to brush up your skills in using this

instrument.

Exercise 1.2

1. Measure the 3 angles in each of the following triangles to the nearest degree,

using a protractor. Then add these measurements.

a) m∠A = ..............................................

m∠B = ..............................................

m∠C = ..............................................

m∠A + m∠B + m∠C = ....................

b) m∠A = ..............................................

m∠2 = ..............................................

m∠C = ..............................................

m∠3 = ..............................................

m∠4 = ..............................................

m∠A + m∠2 + m∠4 = .....................

m∠A + m∠C + (m∠3 + m∠4) = .................................

2. How many minutes are there in 0.6°? .............................................................

C

A B

D

AC B1 2

3 4

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3. Measure each of the sides in these right triangles and verify whether they

satisfy the Pythagorean Theorem. Measurements and calculations should be

given to an accuracy of two decimal places.

a)

F

D

E d

fe

d = .................. e2 = d2 + f2

e = ...................

f = ...................

b)

B

C

b

a

cA

a =................... c2 = a2 + b2

b = ...................

c = ...................

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4. Can the following groups of three numbers be lengths of the sides in a right

triangle? Explain your answer using the Pythagorean Theorem.

a) 17, 144, 145 b) 8, 10, 15

The Pythagorean Theorem allows you to calculate the length of a side in a right

triangle when the lengths of the other 2 sides are known.

Example 2

You travel 25 km east, then turn 90° south and travel a distance of 50 km;

calculate the distance from your departure point A to point B as the crow flies.

Figure 1.7 represents your trip, to a scale of 1 cm =̂ 10 km.

A C

B

N

S

EW

Fig. 1.7 Scale diagram of your trip

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The straight-line distance between points A and B on this trip is the

hypotenuse of a right triangle formed by the dotted line joining the departure

point to the destination point. Thus you can apply the Pythagorean Theorem

to calculate the distance between these two points.

c2 = a2 + b2

c2 = 252 + 502

c2 = 625 + 2 500

c2 = 3 125

c = 3 125

c = 55.901 699...

The straight-line distance is 56 km, to the nearest kilometre.

Now find out whether you have learned the Pythagorean Theorem.

Exercise 1.3

1. A tilting tree is straightened using a guy wire fixed to its trunk 1.5 m above

the ground; the other end of the guy wire is held to the ground by a peg placed

3 m from the trunk.

a) Draw a scale diagram of this situation.

b) Using the Pythagorean Theorem, determine the length of the wire used

to hold the tree. Round your answer to the nearest tenth.

a) b)

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2. A glass door is 3 m high and 1 m wide. Determine the length of its diagonal

to the nearest centimetre.

The hypotenuse is the longest side in a right triangle since it is

equal to the sum of the squares of the other two sides. Thus:

c2 = a2 + b2

c > a

c > b

Also, since the sum of the measures of the interior angles of any

triangle is always equal to 180°, the two non-right angles are

acute angles.

m∠A + m∠B + m∠C = 180°

m∠A + m∠B + 90° = 180°

thus m∠A = 90° – m∠B

m∠B = 90° – m∠A

Fig. 1.8 Calculating angle measurements in a right triangle

When you know the measure of one acute angle in a right triangle, you can derive

the measure of the other acute angle. These two acute angles are said to be

complementary, in other words, the sum of their measures is equal to 90°.

C A

B

180°90°

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Example 3

m∠A = 180° – m∠C – m∠B

m∠A = 90° – m∠B

m∠A = 90° – 39°

m∠A = 51°

Fig. 1.9 Right triangle ABC

Are you able to draw right triangles given the measures of certain sides or certain

angles? The following exercise will show whether you have acquired this ability,

which is essential for further study.

Exercise 1.4

1. a) Sides BC, AC and AB in a right triangle measure 11 cm, 60 cm and 61 cm

respectively. Draw this triangle, using a set-square and a ruler.

Suggested scale: 1 cm =̂ 10 cm.

b) What are the measures of the three angles in the preceding triangle?

m∠A = ............... , m∠B = ............... , m∠C = ...............

