帮我回答一下16.17 题17 要步骤

由已知：b²=ac
设外接圆半径为R
cosA/sinA+cosC/sinC
=(sinCcosA+cosCsinA)/(sinAsinC)
=sin(A+C)/(sinAsinC)
=sinB/(sinAsinC)
=b/(ac/2R)=4√7/7
即2Rb/ac=2R/b=4√7/7
故sinB=b/2R=√7/4
因为b不是最大边
故cosB>0
故cosB=3/4
故b²=a²+c²-2accosB=a²+c²-3ac/2=ac
故a²+c²=5ac/2=(a+c)²-2ac
故ac=2
故三角形ABC的面积=acsinB/2=√7/4