已知三角形ABC,角A,B,C,所对的边分别为a,b,c,且a=4,b+c=5,根号3sinC-cosC=1,求三角形面积
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3sinC-cosC=1
3sinC-1=cosC
9(sinC)^2-6sinC+1=(cosC)^2=1-(sinC)^2
10(sinC)^2-6sinC=0
sinC≠0,
sinC=3/5
cosC=3sinC-1=3*3/5-1=4/5
c^2=a^2+b^2-2abcosC
a=4,c=5-b
(5-b)^2=4^2+b^2-2*4*bcosC
25-10b+b^2=16+b^2-8b*4/5
b=2.5c=5-2.5=2.5
面积S=1/2absinC=1/2*4*2.5*3/5=3