题目
在△ABC中,2C=A+B,1）求角C的大小.2)若AB=1求将△ABC的周长表示为角A的函数关系,并求周长的取值范围..
提问时间：2021-04-01
答案
1、A+B+C=2C+C=3C=180度,故C=60度
2、sinA/BC=sinB/AC=
sinC/AB=sin60度/1=(√ 3)/2
△ABC的周长C=AB+BC+AC
=1+2sinA/(√ 3)+2sinB/(√ 3)
=1+2(sinA+sinB)/(√ 3)
=1+2(sin(120-A)+sinA)/(√ 3)
=1+2sin60度COS(60度-A)/(√ 3)
=1+ COS(60度-A)
因为0
3/2<=1+ COS(60度-A)<=2
即 周长的取值范围为[3/2,2]
2、sinA/BC=sinB/AC=
sinC/AB=sin60度/1=(√ 3)/2
△ABC的周长C=AB+BC+AC
=1+2sinA/(√ 3)+2sinB/(√ 3)
=1+2(sinA+sinB)/(√ 3)
=1+2(sin(120-A)+sinA)/(√ 3)
=1+2sin60度COS(60度-A)/(√ 3)
=1+ COS(60度-A)
因为0
3/2<=1+ COS(60度-A)<=2
即 周长的取值范围为[3/2,2]
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