Given angleA=60∘,c=√3−1,b=√3+1. Solve the triangle
A
a=√6,B=15∘,C=100∘
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B
a=√6,B=15∘,C=105∘
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C
a=√12,B=15∘,C=105∘
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D
a=√6,B=105∘,C=15∘
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Solution
The correct option is Da=√6,B=105∘,C=15∘ A=60∘⇒B+C=120∘Weknow,tanB−C2=b−cb+ccotA2⇒tanB−C2=(√3+1)−(√3−1)(√3+1)−(√3−1)cot(602)=22√3(√3)=1∴B−C2=45∘⇒B−C=90∘B+C=120∘∴B=105∘andC=15∘,Fromsinerule,a=bsinAsinB=√3+1(sin60sin105)sin105=sin(60∘+45∘)=√32(1√2)+12(1√2)=√3+12√2a=(√3+1)(√3)(2)(√2)(√3+1)2=√6∴a=√6,B=105∘,C=15∘ Option d is correct.