Question

高一数学题求解，希望能有步骤依据

Answer 1

(1)根据正弦定理：
2RsinAcosC + √3•2RsinA•sinC - 2RsinB - 2RsinC=0
∵2R≠0
∴sinAcosC + √3sinAsinC - sinB - sinC=0
sinAcosC + √3sinAsinC - sin[π-(A+C)] - sinC=0
sinAcosC + √3sinAsinC - sin(A+C) - sinC=0
sinAcosC + √3sinAsinC - sinAcosC - sinCcosA - sinC=0
√3sinAsinC - sinCcosA - sinC=0
sinC(√3sinA - cosA - 1)=0
∵sinC≠0
∴√3sinA - cosA - 1=0
√3sinA - (1 + cosA)=0
√3•2sin(A/2)cos(A/2) - 2cos²(A/2)=0
2cos(A/2)[√3sin(A/2) - cos(A/2)]=0
∵cos(A/2)≠0
∴√3sin(A/2) - cos(A/2)=0
即：tan(A/2)=√3/3
∵0<A<π
∴0<A/2<π/2
∴A/2=π/6
则A=π/3

追答

(2)由(1)得：sinA=√3/2
cosA=1/2
∵S=(1/2)•bc•sinA
∴√3=(1/2)•bc•(√3/2)
则bc=4.....①
根据余弦定理：
2bc•cosA=b² + c² - a²
2•4•(1/2)=b² + c² - 4
∴b² + c²=8....②
由①②解得：b=c=2