AB

C

8.0 m6.0 m

18˚

20˚

Year 12Mathematics

uLake Ltdu a e tduLake LtdInnovative Publisher of Mathematics Texts

Robert Lakeland & Carl NugentTrigonometric Relationships

Contents• AchievementStandard .................................................. 2

• CircularMeasure ....................................................... 3

• ArcLengthandtheAreaofaSector........................................ 7

• AreaofaTriangleandSegment........................................... 12

• SineRule .............................................................. 17

• CosineRule ........................................................... 22

• ApplicationsusingtheSineandCosineRules............................... 28

• PracticalTrigonometryInvolvingBearings................................. 32

• PracticeInternalAssessment1............................................ 36

• PracticeInternalAssessment2............................................ 40

• PracticeInternalAssessment3............................................ 43

• Answers............................................................... 48

• OrderForm............................................................ 55

IAS 2.4

IAS 2.4 – Year 12 Mathematics and Statistics – Published by NuLake Ltd New Zealand © Robert Lakeland & Carl Nugent

2 IAS 2.4 – Trigonometric Relationships

NCEA 2 Internal Achievement Standard 2.4 – Trigonometric RelationshipsThisachievementstandardinvolvesapplyingtrigonometricrelationshipsinsolvingproblems.

◆ ThisachievementstandardisderivedfromLevel7ofTheNewZealandCurriculum,andis relatedtotheachievementobjective: ❖ applytrigonometricrelationships,includingthesineandcosinerules,intwo andthreedimensions.

◆ Applytrigonometricrelationshipsinsolvingproblemsinvolves: ❖ selectingandusingmethods ❖ demonstratingknowledgeoftrigonometricconceptsandterms ❖ communicatingusingappropriaterepresentations.

◆ Relationalthinkinginvolvesoneormoreof: ❖ selectingandcarryingoutalogicalsequenceofsteps ❖ connectingdifferentconceptsorrepresentations ❖ demonstratingunderstandingofconcepts ❖ formingandusingamodel; andalsorelatingfindingstoacontext,orcommunicatingthinkingusingappropriatemathematical statements.

◆ Extendedabstractthinkinginvolvesoneormoreof: ❖ devisingastrategytoinvestigateorsolveaproblem ❖ identifyingrelevantconceptsincontext ❖ developingachainoflogicalreasoning,orproof ❖ formingageneralisation;

andalsousingcorrectmathematicalstatements,orcommunicatingmathematicalinsight.

◆ Problemsaresituationsthatprovideopportunitiestoapplyknowledgeorunderstandingofmathematicalconceptsandmethods.Situationswillbesetinreal-lifeormathematicalcontexts.

◆ Methodsincludeaselectionfromthoserelatedto: ❖ lengthofanarcofacircle ❖ areaofasectorofacircle ❖ sinerule ❖ cosinerule

❖ areaofatriangle.

Achievement Achievement with Merit Achievement with Excellence• Applytrigonometric

relationshipsinsolvingproblems.

• Applytrigonometricrelationships,usingrelationalthinking,insolvingproblems.

• Applytrigonometricrelationships,usingextendedabstractthinking,insolvingproblems.

60˚

50 m

9.4 m44˚39˚

Zincalume 13.5 m

x y

3IAS 2.4 – Trigonometric Relationships

IAS 2.4 – Year 12 Mathematics and Statistics – Published by NuLake Ltd New Zealand © Robert Lakeland & Carl Nugent

Circular Measure

RadiansTheconventiontodivideacircleinto360˚appearstohavebeenusedfromabout140BC.Itprobablybeganfromdividingthe12signsofthezodiacintosmallerparts.Acycleoftheseasonsofapproximately360dayscouldbemadetocorrespondtothe12signsofthezodiac. Asconvenientasthisis,itdoesnotgiveamathematicalbasisfordegrees.

Seniormathematicsusuallydoesnotusedegreesbutdefinesanothermeasureforanglescalledradians. Radiansatfirstappearclumsy(thereare6.2832radiansinacircle)buttheymakethesolvingofmanytrigonometricproblems,andinparticularcalculus,easier.

Aradianisdefinedasthatanglewhosearcisthesamelengthastheradiusofthearc.

