18 Trigonometry
18AThebasictrigonometricratios
Inthischapterweareconcernedwithright-angledtriangles.Asecond angleinsuchatriangleisoftendenotedbytheGreekletter theta, denotedbythesymbol θ.Thiscanindicateboththenameandthe measureoftheangle.The hypotenuse ofaright-angledtrianglealways referstothesidethatisoppositetherightangle.Theothertwosidesare namedaccordingtotheirpositionsrelativetoangle θ.The opposite side isoppositeangle θ,whilethe adjacentside isnexttoangle θ.
Roughlyspeaking,twotriangleswiththesameangleswillalsohavethesameshape.Bythis,wemeanthat thetwotriangleswillbe similar.Inparticular,theratiosoftheircorrespondingsidesareequal.Weprove thisbelowinthespecialcasethatourtwotrianglesareright-angled.
Theorem1(Thetrigonometricratios)
Iftworight-angledtriangleshaveequalangles,thentheratiosoftheircorrespondingsidesareequal.
Proof
Thetwotrianglesbelowhaveequalangles.Theircorrespondingsideshavebeenlabelledwiththesame letters,onelowercase,theotheruppercase.Wewillprovethat
Taketwocopiesofthefirsttriangleandtwocopiesofthesecond,andarrangetheseintotherectangle shownbelow.Theareabelowthediagonalisthesameastheareaabovethediagonal.Therefore,we caneliminateonecopyofeachtrianglefromaboveandbelowthediagonal,andtheareathatremains abovethediagonalwillequaltheareathatremainsbelowthediagonal.Therefore, aO = oA.Itfollows that o a = O A
Intheexercisesyouwillalsobeaskedtoestablishthat a h = A H and o h = O H
Wenowknowthatforanyright-angledtrianglewithgivenangle θ,theratios O H and A H and O A arefixed. Thesefixedratiosaregiventhefollowingnames.
sin θ = O H The sine of θ equalsoppositeoverhypotenuse.
cos θ = A H The cosine of θ equalsadjacentoverhypotenuse.
tan θ = O A The tangent of θ equalsoppositeoveradjacent.
TheabovethreeratioscanbeeasilyrecalledusingthemnemonicSOHCAHTOA,whoseletterscanbearranged suggestivelyintotrianglesasshown:
Example1
Findsin θ,cos θ andtan θ forthetrianglebelow.
Solution
WefirstusethePythagoreanTheoremtoestablishthelengthofthesideoppositeto θ.Letusdenote thissidelengthby x.Wehave
Drawaright-angledtrianglewithangle θ forwhichtan θ = 2 3 .Findsin θ.
Solution
Sincetan θ = O A = 2 3 ,theright-angledtrianglemustbesimilartothetriangleshownbelow.Weusethe PythagoreanTheoremtoestablishthelengthofthehypotenuse.
18AExercises
1.Findsin θ,cos θ andtan θ foreachofthe right-angledtrianglesshownbelow:
4.Drawaright-angledtrianglewithangle θ for whichsin θ = 5 13 .Find cos θ
5.Drawaright-angledtrianglewithangle θ for whichsin θ = 12 13 .Find tan θ
6.Drawaright-angledtrianglewithangle θ for whichcos θ = 1 2 .Find sin θ.
7.Drawaright-angledtrianglewithangle θ for whichcos θ = 2 3 .Find tan θ
8.Consideraright-angledtriangle.
(a)Provethatthehypotenuseisalwaysthe longestside.
(b)Supposethatthetrianglealsocontainsan angle θ.Provethat0 < cos θ < 1and0 < sin θ < 1.
9.Thetworight-trianglesbelowhaveequalangles. InTheorem1weprovedthat o a = O A .Using this fact,additionallyprovethat o h = O H and a h = A H .
2.Drawaright-angledtrianglewithangle θ for whichtan θ = 3 4 .Find sin θ
3.Drawaright-angledtrianglewithangle θ for whichtan θ = 4 3 .Find cos θ.
