一道正余弦定理的数学题

已知△ABC中,a+c=2b,A-C=60°,求sinB
2025-06-23 05:58:53
推荐回答(2个)
回答1:

A+B+C=180°
A-C=60°
所以C+60°+B+C=180°
C=60°-B/2
所以C小于60°
A=120°-B/2
因为a/sinA=b/sinB=c/sinC (三角形特性)
设a/sinA=b/sinB=c/sinC=t
所以b=t*sinB
c=t*sinC=t*sin(60°-B/2)
a=t*sinA=t*sin(120°-B/2)
因为a+c=2b
所以t*sin(120°-B/2)+t*sin(60°-B/2)=2t*sinB
sin(120°-B/2)+sin(60°-B/2)=2sinB (和差化积公式,2倍角公式)
2sin(90°-B/2)cos30°=4sin(B/2)cos(B/2)
√3cos(B/2)=4sin(B/2)cos(B/2)
sin(B/2)=√3/4
cos(B/2)=√{1-[sin(B/2)]^2}=√13/4 (00)
sinB=2*sin(B/2)cos(B/2)=2*(√3/4)*(√13/4)=√39/8

回答2:

根号下39 /8
由a+c=2b得sinA+sinC=2sinB
2sin(A+C)/2cos(A-C)/2=4sinB/2cosB/2
sinB/2=根号3/4
cosB/2=根号13/4
sinB=2sinB/2*cosB/2=根号下39 /8