A

B C

39°

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2. a) A 6.4 m ladder AB is placed against a wall BC. The foot A of the ladder

is placed 3.2 m from the wall. At what height does the ladder touch the

wall? Using a scale of 1 cm =̂ 1 m, represent the position of the ladder and

measure the angles formed by this right triangle.

m∠A = ............... , m∠B = ............... , m∠C = ...............

3. A kite B is flying 14 m above the ground. The string AB which is attached to

it is 20 m long. By drawing a right triangle representing this situation and

dropping a perpendicular from the kite to the ground, determine the distance

between the point C (where this perpendicular meets the ground) and point

A, the position of the person who is holding the string. Draw the right triangle

that represents this situation.

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4. Two cyclists leave the same point simultaneously. One goes directly south at

10 km/h while the other moves directly east at 14 km/h. What is the distance

between them after 2 12 hours?

N

S

EW

At the beginning of this module you were told that right triangles are found just

about everywhere. Here are two right triangles frequently encountered in

construction.

C

A B30°

Fig. 1.10 Right triangle with an acute angle of 30°

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E

F

D45°

Fig. 1.11 Right triangle with an acute angle of 45°

These right triangles are used so frequently that you will find them in your

geometry sets in the form of set-squares.

Measure the sides and the angles in these triangles carefully.

? m∠A = ............. , m∠B = 30°, m∠C =...................... .

} bc = ..........

? a = ............. , b = .......... , c =..................... .

? m∠D = 45°, m∠E = .......... , m∠F =..................... .

} de = ..........

? d = ............. , e = .......... , f =..................... .

? What can you deduce from looking at these measures?

...........................................................................................................................

...........................................................................................................................

You can verify whether your deductions are correct by doing the following

exercises.

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Exercise 1.5

1. a) Measure the angles in the right triangle in the figure below.

m∠A = ............. , a = ..............

m∠B = 30° , b = ..............

m∠C = ............. , c = ..............

b) What is the ratio of the length of the side opposite the 30° angle to that of

the hypotenuse?

bc =

2. a) Measure the angles and sides in this triangle.

m∠E = ............... e = ...............

m∠F = ............... f = ...............

m∠G = ............... g = ...............

b) Calculate: ge =

3. What connection can you see between the measure of the angle opposite the

30° angle in a right triangle and the measure of the hypotenuse?

...........................................................................................................................

...........................................................................................................................

30°C

A

B

E

F G

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4. a) Measure the angles and sides in this triangle.

m∠A = ............... a = ...............

m∠B = ............... b = ...............

m∠C = ............... c = ...............

b) What connection can you see between the acute angles in this triangle and

the sides opposite these angles?

.......................................................................................................................

.......................................................................................................................

You will now finish this unit by studying two other right triangles whose

properties you will find very useful.

Consider the following equilateral triangle whose sides are 1 unit long.

An equilateral triangle has 3 congruent sides. The 3 congruent

sides are opposite congruent angles each measuring 60°.

a = b = c = 1

m∠A = m∠B = m∠C = 60°

Fig. 1.12 Equilateral triangle ABC

A

CB

B C

A

b = 1

a = 1

c = 1

60°

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From any vertex in equilateral triangle ABC, for example A, drop a perpendicu-

lar to the opposite side. Let this perpendicular be AD. You thereby obtain the

2 right triangles ABD and ADC. These two triangles are congruent under the

A-S-A congruence property.

If two angles and the contained side of one triangle are equal to

two angles and the contained side of another triangle, then the

triangles are congruent.

1. b = c =1;

2. m∠B = 60°, by definition;

3. m∠BAD = 90° – 60° = 30° since acute

angles in a right triangle are comple-

mentary.

Fig. 1.13 Equilateral triangleABC divided into 2 right triangles

Now find the lengths of the sides of right triangle ABD.