10.Thesquarebelowhasasidelengthof1unit,withdiagonalanglesincreasinginincrementsof5○.Use thegridtomeasuretheoppositeandadjacentsidesofthegiventriangles,thenapplythePythagorean Theoremtodetermineeachhypotenuse.Estimatethetrigonometricratiosforthelistedanglesandrecord theminthetable,roundedtotwodecimalplaces.Notethat △XAB canbeusedtoapproximatesin5○ , cos5○,tan5○ aswellassin85○,cos85○,andtan85○ .
11.Namethetrianglefromthetableabovethatissimilartoeachofthetrianglesshownbelow.Thenusethe appropriatetrigonometricratiofromthetabletoestimatethevalueof x.
18BFindinganunknownlength
Trigonometricratioswerehistoricallyfoundusingtables,liketheoneyougeneratedearlier.Amoreaccurate andcomprehensivetableisfoundonthefollowingpage.Usethistableifyou’dpreferatraditionalapproach. Theuseofcalculatorsisnowmuchmorecommon(althoughnotalltrendsareworthfollowing).All approximateanswersinthischapterwillbecalculatedtotwodecimalplaces.
Example1
Find x ineachofthefollowingfigures.
Atableoftrigonometricratios
00.00001.00000.0000450.70710.70711.0000 10.01750.99980.0175460.71930.69471.0355
20.03490.99940.0349470.73140.68201.0724
30.05230.99860.0524480.74310.66911.1106 40.06980.99760.0699490.75470.65611.1504
50.08720.99620.0875500.76600.64281.1918
60.10450.99450.1051510.77710.62931.2349
70.12190.99250.1228520.78800.61571.2799
80.13920.99030.1405530.79860.60181.3269
90.15640.98770.1584540.80900.58781.3764 100.17360.98480.1763550.81920.57361.4281
110.19080.98160.1944560.82900.55921.4826 120.20790.97810.2126570.83870.54461.5399 130.22500.97440.2309580.84800.52991.6003 140.24190.97030.2493590.85720.51501.6643 150.25880.96590.2679600.86600.50001.7321 160.27560.96130.2867610.87460.48481.8040 170.29240.95630.3057620.88290.46951.8807 180.30900.95110.3249630.89100.45401.9626 190.32560.94550.3443640.89880.43842.0503 200.34200.93970.3640650.90630.42262.1445 210.35840.93360.3839660.91350.40672.2460 220.37460.92720.4040670.92050.39072.3559 230.39070.92050.4245680.92720.37462.4751 240.40670.91350.4452690.93360.35842.6051 250.42260.90630.4663700.93970.34202.7475 260.43840.89880.4877710.94550.32562.9042 270.45400.89100.5095720.95110.30903.0777 280.46950.88290.5317730.95630.29243.2709 290.48480.87460.5543740.96130.27563.4874
300.50000.86600.5774750.96590.25883.7321
310.51500.85720.6009760.97030.24194.0108 320.52990.84800.6249770.97440.22504.3315
330.54460.83870.6494780.97810.20794.7046
340.55920.82900.6745790.98160.19085.1446
350.57360.81920.7002800.98480.17365.6713
360.58780.80900.7265810.98770.15646.3138
370.60180.79860.7536820.99030.13927.1154
380.61570.78800.7813830.99250.12198.1443
390.62930.77710.8098840.99450.10459.5144
400.64280.76600.8391850.99620.087211.4301
410.65610.75470.8693860.99760.069814.3007 420.66910.74310.9004870.99860.052319.0811
430.68200.73140.9325880.99940.034928.6363
440.69470.71930.9657890.99980.017557.2899 901.00000.0000undef.
Example2
Aflagpoleisstabilisedbytwocablesaffixedtothetopofthepole.Thefirstofthesemakesa40○ degree angletothegroundandis5.5metreslong.Thesecondmakesanangleof50○ totheground.Howlong isthesecondcable?
