Question

设△ABC的内角A.B.C所对的边分别为abc且a*cosC+1/2c=b若a＝1求三角形周长l取值范围

Answer 1

解:因为a/sinA=b/sinB=c/sinC=2R,且sinB=sin(A C)
所以acosC 1/2c=b可化为
sinAcosC 1/2sinC=sin(A C)
sinAcosC 1/2sinC=sinAcosC cosAsinC
所以cosA=1/2
A=π/3
B C=2π/3, 0<B<2π/3
(2)cosB cosC=cosB cos(2π/3-B)
=cosB [-1/2cosB √3/2sinB)
=√3/2sinB 1/2cosB
=sin(B π/6)
所以1/2<sin(B π/6)≤1
即(2)的取值范围为(1/2, 1]

Answer 2

大于二小于三？