正弦定理问题

显示全部楼层 · 2013-2-20 21:34:55

A=π/3,B=π/6C=π/2a/sinA=c/sinC=b/sinB=12/(1/2)=24a=24 * sinA=24 * √3/2=12√3c=24 * sinC=24(a+b+c)=(12+24+12√3) = 12(3+√3)(sinA+sinB+sinC) = (1/2 + √3/2 + 1) = (3+√3)/2(a+b+c)/(sinA+sinB+sinC) = 12(3+√3) /[ (3+√3)/2] = 24...

千问 · 2013-2-20 21:34:55

sinB=sin(A+C)=sinπ/2=1根据正弦定理求出其他各边...

千问 · 2013-2-20 21:34:55

B=π/2
b/sinB=(a b c)/…=12...