Solution
Let y betheheightoftheflagpoleandlet x bethelengthofthesecondcable.
Usingthesineratioforbothtriangles,wefindthat
Equating (1) and (2) gives
Thesecondcableisapproximately4.62metreslong.
Note. Itisoftenmostefficienttoalgebraicallysolvefortheunknownbeforedoinganyintermediate calculations.Thissavestimeandalsoavoidstheaccumulationofroundingerrors.
Example3
Aropehangsfromthetopofaflagpole.Theropeis2metreslongerthanthepole,andwhenitis pulledtightandheldtotheground,theropemakesanangleof64○ tothehorizontal.Howtallisthe flagpole,correcttothenearestcentimetre?
Solution
Let h mbetheheightoftheflagpole.Then (h + 2) misthelengthofthe rope.Therefore,
θ = h h + 2 (h + 2) sin θ = h
sin θ + 2sin θ = h
h sin θ = 2 sin θ
(1 sin θ) = 2sin θ
Theflagpoleis17 76metrestall.
Example4
Tofindtheheightofamountaintop,asurveyortakesmeasurementsfromtwoaccessiblepoints A and B asshownbelow.Findtheheight h mofthemountain,correcttothenearestmetre.
Solution
Thetwounknowns d and h mustbesolvedusingsimultaneousequations.From∆ACD weseethat
From∆BCD weseethat
Equating (1) and (2) gives
Therefore,
Theheightofthemountainisapproximately273m.
Example5
Aright-angledtrianglehasoneangleof35○ andaperimeterof8.Findthelengthofthehypotenuse.
Solution
Let a,b and c bethesidesofthetriangle,with c thehypotenuse. Wehave
Astheperimeteris8,wefindthat
Thelengthofthehypotenuseis3 34.
18BExercises
1.Find x in eachofthefollowingfigures.
3.Atapoint5 5metresfromthebaseofaflagpole, theangleofelevationtothetopoftheflagpole is40○.Determinetheheight, x metres,ofthe flagpole.
6.Aflagpoleisstabilisedbytwocablesaffixedto thetopofthepole.Oneofthesemakesa38○ degreeangletothegroundandis5 5metreslong. Thesecondcablemakesanangleof48○ tothe ground.Howlongisthesecondcable?
4.Whenfellingatreethatisclosetoahouse,it’s wisetofirstdetermineitsheight.Atapoint 12metresfromthebaseofatree,theangleof elevationtothetopmostpointonthetreeis37○ . Howtallisthetree?
7.Findtheareaofeachofthetrianglesshown below:
5.Find x ineachofthefollowingfigures:
8.Aright-angledtrianglehasoneangleof40○ andaperimeterof10.Findthelengthofthe hypotenuse.
9.Fromthetopofabuildingtheangleofdepression tothetopofaflagpoleis65○.Fromthebaseof thebuilding,theangleofelevationtothetopof theflagpoleis25○.Ifthebuildingis15metres tall,thendeterminetheheightoftheflagpole.
10.Aropehangsfromthetopofaflagpole.The ropeis3metreslongerthanthepole,andwhen itispulledtightandheldtotheground,therope makesanangleof50○ tothehorizontal.Howtall istheflagpole?
11.Tofindtheheightofamountaintop,asurveyor takesmeasurementsfromtwoaccessiblepoints A and B,asshownbelow.Findtheheight h of themountain,tothenearestmetre.
13† RadiusoftheMoon. Let r denotetheradius oftheMoon, θ theanglesubtendedbytheMoon whenitisdirectlyoverhead,and d thedistance fromthesurfaceoftheMoontothesurfaceof Earth.
12† . Aright-angledtrianglehasanangle θ and perimeter P
Findaformulaforthelengthofits hypotenuseintermsof θ and P . (a)
Findaformulafortheareaofthetrianglein termsof θ and P (b)
18CFindinganunknownangle
Giventhat θ = 0 52○ and d = 384000 kilometres,find r,tothenearest10 kilometres.
