问题补充：
在△ABC中,若sinA·sinC=cos^2B,∠B=1/2(∠A+∠C),S△ABC=4根号3,求三边长a,b,c.
答案：
∠B=1/2(∠A+∠C)
A+B+C=180
B=60 A+C=120
sinA*sinC=1/2*[cos(A-C)-cos(A+C)]=cos^2B=1/4
得：cos(A-C)=1/2 + cos(A+C)=1/2 -cosB=0;
则|A-C|=90
不妨设A>C;则A-C=90°.
又由 A+C=120 得
A=105°; C=15°
∴ a/c=sinA/sinC=sin(90°+15°)/sin15°=cos15°/sin15°=tan15°
=(1-cos30°)/sin30°=(1-√3/2)/(1/2)=2-√3.
即a=(2-√3)c.①
由正弦定理得:S=(1/2)·ac·sinB=(√3/4)·ac
即(√3/4)·ac=4√3;
a·c=16 ②
由①②解得:
a=2√6-2√2; c=2√2+2√6;
而b=(sinB/sinC)·c=[(√3/2)/sin(60°-45°)]·(2√2+2√6)
=[(√3/2)/(sin60°·cos45°-cos60°·sin45°)]·(2√2+2√6)
=√3.a=2√6-2√2; b=√3; c=2√2+2√6;
======以下答案可供参考======
供参考答案1:
由于∠B=180-(∠A+∠C)=(∠A+∠C)/2
所以∠A+∠C=120度，∠B=60°
cos^2B=1/4=sinAsinC=(1/2)cos(A-C)-(1/2)COS(A+C)=(1/2)COS(A-C)+1/4
所以(1/2)COS(A-C)=0
A=90+C
解得A=105,B=60,C=15
由正弦定理得:S=(1/2)·ac·sinB=(√3/4)·ac
即(√3/4)·ac=4√3;
(1)a·c=16 ②
①②得a=2√6-2√2; c=2√2+2√6;
b=(sinB/sinC)·c=[(√3/2)/sin(60°-45°)]·(2√2+2√6)
=[(√3/2)/(sin60°·cos45°-cos60°·sin45°)]·(2√2+2√6)
=√3.a=2√6-2√2; b=√3; c=2√2+2√6;
