sinB bsinC csinB 5sin120 7sin120sinB06185895741317Barcsin06185895741317382FinallyA180120382218 Example252SASSolvethetriangleifa3b7andC37SolutionWearegiventwosidesa ID: 486176
Download Pdf The PPT/PDF document "77sinB=5sin120sinB=5" is the property of its rightful owner. Permission is granted to download and print the materials on this web site for personal, non-commercial use only, and to display it on your personal computer provided you do not modify the materials and that you retain all copyright notices contained in the materials. By downloading content from our website, you accept the terms of this agreement.
sinB b=sinC csinB 5=sin120 77sinB=5sin120sinB=5 7sin120sinB=0:6185895741317B=arcsin(0:6185895741317)38:2Finally,A=180(120+38:2)=21:8: Example25.2(SAS)Solvethetriangleifa=3;b=7andC=37:Solution.Wearegiventwosidesandtheincludedangle.Wemustndthethirdside.Themissingsideisc:BytheLawofCosinesc2=a2+b22abcosCc2=9+492(3)(7)cos37c2=5842cos37c=p 5842cos374:9:NowusetheLawofSinesandndthesmallestangle.Thesmallestangleisdenitelyanacuteangle.TheLawofSinescannotdistinquishbetweenacuteandobtusebecausebothanglesgiveapositiveanswer.Thesmallestangleisoppositesidea;theshortestside.sinA 3=sin37 4:94:9sinA=3sin37sinA=3 4:9sin37sinA=:36845817A21:63 ThentheareaofthetriangleisK=1 2heightbase=1 2hb:ButsinA=h corh=csinA:Thus,K=1 2bcsinA:Usingsimilararguments,onecanestablishtheareaformulasK=1 2acsinBandK=1 2absinC:Example25.4GivenA=62;b=12meters;andc=5:0meters,ndtheareaofthetriangleABC:Solution.Usingtheformulaforarea,wehaveK=1 2bcsinA=1 2(12)(5:0)sin6226m2: Example25.5Afarmerhasatriangulareldwithsides120yards,170yards,and220yards.Findtheareaoftheeldinsquareyards.Thenndthenumberofacresif1acre=4840squareyards.Solution.WeneedtondananglesowecanusetheareaformulaSo,leta=120;b=170;c=220:WestartbyndingC.c2=a2+b22abcosC48400=14400+2890040800cosC5100=40800cosC5100 40800=cosCarccos(5100 40800)=C97:2CNowndthearea5 aregiven.Inwhatfollows,welets=a+b+c 2(i.e.sishalftheperimeterofthetriangle).Solet'slookatFigure25.4. Figure25.4UsingthePythagoreantheoremwecanwritea2p2=c2(bp)2ora2p2=c2b2p2+2bp:Solvingforpwendp=a2+b2c2 2b:Asimplearithmeticshowsthefollowingh2=a2p2=(ap)(a+p)=aa2+b2c2 2ba+a2+b2c2 2b=2aba2b2+c2 2b2ab+a2+b2c2 2b=c2(ab)2 2b(a+b)2c2 2b=(ca+b)(c+ab)(a+bc)(a+b+c) 4b2=(a+b+c)(a+b+c)(ab+c)(a+bc) 4b2=(2s)2(sa)2(sb)2(sc) 4b2=4s(sa)(sb)(sc) b2Thus,h=2p s(sa)(sb)(sc) bButtheareaofthetriangleisK=1 2bh=1 2b 2p s(sa)(sb)(sc) b!=p s(sa)(sb)(sc):ThislastformulaisknownasHeron'sformula.7