Considerthetriangleshown.Wearegiventhelengthsoftheadjacent andoppositesides,andsotan θ = 2 5 = 0 4.Ifwewanttofindthe anglethatgivesthisratio,thenwecanfindtheentryinthetable oftrigonometricratiosclosesttothisvalue.
Fromthiswecanseethat θ ≈ 22○.Thesedays,calculatorsaremorecommonlyusedtofindtheseangles.In whatfollowswewillwritetan 1(0.4040) = 22○ toexpressthefactthattan22○ = 0.4040,withsimilarnotation forsinandtan.Inlaterchaptersthischoiceofnotationwillbeplacedinabroadercontext.
Example1
Find θ ineachofthefollowingfigures.
Example2
Findtheangle θ markedinthesquaregridshownbelow.
Solution
Therequiredangle θ isthedifferencebetweenthetwoangles α and β markedonthediagrambelow.Wefindthat
1.Find θ ineachofthefollowingfigures.
2.Find θ ineachofthefollowingfigures:
3.Aflagpoleisstabilisedbytwocablesaffixedto thetopofthepole.Thefirstofthesemakesa40○ degreeangletothegroundandis4 9metreslong. Thesecondcableis3.8metreslong.Determine theangle θ thatthesecondcablemakeswiththe ground.
Thecirclesshownbelowaretangent,and DE istangenttobothcircles.Thelinesegment AC passesthroughthecentresofbothcircles.Find
4.Findtheangle θ markedineachofthesquare gridsbelow.
Explainhowthediagrambelowshowsthat
Explainhowthediagrambelowshowsthat
8† Inanidealisedgameofbilliards,whenaball strikestheedgeofatable,the angleof incidence and angleofreflection areequal.
Considerthebilliardtableshownbelowwhere theballisinitiallylocatedinthelowerleft corner. 2 4 x θ
(a)Theballisstruckatanangleof θ = 40○ to thehorizontal.Findthedistance x fromthe ball’sstartingpositiontothelocationofthe thirdbounce.
(b)Theballisstrucksothat x = 2.Findthe strikeangle θ.
9† . Agoatistiedbyaropetoacornerofagrassy fieldofwidth30metresandlength40metres.
(a)Whatlengthofropewouldensurethatthe goatcanaccessallofthegrassinthefield?
(b)Whatpercentageofthefield’sareacanthe goataccessifthelengthoftheropeis30 metres?
(c)Whatpercentageofthefield’sareacanthe goataccessifthelengthoftheropeis35 metres?
(d)Whatpercentageofthefield’sareacanthe goataccessifthelengthoftheropeis45 meters?
10† Acylindricalglassofheight12cmanddiameter 8cmisfilledtothebrimwithwater.Theglass iscarefullytiltedsothatitssidemakesanangle of θ tothehorizontal.Theglassisthencarefully restoredtoanuprightposition.
(a)Given θ = 50○,whatisthenewheightof waterintheglass?
(b)Giventhenewheightofwaterintheglassis halftheoriginalheight,find θ
18DExactvalues
Upuntilnowwehaveeitherusedatableofvaluesoracalculatortofindtheapproximateanglesandside lengthsofright-angledtriangles.However,therearecertainspecialangles θ forwhichwecanfind exact values ofsin θ,cos θ andtan θ.Thesevaluesareobtainedfromtwospecialtriangleswhoseanglesandsidelengths canbeeasilyfoundusingthePythagoreanTheorem.
The45-45-90Triangle
Cutasquareofsidelength1alongoneofitsdiagonalstogivethe45-45-90triangleshownbelow.
Thetriangleestablishesthefollowingexactvalues:
The30-60-90Triangle
Cutanequilateraltriangleofsidelength2alongoneofitsaltitudestogivethe30-60-90triangleshownbelow.
