已知锐角三角形ABC中,边A,B,C分别是角,B,C所对的边,若满足(a+b+c)(b+c-a)=3bc求 cosB+c

原式=(b+c)^2-a^2=3bc.带入
（sinb+sinc)^2-sin(b+c)^2=3sinb*sinc
即(sinb)^+(sinc)^2-sinb*sinc
=2[(sinb*cosc)^2+(sinc*cosb)^2+2sinb*sinc*cosb*cosc]
=(sinb*cosc)^2+(sinc*cosb)^2+2sinb*sinc*cosb*cosc
(sinb)^+(sinc)^2-sinb*sinc-(sinb*cosc)^2+(sinc*cosb)^2
=(sinb*sinc)^2+(sinc*sinb)^2
=2(sinb*sinc*cosb*cosc)
(sinb*sinc)^2+(sinc*sinb)^2-2(sinb*sinc*cosb*cosc)=0
sinbsinc*sinb*sinc=sinb*sinc*cosb*cosc
我没有时间了,下面交给你自己罗

将（a+b+c)(b+c-a）=3bc 化简，得
b^2+c^2-a^2=bc
cosA=(b^2+c^2-a^2)/2bc=1/2
所以，A=60°
B+C=120°

cosB+cosC=cos(60°-C)+cosC=3/2cosC+√3/2sinC=√3 (sin60°cosC+cos60°sinC)=sin(60°+C)

4455

将（a+b+c)(b+c-a）=3bc 化简，得
b^2+c^2-a^2=bc
cosA=(b^2+c^2-a^2)/2bc=1/2
所以，A=60°
B+C=120°
cosB+cosC=cos(120°-C)+cosC=cosC/2+√3sinC/2=sin30°cosC+cos30°sinC=sin(30°+C)
90°＜C＜120...