一道正余弦定理的数学题

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 (0