Thetriangleestablishesthefollowingexactvalues:
Find x ineachofthefollowingfigures:
Asummaryofexactvalues
Thefollowingtablesummarisestheexactvaluesthatyouarerequiredtocommittomemory.
18DExercises
3.Atrianglewithperimeter10hasangles30○ , 60○ and90○.Findtheareaofthetriangle.
4.Considerthefigureshownbelow.
(a)Provethat α = θ 2
(b)Hence,provethattan
(c)Usetheaboveformulatofindtheexactvalue oftan22 5○
(d)Usethistodeterminetheareaofaregular octagonofsidelength1.
5† Considerrectangle ACDE shownbelow.
(a)Copythediagram,andthendetermineallof themissingangles.
(b)Findlengths AB and DE.
(c)Hence,findlengths BC and CD
(d)Hence,findtheexactvaluesofsin15○ , cos15○,sin75○ andcos75○ .
(e)Apieceofstringoflength1metreistightly wrappedaroundaright-angledtrianglewith angles15○,75○ and90○.Findtheexactlength ofthetriangle’shypotenuse.
(f)Hence,findtheexactareaofaregular dodecagonofsidelength1.
6† Theregularpentagon ABCDE hassidesof length1.Thediagonals AD, BD and AC have beendrawn,anddiagonals AC and BD meetat P
(a)Copythediagram,andthendetermineallof themissingangles.
(b)Showthattriangles∆ADC and∆DCP are similar.
(c)Hence,findthelengthofthediagonal AD.
(d)Findtheexactvaluesofsin18○,cos18○ , sin72○ andcos72○
(e)Findtheexactvaluesofsin36○,cos36○ , sin54○ andcos54○ .
7† . Considerthefiguresshownbelow.Thesedepict anequilateraltriangle,asquare,aregular pentagonandaregularhexagon,andeachis inscribedinacircleofradius1.
(a)Thevertices A,B and C ofanequilateral trianglelieonacircleofradius1.Showthat
AB × AC = 3
(b)Thevertices A,B,C and D ofasquarelieon acircleofradius1.Showthat
AB × AC × AD = 4
(c)Thevertices A,B,C,D and E ofaregular pentagonlieonacircleofradius1.Show that
AB × AC × AD × AE = 5
Hint:Youmayrefertoexercise6.
(d)Thevertices A,B,C,D,E and F ofaregular hexagonlieonacircleofradius1.Showthat
AB × AC × AD × AE × AF = 6.
Note. Theresultsobtainedinthisexercise canbegeneralisedtoaregular n-gon,butwe willnotdothishere.
18EAreaofatriangle
Wewilloftenrefertothethreeanglesinatrianglebytheletters A, B and C.Oppositetheseanglesaresides oflength a, b and c,respectively.Thatis,welabelanangleandthesideoppositetothatangleusingthe sameletter.
Findingtheareaofatriangle
Wewillnowestablishtheareaofatrianglegiventwosidelengths a,b andtheincludedangle C.Weconsider threecases:theincludedanglemaybeacute,obtuseorright.
Case1.Wefirstsuppose C isacute.Drawanaltitudeoflength h fromthevertexat B.Then h = a sin C Therefore,theareaofthetriangleis
Case2.Nowifangle C isobtuse,thenwecanstillfindtheheightofthetrianglebyfindingtheexterior angle:180○ C.Therefore,theareaisgivenby
Thisformulacanbemadetolookexactlylikethepreviousformulaifwe define thesineratioforobtuse angles.Werequirethat
sin(180○ θ) = sin θ.
Case3. Suppose C isarightangle.Ifwedefinesin90○ = 1,thenthesameformulastillworks.Hence,wehave thefollowinggeneralresult.
Theorem1(Theareaofatriangle)
Theareaof∆ABC is
Findtheareaofeachofthesetriangles:
Theareaofthetriangleis
Example2
In∆ABC shown,
∶ DB = 1 ∶ 1
∶ EC = 1 ∶ 3
∶ FA = 1 ∶ 2
Ifarea(∆DEF ) = 5,thenfindarea(∆ABC)
Solution
Welet AF = x, AD = y, AF = z asshownbelow.
