Year 11Mathematics

Contents

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IAS 1.7Robert Lakeland & Carl Nugent

Right-Angled Triangles

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

• TheTheoremofPythagoras.............................................. 3

• CalculatingLengths..................................................... 7

• CalculatingAngles...................................................... 14

• Three-DimensionalFigures............................................... 17

• SimilarTriangles........................................................ 27

• LimitsofAccuracy...................................................... 32

• MeasurementError ..................................................... 36

• PracticeInternalAssessment1............................................ 37

• PracticeInternalAssessment2............................................ 41

• Answers............................................................... 45

IAS 1.7 – Year 11 Mathematics and Statistics – Published by NuLake Ltd New Zealand © Robert Lakeland & Carl Nugent

2 IAS 1.7 – Right-Angled Triangles

NCEA 1 Internal Achievement Standard 1.7 – Trigonometry Thisachievementstandardinvolvesapplyingright-angledtrianglesinsolvingmeasurementproblems.

Achievement Achievement with Merit Achievement with Excellence• Applyright-angledtriangles

insolvingmeasurementproblems.

• Applyright-angledtriangles,usingrelationalthinking,insolvingmeasurementproblems.

• Applyright-angledtriangles,usingextendedabstractthinking,insolvingmeasurementproblems.

◆ ThisachievementstandardisderivedfromLevel6ofTheNewZealandCurriculum,Learning Media.ThefollowingachievementobjectivestakenfromtheShapeandMeasurementthreadsofthe MathematicsandStatisticslearningareaarerelatedtothisachievementstandard: ❖ usetrigonometricratiosandPythagoras’theoremintwoandthreedimensions ❖ recognisewhenshapesaresimilaranduseproportionalreasoningtofindanunknown length ❖ selectanduseappropriatemetricunitsforlengthandarea ❖ measureatalevelofprecisionappropriatetothetask.

◆ Applyright-angledtrianglesinvolves: ❖ selectingandusingarangeofmethodsinsolvingmeasurementproblems ❖ demonstratingknowledgeofmeasurementandgeometricconceptsandterms ❖ communicatingsolutionswhichwouldusuallyrequireonlyoneortwosteps.

◆ Relationalthinkinginvolvesoneormoreof: ❖ selectingandcarryingoutalogicalsequenceofsteps ❖ connectingdifferentconceptsandrepresentations ❖ demonstratingunderstandingofconcepts ❖ formingandusingamodel; andalsorelatingfindingstoacontext,orcommunicatingthinkingusingappropriate mathematicalstatements.

◆ Extendedabstractthinkinginvolvesoneormoreof: ❖ devisingastrategytoinvestigateorsolveaproblem ❖ identifyingrelevantconceptsincontext ❖ developingachainoflogicalreasoning,orproof ❖ formingageneralisation; andalsousingcorrectmathematicalstatements,orcommunicatingmathematicalinsight.

◆ Problemsaresituationssetinareal-lifecontextwhichprovideopportunitiestoapplyknowledge orunderstandingofmathematicalconceptsandmethods.Forassessment,situationsmayinvolve nonright-angledtriangleswhichcanbedividedintoright-angledtriangles.

◆ Thephrase‘arangeofmethods’indicatesthatevidenceoftheapplicationofatleastthree differentmethodsisrequired.

◆ Studentsneedtobefamiliarwithmethodsrelatedto: ❖ Pythagoras’theorem ❖ trigonometricratios(sine,cosine,tangent) ❖ similarshapes ❖ measuringatalevelofprecisionappropriatetothetask.

3IAS 1.7 – Right-Angled Triangles

IAS 1.7 – Year 11 Mathematics and Statistics – Published by NuLake Ltd New Zealand © Robert Lakeland & Carl Nugent

The Theorem of Pythagoras

Pythagoras’ TheoremPythagoras’Theoremgivesarelationshipbetweenthesidesofaright-angledtriangle.Wealwayslabelthesidesa,bandhwiththelongestside(thehypotenuse)labelledhasinthediagrambelow.

TheTheoremofPythagorasis

h2=a2+b2

The hypotenuse is always opposite the right angle symbol. Always label it h.

ExampleFindthelengthxintheright-angledtriangle.

Webeginbylabellingthesidesofthetrianglea,bandh.

Pythagoras h2 =a2+b2

Substitute x2 =7.652+11.32

x2 =186.2125

Squareroot x = 186.2125

x =13.64597...

x =13.6cm(1dp)

ExampleFindthelengthyintheright-angledtriangle.

Webeginbylabellingthesidesofthetrianglea,bandh.

Pythagoras h2 =a2+b2

Substitute 9.42 =y2+8.62 88.36 =y2+73.96 y2+73.96 =88.36 y2 =14.4Squareroot y = 14.4

y =3.8cm(1dp)

7.65cm

11.3cm

x

(a)

(b)

(h)y

8.6cm

9.4cm

(a)

(b)

(h)

WeusetheTheoremofPythagoraswhenwehavearight-angledtriangleandweknowtwolengthsofthetriangleandneedtofindthethirdlength.

hypotenuse (h)

right angle symbol

Using the Casio 9750GII. From the MENU select EQUA then the Solver, deleting any existing equations

Enter in Pythagoras’ theorem h2 = a2 + b2

Enter the known values for H, A or B and place the cursor next to the unknown (in this case H) and select F6 to solve.

