ntchenjie
幼苗
共回答了17个问题采纳率:94.1% 举报
△ABC的面积S满足S=√3/2bccosA=bcsinA/2,cosA=1/2,sinA=√3/2,A=60°
若a=√3,设角B的大小为x,b=asinB/sinA=√3sinx/(√3/2)=2sinx;
c^2=a^2+b^2-2abcosC=3+4sin^2x-4√3sinxcos(180°-A-B)=4sin^2x-4√3sinxcos(120°-x)+3
=4sin^2x-4√3sinx(cos120°cosx+sin120°sinx)+3=-2sin^2x+2√3sinxcosx+3
=1-2sin^2x+2√3sinxcosx+2=cos(2x)+√3sin(2x)+2=2cos(2x-60°)+2
0°
1年前
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