(1)解：∵cosB=BC/AB=1/2∴∠B=60°∴AC=BC×tanB=3根号3(2)解：∵sinA=BC/AC=3/2根号3=根号3/2∴∠A=60°∴AC=AB×cosA=根号3
(1)∵cos∠B=a/c=1/2 ∴∠B=60° ∴b=a×tan∠B=3(2)∵sin∠A=BC/AB=3/2√3=√3/2 ∴∠A=60° ∴AC=AB×cos∠A=√3
