/
77sinB=5sin120sinB=5 77sinB=5sin120sinB=5

77sinB=5sin120sinB=5 - PDF document

pasty-toler
pasty-toler . @pasty-toler
Follow
397 views
Uploaded On 2016-11-08

77sinB=5sin120sinB=5 - PPT Presentation

sinB bsinC csinB 5sin120 7sin120sinB06185895741317Barcsin06185895741317382FinallyA180120382218 Example252SASSolvethetriangleifa3b7andC37SolutionWearegiventwosidesa ID: 486176

Share:

Link:

Embed:

Download Presentation from below link

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.


Presentation Transcript

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.Wemust ndthethirdside.Themissingsideisc:BytheLawofCosinesc2=a2+b2�2abcosCc2=9+49�2(3)(7)cos37c2=58�42cos37c=p 58�42cos374:9:NowusetheLawofSinesand ndthesmallestangle.Thesmallestangleisde nitelyanacuteangle.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: Example25.5Afarmerhasatriangular eldwithsides120yards,170yards,and220yards.Findtheareaofthe eldinsquareyards.Then ndthenumberofacresif1acre=4840squareyards.Solution.Weneedto ndananglesowecanusetheareaformulaSo,leta=120;b=170;c=220:Westartby ndingC.c2=a2+b2�2abcosC48400=14400+28900�40800cosC5100=�40800cosC�5100 40800=cosCarccos(�5100 40800)=C97:2CNow ndthearea5 aregiven.Inwhatfollows,welets=a+b+c 2(i.e.sishalftheperimeterofthetriangle).Solet'slookatFigure25.4. Figure25.4UsingthePythagoreantheoremwecanwritea2�p2=c2�(b�p)2ora2�p2=c2�b2�p2+2bp:Solvingforpwe ndp=a2+b2�c2 2b:Asimplearithmeticshowsthefollowingh2=a2�p2=(a�p)(a+p)=a�a2+b2�c2 2ba+a2+b2�c2 2b=2ab�a2�b2+c2 2b2ab+a2+b2�c2 2b=c2�(a�b)2 2b(a+b)2�c2 2b=(c�a+b)(c+a�b)(a+b�c)(a+b+c) 4b2=(a+b+c)(�a+b+c)(a�b+c)(a+b�c) 4b2=(2s)2(s�a)2(s�b)2(s�c) 4b2=4s(s�a)(s�b)(s�c) b2Thus,h=2p s(s�a)(s�b)(s�c) bButtheareaofthetriangleisK=1 2bh=1 2b 2p s(s�a)(s�b)(s�c) b!=p s(s�a)(s�b)(s�c):ThislastformulaisknownasHeron'sformula.7

Related Contents


Next Show more