1. AB is the hypotenuse of the right triangle and mAB = 1;

2. mBD = 12 , since mBD = mDC = 1

2 mBC and mBC = 1;

3. mAD2 = mAB2 – mBD2 (Pythagorean Theorem)

mAD2 = 12 – 12

2

mAD2 = 1 – 14

mAD2 = 34

mAD = 34 = 3

4

mAD = 32 or 0.866

b = 1

A

30°

B D

60°

h

C

c = 1

60°

30°

Fig. 1.14 Right triangleABD derived from equilateral

triangle ABC

a D

hd

A

30°

B

60°

C

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The lengths of the sides in these two right triangles derived from the decompo-

sition of the equilateral triangle are 1, 12 and 3

2 .

• The side opposite the 30° angle measures 12 .

• The side adjacent to the 30° angle measures 32 .

• The side opposite the 60° angle measures 32 .

• The side adjacent to the 60° angle measures 12 .

Thus, after examining the results obtained by partitioning this equilateral

triangle, a first theorem can be stated about triangles.

Theorem 1

In any right triangle, the length of the side opposite the 30°

angle is equal to half the length of the hypotenuse and the

length of the side adjacent to it is equal to 32 × the length of

the hypotenuse.

Apply this theorem to solve the following problems.

Exercise 1.6

1. When Martin is standing 2.2 m from

the foot of the mast on his sailboat, he

measures a 60° angle from his posi-

tion relative to the top of the mast. 60°

Fig. 1.15 Measures of the sidesof right triangle ABD derivedfrom equilateral triangle ABC

D

30°

B

60°

A

1

12

32

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a) What is the measure of the third angle formed at the top of the mast

relative to Martin's position? .....................................................................

b) What is the length of the hypotenuse in this right triangle?

c) How long is the sailboat's mast?

2. Calculate the length of the third side BC in the triangle below.

Hint: You may decompose the figure into two right triangles.

B C

A

30°

2519

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3. According to safety standards, the slope of a slide must not exceed a 30° angle.

If these standards were applied, what would the height of the ladder E be if

the slide were 50 m long?

S

30°

E

Consider now the isosceles right triangle in Figure 1.16 whose hypotenuse

measures c.

All isosceles triangles have 2 congruent sides.

The two congruent sides are opposite congruent angles.

a = b

m∠A = m∠B = 180° – 90°2 = 45°

Fig. 1.16 Isosceles right triangle ABC

A B

C

ab

c

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Apply the Pythagorean Theorem to this triangle:

a2 + b2 = c2

a2 + a2 = c2

2a2 = c2

a2 = c2

2

a = c2

2

a = c2

a = c2

× 22

a = c 22 or 2

2 c

N.B. In order to make the denominator a rational number, the fraction was

multiplied by 22

= 1 since it is easier to divide by 2 than by 2 , which is equal

to 1.414... .

• The side opposite a 45° angle measures 22 c .

The same procedure can be used to derive the length of the hypotenuse. Thus:

c2 = a2 + b2

c2 = a2 + a2

c2 = 2a2

c = 2a2

c = 2 × a

• The hypotenuse measures 2 × a or 2 × b .

Theorem 2

In any isosceles right triangle, the length of the side opposite

the 45° angle is equal to 22 × the length of the hypotenuse

and the length of the hypotenuse is equal to 2 × length of the

side opposite the 45° angle.

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Here now is a little problem which will allow you to apply this second theorem

about right triangles immediately!

Exercise 1.7

Your house has a gabled roof. The gable

has a 45° angle of inclination. If the

gable is 4 m high, how long is slope

m and base b of the roof?

Thus we can conclude that when we are confronted with a right triangle and

know the length of one of its sides, it is possible to determine the length of its

other two sides if one of the acute angles measures 30° or 45°. Look at the next

example.

Example 4

Derive the missing dimensions in

the adjacent triangle.

Fig. 1.17 Right triangle, one ofwhose acute angles measures 30°

4m

45°

b

30°

Bc = 8A

b

C

a

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1. You are given that m∠B = 30° and m∠C = 90°; thus

m∠A = 90° – 30° = 60°.

