Question

△ABC中A=π/3,BC=3,则三角形ABC的周长为？

由正弦定理：a/sinA=b/sinB=c/sinC
∵BC=a=3，A=π/3
∴a/sinA=b/sinB=c/sinC=3/√3/2=2√3
∴b=2√3sinB，c=2√3sinC
∵A+B+C=π，A=π/3
∴C=2π/3-B
∴c=2√3sin(2π/3-B)
三角形ABC周长=a+b+c
=3+2√3sinB+2√3sin(2π/3-B)
=3+6sin(B+π/6) 在过程中∴a/sinA=b/sinB=c/sinC=3/√3/2=2√3
√3/2是怎样来的

Answer 1

a就是BC，所以a=3
A是π/3，那sinA=√3/2
a/sinA=3/√3/2

Answer 2

π/3,即60°
sinA=sin(π/3)=sin60°＝√3/2

Answer 3

Answer 4

sinπ/3=sin60°=√3/2