(1)a//b
sinO=√3cosO
tanO=√3
O∈(0,π/2),
O=π/3
sinO=√3/2,cosA=1/2
(2)f(o)=a²+b²+2ab
=1+4+2(√3sinO+cosO)
=5+4sin(O+π/6)
sin(o+π/6)∈[-1,1]
所以 f(o)∈[-1,9]
(1)a//b
sinO=√3cosO
tanO=√3
O∈(0,π/2),
O=π/3
sinO=√3/2,cosA=1/2
(2)f(o)=a²+b²+2ab
=1+4+2(√3sinO+cosO)
=5+4sin(O+π/6)
sin(o+π/6)∈[-1,1]
所以 f(o)∈[-1,9]