Afullcirclehasanarclengthof2πrandaradiusofrsothenumberofradiansinacircleis r

r2r or2π.Thereare2πorapproximately6.2832radiansinacircleso

2πradians ≡360˚

approximately

6.2832radians ≡360˚

1radian ≡ .6 2832360

≡57.3˚ (1dp)

ConversionsDegrees to RadiansWeusetheexactconversion 2πrad. ≡360˚or πrad. ≡180˚toconvertfromdegreestoradiansandviceversa.Toconvertθdegreestoradiansallwehavetofindiswhatfractionof180˚itisandmultiplybyπ.

Angleinradians = 180xi r

Radians to Degrees

Toconvertθradianstodegreeswedividebyπandmultiplyby180.

Angleindegrees = 180xr

i

The symbol ≡ means equivalent to. It is used when two quantities are the same but with different units.

If you are given an angle in radians which includes the symbol π (e.g. 6

π ) it is easier to convert it to degrees by just replacing the π symbol with 180˚ (See the Example on the next page).

You can use the fraction button on your calculator to simplify this fraction quickly.

Anangleofoneradianformsanarclengthequalinlengthtotheradiusofthesector.

θ=oneradian

a) 71.4˚

b) 135˚leavingyouranswerintermsofπ.

Converttheanglesindegreestoradians.

a) θrad. = 180xi r

= 71.4180

xr

=1.25 (2dp)

b) = 180xi r

= 135180

xr

andsimplifythefraction.

= 43r

Degrees to radian conversion is also covered in the Graphical Models Achievement Standard. If you have already done this there is no need to repeat it.

r

r

r

θ

Example

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6 IAS 2.4 – Trigonometric Relationships

Merit–Answerthefollowing.

Acarhasaweakspotononeofitstyres.Itwilleventuallyfailandyouaretoexplorehowmanytimestheweakspotisincontactwiththeroad.

Thecaristravellingatanaveragespeedof90km/handeachwheelis37cminradius.Youwillneedtoconsiderthefollowing:

• thespeedofthecarinm/s

• thedistancetravelledbythecarinonerevolutionofthewheel

• thenumberofrevolutionsofthewheelinonesecond.

ThecarisbeingdrivenfromNapiertoWellingtonwhichwilltake3.75hours.

Howmanytimesdoyouexpecttheweakspottobeincontactwiththeroad?

Makesureyouexplaineachstepofyourworkingwhensolvingthisproblem.

41. 42. Asolidcylinder(radius652mm)hasawedgeof51.0˚cutoutofitsoitsortoflookssimilartoaPac-man.Thecylinderis0.845mindepth.

Thesidesurfacearea(ignoringbothends)istobecoveredinexpensivegoldfoil.Youwillneedto

• findthegreenangleAinradians

• findthearclengthofthegreencircularsector

• thetotaldistancearoundthegreencircularsector

• theareaofthegoldfoil.

Thefoilcosts$75asquaremetre.Whatistheestimatedcostofthefoilrequired?

Makesureyouexplaineachstepofyourworkingwhensolvingthisproblem.

0.845m

51˚

0.652mA©

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IAS 2.4 – Year 12 Mathematics and Statistics – Published by NuLake Ltd New Zealand © Robert Lakeland & Carl Nugent

Excellence–Solvethesemorecomplexproblemsexplainingwhatyouaredoingateachstep.

56. Aconicalhatismadebycuttingasectorfromacircleofcardandjoiningtheedges.Thehathasaslantheightof32cmandtheresultinghathasadiameterof22cm.

Slant height 32 cm

32 cm

22 cm in diameter

Note:Notdrawntoscale.

57. AtasideshowatanA&PShowthereisamodifieddartboardwithfoursectionsshadedgreen.Thesideshowoffersa$100prizeifonedart(costing$2)landsinanyofthefourgreensections.Youwanttofindtheprobabilitythatbychanceadartcouldlandinawinningarea.Thedartboardhasaradiusof312mmandanareaof305800mm2(4sf).

Acloseupofoneofthesectorsgivesyouthefollowingdimensions.Theinnergreensection(ofwhichtwoaregreen)goesfrom132mmto143mmwhiletheoutersection(ofwhichtwoaregreen)goesfrom212mmto223mm.