2. b is equal to half the length of the hypotenuse, thus

b = 12 c = 1

8 × 8 = 4.

3. a is equal to 32 × the length of the hypotenuse, thus

a = 32 × 8 = 6.93

a = 6.93

N.B. You should always round off the result according to the context of the

problem. For example, if the length of a side is in metres, you generally round

off your result to the nearest centimetre (i.e., to the nearest hundredth). If the

length is in centimetres, you generally round off your result to the nearest

millimetre (i.e., to the nearest thousandth). However, when a great deal of

precision is not required, you can round off your result to the nearest unit.

The same procedure can be applied to a triangle containing an acute angle of 45°.

Now you have a general idea about triangles and are ready to tackle more meaty

problems. If you are uncertain about anything in the following practice

exercises, do not hesitate to review the preceding concepts.

Did you know that...

Egyptians were the first accurate surveyors? The taxes

collected by the pharaoh were set according to the size of the

cultivated land, so lands had to be marked precisely and

areas had to be calculated. It was therefore through surveying that the

Egyptians learned the fundamentals of geometry.

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?1.2 PRACTICE EXERCISES

1. Here is a kite whose shape is given by

quadrilateral ABCD. If the length of

segment AB in the kite is 36 cm,

determine the lengths of the other

sides in this quadrilateral.

N.B. Round your answers to the

nearest cm.

mAE = ...............................................................................................................

mBE = ...............................................................................................................

mBC = ...............................................................................................................

mEC = ...............................................................................................................

2. The angle from the feet of an ob-

server to the top of a tower is 45° and

the distance between the observer

and the base of the tower is 140 m.


nearest tenth metre.

a) How high is this tower? ..............................................................................

b) How far is the eye of the observer from the top of the tower?

.......................................................................................................................

30°

C

A

B45° E 45°

D

36 cm

O B140 m

45°

A

C1,70 m{

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3. A carpenter is sawing rafters for the

roof of a shed which has a 30° angle

of inclination.


nearest tenth of a metre.

a) What is the height AC of this roof?

.......................................................................................................................

b) What is the width BD of the shed?

.......................................................................................................................

.......................................................................................................................

c) How much does CE measure?

.......................................................................................................................

4. A student in surveying wants to

know the width RA of a marsh. To

find it, she goes to a point A opposite

a reference point R located on the

opposite shore of the marsh. She

then backs up 37 m to B, turns 90°

and continues for another 37 m. She

is now at point C, where she mea-

sures angle RCB. It is 60°. Deter-

mine the width of the marsh to the

nearest metre.

mRA = ..............................................

2.8 m

30°

E

DCB

A

37 m BA

37 m

60°

C

R

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30°

1.3 SUMMARY ACTIVITY

1. What has to be done to solve problems concerning triangles?

...........................................................................................................................

...........................................................................................................................

2. A tree casts a 6 m shadow when the

angle of elevation of the sun is 30°.

How high is this tree?

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3. An antenna that is 2.3 m long is

placed 1 m from the edge of the roof

of a building. An observer A is stan-

ding 50 m from the building. Given

this information, can you:

a) Determine the height of this building?

b) Determine how far the observer is from the top of the antenna, that is, the

length of AC, to the nearest tenth of a metre?

A 50 m

1 m

B F

C

45°

2.3 m

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1.4 THE MATH WHIZ PAGE

Ready for a Challenge?

Here is a little challenge for surveying enthusiasts.

Two semi-circles are drawn on the sides of the right angle in a right

triangle.

What is the area of the shaded region?

S

ab

c

S1

S2

Fig. 1.18 Semi-circles drawn on the sides of the right angle in

a right triangle

MTH-4103-1 Trigonometry I - SOFAD · 2014-05-15 · MTH-4103-1 C1-C4 Tige I_La 1 10-10-18 13:57...

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Transcript of MTH-4103-1 Trigonometry I - SOFAD · 2014-05-15 · MTH-4103-1 C1-C4 Tige I_La 1 10-10-18 13:57...