解三角形已知△ABC的三边a＞b
已知△ABC的三边a＞b＞c且a+c=2b,A-C=90度，求a:b:c
解:
∵a+c=2b
∴sinA+sinC=2sinB
sinA+sinC=2sin[(A+C)/2]cos[(A-C)/2]
=2sin[90°-(B/2)]cos45°=(√2)cos(B/2)=2sinB
=4sin(B/2)cos(B/2)
0＜B/2＜90° cos(B/2)≠o
sin(B/2)= (√2)/4
cosB=1-2[sin(B/2)]＾=1-1/4=3/4
b＾=a＾+c＾-2accosB=a＾+c＾-(3ac/2)
a+c=2b a＞b＞c
∴a/c=(4+√7)...全部
已知△ABC的三边a＞b＞c且a+c=2b,A-C=90度，求a:b:c
解:
∵a+c=2b
∴sinA+sinC=2sinB
sinA+sinC=2sin[(A+C)/2]cos[(A-C)/2]
=2sin[90°-(B/2)]cos45°=(√2)cos(B/2)=2sinB
=4sin(B/2)cos(B/2)
0＜B/2＜90° cos(B/2)≠o
sin(B/2)= (√2)/4
cosB=1-2[sin(B/2)]＾=1-1/4=3/4
b＾=a＾+c＾-2accosB=a＾+c＾-(3ac/2)
a+c=2b a＞b＞c
∴a/c=(4+√7)/3
令: a=(4+√7)u 则c=3u
b=(7+√7)u/2
a:b:c=2(4+√7):(7+√7):6
。
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