Wethenfindtheareaoftriangle ABC threedifferentways:
Similarly,wecanfindthatarea(∆ABC) = 6 area∆(BED) andarea(∆ABC) = 2 area∆(CEF ).The areaof∆ABC canbewrittenasthesumofareasoffourtriangles.Thatis,
1.Findtheareaofeachofthesetriangles:
2.Findtheareaofanequilateraltriangleofside length a.
3.For∆ABC wearegiventhat a = 2, b = 3√2and ∠C = 45○.Findtheareaofthetriangle.
4.For∆ABC wearegiventhat a = 6and b = 5. Giventhattheareaofthetriangleis10,find sin C
5.Findalltriangleswitharea1andtwosidesof length2.
6.Findalltriangleswitharea √2andtwosidesof length2.
7.For∆ABC wearegiventhat a = 6, ∠A = 30○ and ∠B = 45○.Findtheareaofthetriangle.
8.Considerthetriangleshownbelow.
A B
(a)Theformulaarea(∆ABC) = 1 2 ab sin C impliestheareaofthetriangleaboveis 1 2 sin(2θ.Splitthetriangleinhalfandobtain analternativeformulaforitsareainterms of θ
(b)Hence,showthat sin2θ = 2sin θ cos θ.
9.Twosidelengthsofatriangleare a and b.
18FThesinerule
(a)Whatisthemaximumpossibleareaofthe triangle?
(b)Whattypeoftrianglegivesthismaximum area?
10† For∆ABC shownbelow,
Giventhatarea(∆DEF ) = 5,findarea(∆ABC)
The sinerule,or lawofsines,statesthattheratioofeachsidetothesineofitscorrespondingangleisa constant.Giventhemeasuresoftwoanglesandonesideofatriangle,thesineruleallowsustodeterminethe lengthoftheothertwosides.
Theorem1(Thesinerule)
If∆ABC hassidesoflength a, b and c,then
Proof
Weonlyprovethefirstequality,asthesecondfollowssimilarly.
Case1.Firstsuppose A isacute.Drawanaltitudeoflength h fromthevertexat C.Thisaltitudelies insidethetriangle.
Then h canbefoundtwodifferentways:
Weequatethesetogive b sin A = a sin B,whichisequivalenttotherequiredexpression.
Case2.Nowsuppose A isobtuse.Thealtitudeliesoutside∆ABC
Weagainfind h twodifferentways.Notingthatsin(180○ A) = sin A wefindthat
Weequatetheseagaintogive
Example1
Find x inthetriangleshownbelow.
Usingthesinerulewefindthat
Example2
Findtheareaof∆
Notethat ∠B = 80○.Wethenfind b usingthesinerule:
Theareaofthetrianglecannowbefound:
WenowshowthatthesineruleallowsustoefficientlysolvetheproblemgiveninExample4inSection18B.
Example3
Tofindtheheightofamountaintop,asurveyortakesmeasurementsfromtwoaccessiblepoints A and B asshownbelow.Findtheheight h ofthemountain,correcttothenearestmetre.
Solution
Let α and β bethemeasureoftheanglesindicatedinthediagrambelow.
Wefindthat
Wethenfindlength BD = x usingthesinerule:
Finally,wefindthat
Theheightofthemountainisapproximately273m.
Example4
Consider∆ABC below.Findthemeasureof ∠C,correcttothenearestdegree.
Solution
Whenfindinganunknownangle,ithelpstoinvertthesinerule.Wefindthat
Theambiguouscase
Usingthesineruletofindanunknownanglecangiverisetothe possibilityoftwosolutions.Thiscanoccurwhenwearegiventwo sidelengthsandanon-included,acuteangle.Thatis,whenthe givenangle ∠A isacuteandcorrespondstotheshorterofthetwo givensidelengths,so a < c,andwealsohavethat h < a.Inthis case,twopossibletrianglesexist:∆
Example5
Showthattherearetwopossibletriangles∆ABC forwhich a = 5cm, c = 8cmand ∠A = 35○.Findthe twopossiblemeasuresof ∠C,correcttothenearestdegree.
