Question

在三角形ABC中，角A,B,C所对的边分别为a，b，c，且满足√3sinA-cosA=0，cosB=4/5，b=2√3。求sinC

再求三角形ABC的面积

Answer 1

解：
1.
√3sinA-cosA=0
√3sinA=cosA
sinA/cosA=1/√3
tanA=√3/3
则∠A=30°
sinA=1/2,cosA=√3/2(运用sin^2x+cos^2x=1)
cosB=4/5,sinB=3/5
sinc
=sin(π-A-B)
=Sin(A+B)
=SinA*cosB+cosA*sinB
=(1/2)*(4/5)+(√3/2)*(3/5)
=2/5+(3√3/10)

2.S三角形ABC=bc*sinA/2............(1)
S三角形ABC=ac*sinB/2...............(2)
(1):(2)有b*sinA/(a*sinB)=1/1
B=2√3,sinA=1/2,sinB=3/5
带入得a=5√3/3
S三角形ABC=ab*sinC/2=(5√3/3)*2√3*[2/5+(3√3/10)]/2=2+(√3/3/2)