Question

在三角形ABC中,角A,B,C的对边分别为a,b,c,且2cos(B–C)–1=4cosBcosC 求角A的大小 若a=

Answer 1

2cos(B-C)=4cosBcosC+1

2（cosBcosC+sinBsinC）=4cosBcosC+1

2cosBcosC+2sinBsinC=4cosBcosC+1

2cosBcosC-2sinBsinC= -1

2cos(B+C)= -1

cos[π-(B+C)]=1/2

cosA=1/2

A=π/3

a=3,2sinB=sinC,即有2b=c

a^2=b^2+c^2-2bccosA

9=b^2+4b^2-2b*2b*1/2=3b^2

b=根号3

c=2b=2根号3