Question

在三角形ABC中,角A,B,C所对应的边分别为a,b,c,A=2B,sinB=根号3/3.(1)求cosA及sinC的值

Answer 1

（1）sinB=√3/3 cosB=√6/3

A=2B

cosA=cos2B=1-2sin²B=1-2/3=1/3

sinA=2√2/3

正弦定理a/sinA=b/sinB

a*√3/3=2*2√2/3 a=4√6/3

sinC=sin(A+B)=sinAcosB+cosAsinB=4√3/9+√3/9=5√3/9

（2）c/sinC=b/sinB

c*√3/3=2*5√3/9

c=10/3

∵S=1/2sinAbc

=1/2x2√2/3x10/3

=10√2/9

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