Solution
Wearegiventwosidelengthsandanon-includedangle.We canfindtwosolutionstothisproblem.Bythesinerule,
18FExercises
1.Find x ineachofthetrianglesshownbelow, correcttotwodecimalplaces:
7.For∆ABC wearegiventhat c = 6, a = 4and ∠A = 25○
(a)Drawonecleardiagramthatillustrateshow theinformationabovedescribestwopossible triangles.
(b)Find ∠C inbothtriangles,correcttothe nearestdegree.
8.Showthattherearetwopossibletriangles
∆ABC forwhich a = 6, c = 9and ∠A = 30○ Findthetwopossiblemeasuresof ∠C,correct tothenearestdegree.
2.Find θ ineachofthetrianglesshownbelow, correcttothenearestdegree.
3.Findtheareaofeachofthetrianglesshown below,correcttotwodecimalplaces:
9.Showthattherearetwopossibletriangles
∆ABC forwhich a = 5, c = 10and ∠A = 20○ . Findthetwopossiblemeasuresof ∠C,correct tothenearestdegree.
10.Tofindtheheightofamountaintop,asurveyor takesmeasurementsfromtwoaccessiblepoints A and B asshownbelow.Findtheheight h of themountain,correcttothenearestmetre.
4.For∆ABC wearegiventhat a = 1, ∠A = 30○ and ∠B = 45○.Find b.
5.For∆ABC wearegiventhat ∠A = 43○ , ∠B = 72○ and a = 5 4.Find b and c,correcttoone decimalplace.
6.For∆ABC wearegiventhatsin A = 3 4 ,sin B = 2 3 and a = 3.Find b
11.Averticaltower AB islocatedonahillthatis inclinedat15○.FrompointC,85metresdownhill fromthebase A ofthetower, ∠BAC ismeasured andfoundtobe25○.Findtheheight h ofthe tower,correcttothenearestmetre.
12.Toestablishtheheightofahotairballoonabove levelground,twoobserversmeasuretheangleof inclinationtotheballoonatpoints A and B, located150mapart.Findtheheight h ofthe hotairballoon,correcttothenearestmetre.
13.Showthatonecannotconstructatriangle ∆ABC forwhich ∠A = 45○ , a = 2cmand c = 4cm.
14.For∆ABC wearegiventhat c = 6and ∠A = 25○ Findthevalue(s)of a forwhichthereisonlyone suchtriangle.
15.Findallpossibleareasofatrianglewithtwosides oflength10and11,andoneangleof60○,correct totwodecimalplaces.
16.Inthisexercisewe’llgiveanalternativeproof ofthesinerule.Usetheformulafortheareaof atriangleintroducedintheprevioussectionto provethat
18† The extendedsinerule establishesaconnectionbetweenthesides,anglesandcircumradius R ofatriangle. R a b c B
(a)Assumingthesinerule,provethat a sin A = b sin B = c sin C = 2R
(b)Hence,showthat area(∆ABC) = abc 4R
(c)If r istheinradiusof∆ABC,thenshowthat area(∆ABC) = (a + b + c)r 2
(d)Hence,provethat rR = abc 2(a + b + c) .
19† Showthatinany∆ABC, a (sin B sin C) + b (sin C sin A) + c (sin A sin B) = 0
17.(a)Suppose ∠A, ∠B ≤ 90○.Withtheaidofa diagramexplainwhy ∠A ≤ ∠B ⇐⇒ sin(A) ≤ sin(B)
(b)Hence,provethat ∠A ≤ ∠B ⇐⇒ sin(A) ≤ sin(B) whenever ∠A and ∠B are(possiblyobtuse) anglesinatriangle.
