Geometry Lines and Angles - Mr-Kuijpers-Math：幾何線和angles-kuijpers數(shù)學(xué)

GEOMETRYLinesandAnglesRightAngles-Are90°oraquarterturnthroughacircle

e.g.1)Acute:AngleTypes-Anglescanbenamedaccordingtotheirsizes2)Obtuse:3)Reflex:-Straightlinesare180°

Areangleslessthan90°

Areanglesbetween90°and180°

Areanglesbetween180°and360°

NamingAngles-Wheretheraysmeet(vertex)givesthemiddleletteroftheanglename.Namethefollowinganglesa)b)ABCZYXABCorCBA

XYZorZYX

MeasuringAngles.Measurethefollowinganglesa)b)ABCABC=

43°

Makesureyoureadfromthescalestartingfrom0ontheline!ZYXXYZ=

141°

-Whenmeasuringreflexangles,measuresmallerangleandsubtractfrom360°.MeasurethefollowingangleABCABC=

360

-65

=295°DrawingAngles.Drawthefollowinganglesa)b)ABC=62°

XYZ=156°

ABCXYZ-Whendrawingreflexangles,subtractanglefrom360°anddrawthisnewangle.Drawthefollowinganglea)ABC=274°

360-274=86ABCDon’tforgettoaddanarctoshowthecorrectangle!AddthearctotheoutsidetoindicateareflexangleEstimatingAngles-Involvesguessinghowbiganangleis-Firstlydecidewhethertheangleisacute,obtuseorreflex.EstimatethesizeofthefollowinganglesABC45°

ABC=

a)b)ZYXXYZ=

220°

SpecialTrianglesEquilateralTriangleAllsidesareequalAllanglesareequal(60°)IsoscelesTriangleTwosidesareequalBaseanglesareequalRightAngleTriangleContainsa90°angleScaleneTriangleNosidesareequalNoanglesareequalDescribingTrianglesByAngles1)AcuteTriangle:2)RightAngleTriangle:3)ObtuseTriangle:Allanglesarelessthan90°Containsa90°angleContainsananglegreaterthan90°Quadrilaterals-Arefoursidedfigures-HavetwodiagonalsDiagonalsgobetweenoppositecornersSpecialSpacesandtheirPropertiesSquareRectangle/OblongParallelogramKite/DiamondArrowheadTrapeziumIsoscelesTrapeziumRhombusDashesonlinesindicateequallengthandarrowsindicateparallellinesANGLESTATEMENTSRemember:Youmustsupplyageometricalreasonwhencalculatingangles!AdjacentAnglesOnAStraightLineAddTo180°x37°x+37=180(adj.’sonastr.line=180°)-37-37x=143°x119°x+119=180(adj.’sonastr.line=180°)-119-119x=61°1.Findx2.FindxComplementaryAnglesAddTo90°Whentwoanglesmakeuparightangle(i.e.48°and42°arecomplementaryangles)e.g.Findxx50°x+50=90(complementaryangles)-50-50x=40°(therefore40°isthecomplementof50°)SupplementaryAnglesAddTo180°Whentwoanglesmakeupastraightangle(i.e.125°and55°aresupplementaryangles)e.g.Whatangleisthesupplementof10°?x+10=180(supplementaryangles)-10-10x=170°(therefore170°isthesupplementof10°)VerticallyOppositeAnglesAreEqualVerticallyoppositeanglesareformedbytwostraightlines1.Findxx58°x=58°(vert.opp.’sare=)2.Findxx38°12°x=38+12(vert.opp.’sare=)x=50°AnglesAtAPointAddTo360°1.Findx34°xx+34=360(’satapoint=360°)-34-34x=326°2.Findxx71°59°82°x+90+82+71+59=360(’satapoint=360°)-302-302x=58°x+302=360InteriorAnglesInATriangleAddTo180°x52°85°1.Findxx+85+52=180(’sinatriangleaddto180°)x+137=180-137-137x=43°2.Findxx46°x+90+46=180(’sinatriangleaddto180°)x+136=180-136-136x=44°BaseAnglesInAnIsoscelesTriangleAreEqual1.Findxx40°x+x+40=1802x+40=180-40-402x=140÷2÷2x=70°(base’sofanisoscelestriangle)(’sinatriangleaddto180°)ExteriorAnglesOfAPolygonAddTo360°1.Findx68°68°56°55°55°xx+68+55+56+68+55=360(ext.’sofapolygonaddto360°)x+298=360-298-298x=62°2.FindxinthisregularpentagonRegularmeansequalsidesandanglesx5x=360(ext.’sofapolygonaddto360°)÷5÷5x=72°3.Aregularpolygonhasexterioranglesof36°.Howmanysidesdoesithave?Thenumberofsides=thenumberofangles36x=360÷36÷36x=10TheSumOfTheInteriorAnglesOfAPolygonIs(n–2)180n=numberofsidesofapolygon1.FindtheanglesumofthisregularhexagondivideintotrianglesfromonecornerInterioranglesum=(6–2)x180n=6(or4triangles)(interioranglesumofapolygon)Interioranglesum=720°2.FindxxInterioranglesum=(5–2)x180n=5or3triangles(interioranglesumofapolygon)Interioranglesum=540°x=540÷5x=108°Anothermethodistocalculateanexterioranglefirstthenuseadjacentanglesonastraightlinetocalculateinteriorangle72°Exteriorangle=72°x+72=180x=108°(adjacent.anglesonastraightline=180°)PerpendicularLines-Alwayscrossatrightangles.e.g.ABCDABisperpendiculartoCDorABCD