Youwillneedtoconsider: • theangularmeasureof eachwedge • theareaofthetwoinnerandoutergreen

sections • theratioofthewinningareatothetotal

boardarea indecidingwhetherthisgameoffersagood

randomchanceofwinning.

Findtheangleofthesectorcutfromthecircleandtheareaoftheresultinghat.

132 mm

212 mm

143 mm

223 mm©uLake LtduLake Ltd Innovative Publisher of Mathematics Texts

17IAS 2.4 – Trigonometric Relationships

IAS 2.4 – Year 12 Mathematics and Statistics – Published by NuLake Ltd New Zealand © Robert Lakeland & Carl Nugent

The Sine Rule Formula

sin sin sinAa

Bb

Cc= = = sin sin sinA

aB

bC

c= = = sin sin sinAa

Bb

Cc= =

wherea,bandcarethelengthsofthethreesidesandA,BandCaretheanglesoppositeeachofthecorrespondingsides.

Thesinerulecanbeexpressedintwoforms.Theformbelowisusefulwhenwearerequiredtocalculateanangle.

in in ins s sa

Ab

Bc

C= = = in in ins s sa

Ab

Bc

C= = = in in ins s sa

Ab

Bc

C= =

Onlytwoofthethreepartsoftheseformulaeareeverusedtosolveasingleproblem.

Sine Rule

Deriving the Sine Rule

Toderivethesineruleanaltitudeistemporarilyconstructedtooneofthesides.

FromtriangleBCXwefindthat

sinC= ah

h=asinC

andfromtriangleABXweget

sinA= ch

h=csinA

Therefore

asinC=csinA

or ins Aa = ins C

c

AsimilarapproachisusedtoextendtherelationtosidebandsinB.

Giving ins Bb = ins C

c

Theanglesofanon-right-angledtrianglearelabelledwithuppercaselettersandeachangle’soppositesidewiththecorrespondinglowercaseletter.

Itdoesnotmatterhowwelabeltheothersidesaslongaswehavesidesoppositetheircorrespondingangle.

Youcanuseyourgraphicscalculatorforsineandcosineruleproblemsbyenteringtheprobleminthesolverandgettingthecalculatortosolvefortheunknown.

It is easier if we always represent our unknown with the letter a or if the unknown is an angle, the letter A.

Forexample,tocalculatexinthistrianglewestartbylabellingsidexasa.

WethenlabelitsoppositeangleAandtheknownsidebanditsangleB.

C

a

b

B

A

c

C

a

b

B

A

ch

X

34°

34 mm

41°

x

34°

34 mm

41°

xab

B

A

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27IAS 2.4 – Trigonometric Relationships

IAS 2.4 – Year 12 Mathematics and Statistics – Published by NuLake Ltd New Zealand © Robert Lakeland & Carl Nugent

117. ClintandJennyenjoylongdistancecycling.Onaparticularweekendtheybothheadoffatrightanglestoeachother.Theyagreetostopandringeachotherafterabout75minutes.

Clintcyclesat4.1m/sandafter32.3minutesmustturn81.0˚rightandthenstraightfor43.5minutes.

Jenny(withMollyontheback)cyclesstraightat3.9m/sfor21.6minutesthenturns42.3˚leftandcontinuesfor54.8minutes.

a) HowfarareClintandJennyfromthestartpointwhentheystop?

Excellence–Solvethistrigonometricproblemexplainingwhatyouaredoingateachstep.

Not to scale

81.0° 42.3°

J1

J2

C1

C2

S

D

EL

K

b) Howfararetheyfromeachotherwhentheystop?

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40 IAS 2.4 – Trigonometric Relationships

Practice Internal Assessment Task 2Trigonometric Relationships 2.4

INTRODUCTION TheperimeterofaquadrilateralshapedfitnesstrackataprimaryschoolismarkedoutbythegroundsmanwithfourpegsA,B,CandD.Aplanofthetrackisdrawnbelowwithalltheboundarylengthsandoneanglemarked.Eachschoollevel(junior,middleandsenior)completedifferentpathsaroundthepegs.Theschoolwouldliketocalculatethedistancesforeachsectionandtoalsocalculatetheareaenclosedbythepegssotheycanorderthecorrectamountoflawnfertiliser.