F3MENU F2 F1

DEL YesSolver

8

EQUA

ALPHA F)D x2 ALPHASHIFT .H A

x, ø, T

x2 ALPHA+Blog x2 EXE

=

On the TI-84 Plus select the MATH menu. Press 0 or scroll down until the cursor is on Solver and press ENTER. To delete any existing equation press the up arrow and then the CLEAR key.

Enter h2 = a2 + b2 as a2 + b2 – h2 as it needs to = 0.

Enter the values for the known variables and enter an initial value for the unknown and select

to solve. ALPHA ENTERSOLVE

ALPHA x2MATHa

+ x2APPSb

ALPHA

h– ALPHA x2 ENTER

You can use the equation function on your graphics calculator to solve Pythagoras problems (see right). Once you have the equation entered you can solve multiple problems.

ha

b ©uLake LtduLake Ltd Innovative Publisher of Mathematics Texts

IAS 1.7 – Year 11 Mathematics and Statistics – Published by NuLake Ltd New Zealand © Robert Lakeland & Carl Nugent

8 IAS 1.7 – Right-Angled Triangles

Example

Webeginbylabellingthesidesofthetriangleopp,hypandadj.

Becausewearerequiredtofindw(opposite)andhaveinformationontheadjacentweusethetrigonometricratiothatinvolvesthesetwo,i.e.tangent.

Findthedistancetheladderreachesupthewall,labelledwinthediagram.

tanA =oppadj

Substitute tan43˚ = w6 5.

Multiplyingby6.5 w =6.5xtan43˚

=6.1m(2sf)

Achievement–Findthelengthsoftheunknownsidesintheright-angledtrianglesbelow.

19. 20.

6.5m

w

43˚

w

6.5madj

opphyp

43˚

y23.4mm 43˚

z36.4m

36˚

ExampleA9.8mladderisleaningagainstawall.Theladdermakesanangleof74˚withtheground.Howfaristhebasefromthewall?

Webeginbydrawingadiagram,addingtheinformationwehaveandlabellingthesidesofthetriangleopp,hypandadj.

9.8mhyp opp

adj g74˚

Becausewearerequiredtofindg(adjacent)andhaveinformationonthehypotenuseweusethetrigonometricratiothatinvolvesthesetwo, i.e.cosine.

cosA=adjhyp

Substitute cos74˚ = g9.8

Multiplyingby9.8 g =9.8xcos74˚ =2.7m(2sf)

21. 22.

31˚

v13.4kmw

12.8m29˚

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13IAS 1.7 – Right-Angled Triangles

IAS 1.7 – Year 11 Mathematics and Statistics – Published by NuLake Ltd New Zealand © Robert Lakeland & Carl Nugent

46. A hill has an angle of climb of 8.5˚ to the horizontal. How far would it be necessary to walk to increase one’s altitude by 200 m?

47. The angle of pitch of an A frame house roof is 63.0˚ to the horizontal. The vertical rise of the roof is 7.40 m. Find the slanted length of the roof, and the width of the house at ground level.

48. Two support cables each 262 m long run from the top of a radio tower to the ground. Each cable makes an angle of 37˚ with the ground.

a) Find, to the nearest metre, the height of the tower.

Another cable is run from the top of the tower to the ground and makes an angle of 54˚ with the ground.

b) Find the length of this cable to the nearest metre.

c) Calculate the distance between the two cables on the ground to the nearest metre.

49. An escalator has an angle of elevation of 36˚ and climbs a vertical height of 9.6 m. What is the horizontal movement of the escalator?

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17IAS 1.7 – Right-Angled Triangles

IAS 1.7 – Year 11 Mathematics and Statistics – Published by NuLake Ltd New Zealand © Robert Lakeland & Carl Nugent

Three-Dimensional Figures

Find Lengths and Angles in Three-Dimensional Figures

Tosolvelengthandangleproblemsinthree-dimensionsweidentifythetwo-dimensionalright-angledtrianglethatcontainsourunknownandhastwosidemeasurementsorasideandananglemeasurement.WethenuseourknowledgeofPythagorasand/ortrigonometricratiostofindtherequiredangleorlength.

A

C BD

H

EF

G

6.4 m

2.85 m

w

x

7.2 mTofindthelengthxortheangleGFHweusetheright-angledtriangleFGH.

G

F

H

x

Angle GFH

TofindthelengthwortheangleFHCweusetheright-angledtriangleFCH.

Onceyouhaveidentifiedthetwo-dimensionaltrianglefromthe3Dfigure,drawitseparately,carefullymarkinginalltheappropriatelengthsandangles.

