40Chapter 2•General Triangles§2.1Example2.1ba=10cA=41◦BC=75◦Case 1: One side and two angles.Solve the triangle△ABCgivena=10,A=41◦, andC=75◦.Solution:We can find the third angle by subtracting the other two anglesfrom 180◦, then use the law of sines to find the two unknown sides. In thisexample we need to findB,b, andc. First, we see thatB=180◦−A−C=180◦−41◦−75◦⇒B=64◦.So by the Law of Sines we havebsinB=asinA⇒b=asinBsinA=10 sin 64◦sin 41◦⇒b=13.7,andcsinC=asinA⇒c=asinCsinA=10 sin 75◦sin 41◦⇒c=14.7.Example2.2b=30a=18cA=25◦BCCase 2: Two sides and one opposite angle.Solve the triangle△ABCgivena=18,A=25◦, andb=30.Solution:In this example we know the sideaand its opposite angleA,and we know the sideb. We can use the Law of Sines to find the otheropposite angleB, then find the third angleCby subtractingAandBfrom180◦, then use the law of sines to find the third sidec. By the Law of Sines, we havesinBb=sinAa⇒sinB=bsinAa=30 sin 25◦18⇒sinB=0.7044.Using the✄✂a0✁sin−1button on a calculator givesB=44.8◦.However, recall from Section 1.5 thatsin (180◦−B)=sinB.So there is a second possible solution for
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