Question

在△ABC中 A∶B∶C=1∶2∶3 a=1 求a-2b+c/(sinA-2sinB+sinC)

在△ABC中 A∶B∶C=1∶2∶3 a=1 求a-2b+c/(sinA-2sinB+sinC)

Answer 1

A:B:C=1:2:3
设A=t,B=2t,C=3t
t+2t+3t=180°
6t=180°
t=30°
即A=30 °

由正弦定理
a/sinA=b/sinB=c/sinC=2R
a-2b+c/(sinA-2sinB+sinC)
=(2RsinA-4RsinB+2RsinC)/(sinA-2sinB+sinC)
=2R
=a/sinA
=1/sin30°
=2

Answer 2

A∶B∶C=1∶2∶3
所以
A=30 B=60 C=90
a=1 b=√3 c=2
sinA=1/2 sinB=√3/2 sinC=1
(a-2b+c)/(sinA-2sinB+sinC)
=(1-2√3+2)/(1/2-√3+1)
=(3-2√3)/(3/2-√3)
=2(3-2√3)/(3-2√3)
=2