Question

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

Answer 1

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=2.5

c=5-2.5=2.5

面积S=1/2absinC=1/2*4*2.5*3/5=3