TheYear7and8students(seniors)runaroundABCD(adistanceof496.5m),themiddleschoolaroundABDandthejuniorsrunaroundABC.Workingindependentlyyouarerequiredto: • findthedistancethemiddleandjuniorschoolstudentsruninonelap. • calculatetheareaenclosedbythefourpegsABCD.ThestaffwouldliketoadjustpegBsotheseniorscompletea500mlapbuttheyareunabletochangeangleBADasitisthelimitedbytheboundaryfenceoftheschool. • FindanewpositionforpegBandtheinstructionsthatshouldbegiventothegroundsman.Thequalityofyourreasoning,usingarangeofmethods,andhowwellyoulinkthiscontexttoyoursolutionswilldetermineyouroverallgrade.Clearlycommunicateyourmethodusingappropriatemathematicalstatementsandworking.

Diagram NOT to scale

A

BC

D

183 m

67.5 m

102 m

144 m

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48 IAS 2.4 – Trigonometric Relationships

AnswersPage 41. 1.070(3dp) 2. 0.785(3dp)3. 3.547(3dp)4. 6.283(3dp)5. 0.838 (3dp)6. 4.294 (3dp)7. 10.036(3dp)8. 18.047(3dp)

9. 2r

10. r

11. 6r

12. 12r

Page 5

13. 4r

14. 76r

15. 3r

16. 512r

17. 103π

18. 43π

19. 926π

20.14π3

21. 114.6˚ (1dp)22. 31.0˚ (1dp)23. 120.0˚(1dp)24. –71.0˚ (1dp)25. 235.0˚(1dp)26. 345.0˚(1dp)27. 470.0˚(1dp)28. 896.7˚(1dp)29. 180˚30. 45˚31. 90˚32. 60˚33. 30˚34. 225˚

Page 5 cont...35. 240˚36. 270˚37. 540˚38. 210˚39. 80˚40. 162˚Page 641. Thespeedofthecarinm/s Speed =90000/3600 =25m/s Circumferenceofwheel C =2π x 0.37 =2.3248m Revolutionspersecond Revs =25/2.3248 =10.754revs/s TimetoWellington Time =3.75 x 3600 =13500s RevolutionstoWellington Revs =13500 x 10.754 =145200 (4sf) Theweakspotwilltouchthe

roadabout145200times(4sf).42. Weneedtheangleinradiansto

calculatethearclength.Anglecutout

Angleout =0.890 AngleA =2π–0.890 =5.393radians Arclengthofgreen Arc =5.393 x 0.652 =3.5163m(5sf) Perimeter+wedge Dist. =3.5163+2 x 0.652 =4.8203m(5sf) Areaofgold Area =4.8203 x 0.845 =4.0731m2(5sf) Costoffoil Cost =$305.49 (2dp)

Page 843. 5.97m(3sf)44. 1.235radians(3dp)

70.8˚(1dp)

Page 945. a=6.3m(2sf)46. 1.47radiansor84.0˚(3sf)

Page 9 cont...47. Arc= 5.52m(3sf)

Sector=6.84m2(3sf)48. Arc=153cm(3sf)

Sector=1940cm2(3sf)49. Perimeter= 545m(3sf)50. r=9.80m(2sf)51. 72000mm2(3sf)52. Arc=3.29m(3sf) Area=2.39m2(3sf)Page 1053. a) 2.71radians b) 0.61m2(2sf)54. Angle=4.5radians

or =256.9˚Area=6800mm2(2sf)

55. a) 1300m2(2sf) b) 152m c) 105m

Page 1156. Letr=radiushat,

R=radiuscard. θR =2πr 32θ =2π11

θ = 1611 π(2.16)

cutout = 1621 π(4.12)

Area =1100cm2

57. Angleofonesector. Angle =0.31416rad. Areain =0.5π(1432–1322) =475mm2(0dp) Areaout =0.5π(2232–2122) =752mm2(0dp) Totalwinningarea =2(475+752) =2454mm2

Ratiowintototal =305800:2454 =125:1 Thereforegivenoddsof

$100:$2or50:1arenotgoodparticularlyasyoumaymisstheentireboard.

Page 1458. Area=32m2(2sf)59. Area=100cm2(2sf)60. Area=75.9m2(1dp)61. Area=427cm2(3sf)62. Area=26.8m2(3sf)

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