△ABC面积为根号15,且a+c=6,1- c/2a=sin(B-C)/sin(B+C),求三角形三

网友回答

1-c/2a=(2a-c)/2a=(2sinA-sinC)/2sinA 又sin(B+C)=sinA 原式(2sinA-sinC)/2sinA=sin(B-C)/sinA
====>2sinA-sinC=2sin(B-C)
2[sinA-sin(B-C)]=sinC
2*2cos[(A+B-C)/2]sin[(A-B+C)/2]=sinC
4cos[(180-2C)/2]sin[(180-2B)/2]
=sinC 4cos(90-C)sin(90-B)=sinC
4sinCcosB=sinC ===>cosB=1/4==>sinB=(根号15)/4 S=(1/2)acsinB=根号15==>ac=8 a+c=6 aa=2c=4 b^=c^+a^-2accosB====>b=4所以a=4,b=4,c=2