高中数学题

Question

要详细过程

Answer 1

由正弦定理√3sinAcosC-√3cosAsinC=sinB
易得B=π/3
则√3sin（A-C）=√3/2
sin（A-C）=1/2
A-C=π/6又A+C=2π/3
所以A=5π/12,C=π/4
由a/sinA=c/sinC可求c

Answer 2

由已知：B=π-(A+C)=π - 2π/3=π/3
根据正弦定理：
√3sinAcosC=sinB + √3sinCcosA
√3sinAcosC - √3sinCcosA=sin(π/3)
√3sin(A - C)=√3/2
∴sin(A - C)=1/2
则A-C=π/6或A-C=5π/6
∵A+C=2π/3
∴A=5π/12，C=π/4或A=3π/4，C=-π/12(舍)
根据正弦定理：a/sinA=c/sinC
2/sin(5π/12)=c/sin(π/4)
解得：c=2√3 - 2

Answer 3

本回答被提问者采纳

Answer 4

Answer 5

根号3sinAcosC=sinB+根号3sinCcosA
根号3(sinAcosC-sinCcosA)=sinB
根号3sin(A-C)=sinB
A+C=2π/3
B=π/3
sin(A-C)=1/2
A-C=π/6或5π/6(舍去)
A=5π/12 C=π/4
（1）a/sinA=c/sinC a=2 c=asinC/sinA=2根号3-2
（2）S△ABC=3+根号3 1/2acsinB=3+根号3 ac=4+4根号3
a/c=sinA:sinC=[（根号6+根号2）]/4:(根号2/2)=根号3+1

a>0,c>0
a=2+2根号3，c=2
b/c=sinB:sinC=(根号3/2):(根号2/2)
b=2*根号3/根号2=根号6
周长a+b+c=4+根号6+2根号3本回答被网友采纳