ParallelLines-Nevermeetandarealwaysthesamedistanceapart.e.g.ABCDFEABisparalleltoCDorAB??CD

EFisknownasatransversalAngleStatementsandParallelLinesAlternateAnglesOnParallelLinesAreEqual-Therearetwopairsofalternateanglesbetweenparallellinesandatransversal.e.g.e.g.Findx113°xx=113°(Alternateanglesonparallellinesareequal)CorrespondingAnglesOnParallelLinesAreEqual-Therearefourpairsofcorrespondinganglesbetweenparallellinesandatransversal.e.g.e.g.Findx122°xx=122°(Correspondinganglesonparallellinesareequal)Co-InteriorAnglesOnParallelLinesAddTo180-Therearetwopairsofco-interioranglesbetweenparallellinesandatransversal.e.g.e.g.Findx77°xx+77=180(Co-interioranglesinparallellinesaddto180°)-77-77x=103°Remembertoalwaysadd‘onparallellines’withyouranglestatementsBearings-Bearingsareusedtoindicatedirections-AremeasuredclockwisefromNorth-Mustbeexpressedusing3digits(i.e.000°to360°)-CompassdirectionssuchasNWgivedirectionsbutarenotbearingse.g.Thecompasspointsandtheirbearings:ESWNESWSENW000°090°180°270°045°135°225°315°e.g.Drawabearingof051°:N51°e.g.WhatisthebearingofRfromN?N37°Bearing=180+37=217°SimilarTrianglesAndOtherShapes-Oneshapeissimilartoanotheriftheyhaveexactlythesameshape.Theratiosofthecorrespondingsidesarethereforethesame.-Trianglesaresimilariftheyhavethesameanglese.g.Thefollowingtwotrianglesaresimilar.Workoutthelengthsxandy

615204x

y

FirstcalculateratiobetweencorrespondingsidesABCFEGAC=15EG6Tofindxweneedtomultiplythecorrespondingsidebytheratio:Tofindyweneedtodividethecorrespondingsidebytheratio:x=10y=8(SimilarTriangles)(SimilarTriangles)PartsOfACircleRadiusDiameterChordArcSectorSegmentCircumferenceTangentAnglePropertiesofCirclesBaseAnglesOfAnIsoscelesTriangleAreEquale.g.x40°Becausetwosidesofthetriangleareradii,anisoscelestriangleisformedx=40°(base’sofanisoscelestriangle)TheAngleAtCentreIsTwiceTheAngleAtTheCircumferenceProof:AABx=180–2ACx+C=180180–2A+C=180C=2ADD=2BC+D=2A+2BC+D=2(A+B)xe.g.Findxx42°x=2×42x=84°(atcentre=2×atcircumf.)rrAngleInASemiCircleIsARightAngle-Thiscaseisaspecialversionofthepreviousrulexx=90°(inasemi-circle)AnglesOnTheSameArcAreEquale.g.Proof:ACBC=2AC=