在三角形ABC中.A.B.C成等差数列,且COSC=12/13 求sinA
问题描述：
在三角形ABC中.A.B.C成等差数列,且COSC=12/13 求sinA
答
A.B.C成等差数列,则
2B=A+C
而A+B+C=π
则B=π/3
A+C=2π/3
A=2π/3-C
sinA=sin(2π/3-C)=sin2π/3cosC-cos2π/3sinC
COSC=12/13
则sinA=√(1-(cosC)^2)=5/13 ,（A