Question

高一数学题。求过程

Answer 1

acosC+√3asinC-b-c=0
sinAcosC+√3sinAsinC-sinB-sinC=0
sinAcosC+√3sinAsinC-sin(A+C)-sinC=0
sinAcosC+√3sinAsinC-sinAcosC-cosAsinC-sinC=0
√3sinAsinC-cosAsinC-sinC=0
sinC(√3sinA-cosA-1)=0
√3sinA-cosA-1=0
√3sinA-cosA=1
√3/2sinA-1/2cosA=1/2
sin(A-π/6)=1/2

A-π/6=π/6 或 A-π/6=5π/6
A=π/3 或A=2π/3
2)a=4 ,2R=b/sinB=c/sinc=a/sinA=4/sinπ/3=8√3/3
a+b+c=4+2R(sinB+sinC)=4+8√3/3(sinB+sinC)
=4+8√3/3[sinB+sin(2π/3-B)
=4+8√3/3(sinB+√3/2cosB+1/2sinB)
=4+8√3/3(3/2sinB+√3/2cosB)
=4+8(√3/2sinB+1/2cosB)
=4+8sin(B+π/6)
B在(0,2π/3),B+π/6在(π/6,5π/6)
sin(B+π/6)在(π/6,5π/6)上值域为:(1/2,1]
a+b+c=4+8sin(B+π/6)值域为:(8,12]
a+b+c的最大值=12

Answer 2