F

C

Hx

Angle FHC

w

7.2m

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21IAS 1.7 – Right-Angled Triangles

IAS 1.7 – Year 11 Mathematics and Statistics – Published by NuLake Ltd New Zealand © Robert Lakeland & Carl Nugent

72. 73.Anail9cmlongislyinginsideanemptybakedbeanscanwhichhasradius3.75cmandheight10.5cm.Seethediagrambelow.

a) Whatangledoesthenailmakewiththe bottomofthecan?b) Howfarupfromthebaseofthecandoesthe nailreach?

Ateepeeisconeshaped.Ithasdiameter2.4mandslantheight2.2m.Seethediagrambelow.

a) Couldaperson1.85mtallstandinthemiddle oftheteepeewiththeirheadnottouchingthe top?Justifyyouranswer.b) Calculatetheapexangleoftheteepee labelledAinthediagram.

2.4 m

2.2 m

A

10.5 cm

3.75 cm

9.0 cm

Excellence–Deviseastrategytosolveeachproblem.Explainyourapproachwiththeaidofa diagram

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IAS 1.7 – Year 11 Mathematics and Statistics – Published by NuLake Ltd New Zealand © Robert Lakeland & Carl Nugent

46 IAS 1.7 – Right-Angled Triangles

Page 2070. r=1.30m (3sf) A=15.6˚ (1dp) steel=11.5m (3sf)

71. SQ=5.94m (3sf) WQ=7.87m (3sf) VQ=6.87m (3sf) SQP=49.6˚ (1dp) WQT=35.0˚ (1dp) VQU=41.2˚ (1dp)

Page 2172.

a) AngleX=33.6˚(1dp) b) h=5.0cm (1dp)

73.

a) Ht.=1.84m (3sf) Wouldtouchthetop. b) A=66.1˚ (1dp)

Page 2474. a)FirstcalculateBDusing

Pythagoras.ThenworkwithtriangleFBD.

FDB=39.2˚ (1dp) b)FirstcalculateED.Then

workwithtriangleFDE.

FDE=38.6˚ (1dp)

Page 24 cont...75. a)CalculateBXashalfofBG.

ThenworkwithtriangleABX.

AXB=45.1˚ (1dp) b)LetKbethemidpointofBC.

ThenwecalculateAKandKX.

XAK=34.5˚ (1dp)Page 2576. a)Theprojectionofplane

ABGHonplaneABCDhasHinfrontofD.SoweworkwithtriangleADH.

HAD=57.9˚ (1dp) b)Theprojectionofplane

ABGHonplaneABFEhasHinfrontofE.SoweworkwithtriangleAEH.

HAE=32.1˚ (1dp)77. a)Theraysatrightanglesto

thelineofintersectionpassthroughZandXfromthemidpointofADlabelledE.

XEZ=50.4˚ (1dp) b)Theraysatrightanglesto

thelineofintersectionpassthroughZandXfromthemidpointofABlabelledF.

XFZ=67.2˚ (1dp)

Page 2678.a)

DAH=61.5˚ (1dp)b)

HBC=65.0˚ (1dp)79. a)

AC=8.5m (2sf) b)

Ropes=5.3m (2sf) c) Sametriangle. Angle=37.1˚ (1dp)

Page 2880.x=12.0cm (1dp)81. x=6.0cm (1dp)82. x=3.5cm (1dp) y=3.6cm (1dp)83. x=8.0cm (2sf) y=17cm (2sf)84. x=12.0cm (1dp) y=10.9cm (1dp)85. x=10.2cm (1dp) y=28.8cm (1dp)

Page 29 86.x=7.5m (1dp) y=13.0m (1dp)87. x=9.0m (1dp) y=15.0m (1dp)88. x=28.0cm (1dp) y=24.2cm (1dp) z=6.9cm (1dp)89. x=2.8m (1dp) y=17.2m (1dp)90. x=26.8mm (1dp) y=10.7mm (1dp) z=14.4mm (1dp)91. x=3.6m (1dp) y=8.0m (1dp)

7.50cmdiameter

9.0cm

N

BA

h

X

Approach.Thetriangleisformedbythenail,diameterandheightitreachesuptheside.AngleXistheanglethenailmakeswiththebottomofthecan.

2.2m

1.2m

h

A/2

Approachistoformatrianglewiththecentreoftheteepee.Halftheangleattheapexandtheheightcanbecalculated.

B

D

F23.7mm

29.1mm

F

E D29.3mm

23.4mm

B

A

X

4.85m

4.825m

K

A

X Midpointof BC5.64m

3.875m

D

A

H

4.67m

7.45m

E

H

A7.45m

4.67m

X

Z Midpointof AD5.68m

6.86m

E

X

ZMidpointof AB

F 2.89m

6.86m

H

A

D

3.8m

7.0m

H

B

C

3.8m

8.163m

A B

C

6.5m

5.4m

E

BF

3.2m

4.25m©uLake LtduLake Ltd Innovative Publisher of Mathematics Texts