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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IAS 2.4 – Year 12 Mathematics and Statistics – Published by NuLake Ltd New Zealand © Robert Lakeland & Carl Nugent
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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11IAS 2.4 – Trigonometric Relationships
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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IAS 2.4 – Year 12 Mathematics and Statistics – Published by NuLake Ltd New Zealand © Robert Lakeland & Carl Nugent
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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IAS 2.4 – Year 12 Mathematics and Statistics – Published by NuLake Ltd New Zealand © Robert Lakeland & Carl Nugent
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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