(c)Given∆ABC,provethat a ≤ b ≤ c ifand onlyif ∠A ≤ ∠B ≤ ∠C.
20† Showthatinany∆ABC, (b c) sin A + (c a) sin B + (a b) sin C = 0.
21† . Findaformulafortheareaofthetrianglein termsofitsthreeangles A, B and C,andits perimeter P .Hint:startbywriting P = a + b + c, andthenusethesinerule.
18GThecosinerule
Supposewearegiventwosidelengthsoftriangle, a and b,andangle C betweenthem.Asillustratedbelow, wearenotfreetochoosethelengthoftheremainingside.Therefore,weshouldbeabletofindarelationship betweenunknownsidelength c andsides a,b andangle C.Thisrelationshipiscalledthe cosinerule
Theorem1(Thecosinerule)
If∆ABC hassidesoflength
Weneedtoconsiderconsiderthreecases.Theangleat C willeitherbeacute,obtuseorright.
Case1.Suppose C isacute.Drawanaltitude BD oflength h.Thealtitudemustliewithinthetriangle. ApplyingthePythagoreanTheoremtobothtriangles BCD and ABD gives
Weequatethesetogive
.Therefore,
Case2.Nowsuppose C isobtuse.Thealtitude BD liesoutside∆ABC.ApplyingthePythagoreanTheorem tobothtriangles BCD and ABD gives
Asbefore,weequatethesetogive
Thisformulacanbemadetolookexactlylikethepreviousformulaifwe define thecosineratioforobtuse angles.Werequirethat cos(180○ θ) = cos θ.
Case3. Suppose C isarightangle.Ifwedefinecos90○ = 0,thenthesameformulastillworks.Infact,it reducestotheveryfamiliarformula: c2 = a2 + b2 .
Example1
Findtheunknownsidelengthofeachofthetrianglesshownbelow.
Let x cmbetheunknownsidelength.Then
x 2 = 42 + 52 2(4)(5) cos88○ = 41 40cos88○ x = √41 40cos 88○ ≈ 6 29
Thethirdsidelengthisapproximately6 29cm. (a)
Thethird sidelengthis √52 + 24√2cm. (b)
Let x cmbetheunknownsidelength.Then x 2 = 42 + 62 2(4)(6) cos135○ = 16 + 36 + 48cos45○
52 + 24√2 x = √52 + 24√2
Example2
For∆ABC wearegiventhat ∠C = 30○ , a = √3and c = 1. Find b. Solution
We findthat
Therefore, b = 1or b = 2.
Ifweknowallthreesidelengthsofatriangle,thenweshouldbeabletodetermineeachofitsangles.Thisis duetothefactthatthreesideswilluniquelydeterminethetriangle,uptocongruence.Thecosineruleallows ustodojustthat:
Example3
Find ∠C ofthetriangleshownbelow.
Usingthecosinerulewefindthat
1.Find x ineachofthetrianglesshownbelow,correcttotwodecimalplaces.
2.Find θ ineachofthetrianglesshownbelow, correcttothenearestdegree.
θ 8cm 5cm (a)
(b)
(c)
(d)
3.Consider∆ABC shownbelow.
(a)Findthelengthsofsides BC and CA
(b)Usingthecosinerule,findtheexactvalueof cos75○ .
4.(a)Consider∆ABC shownbelow.
Usethe cosineruletodetermineaformula for AC intermsof θ.
(b)Drawtheanglebisectorat B.Findasecond formulafor AC intermsof θ 2 . 1
6.Kate walksdueeastforadistanceof3 0km, turns120○ toherleftandthenwalks4.0kmin thenewdirection.HowfarisKatethenfromher startingpoint?
7.Apipelineneedstobeconstructedfrompoints A to B acrossamarshland.Theengineers cannotmeasurelength AB directly,sothey markouttwoadditionalpoints D and C.Their measurementsaresummarisedonthediagram below.
(c)Hence, provethatsin2