Question

如图,在△ABC中,∠B,∠C的角平分线交于点O,若∠A=70°求∠BOD的度数,若∠A=n°求∠BOC的度数。

Answer 1

解：(1)∵∠A=70°
∴∠ABC+∠ACB=110°
∴1/2∠ABC+1/2∠ACB=55°
∵∠OBC=1/2∠ABC ∠OCB=1/2∠ACB
∴∠OBC+∠OCB=55°
∴∠BOC=125°
(2))∵∠A=n°
∴∠ABC+∠ACB=(180-n)°
∴1/2∠ABC+1/2∠ACB=[(180-n)/2]°
∵∠OBC=1/2∠ABC ∠OCB=1/2∠ACB
∴∠OBC+∠OCB=[(180-n)/2]°
∴∠BOC=90°+n°/2

追问

还是这个问题，∠BOC=3∠A

追答

∵∠BOC=90°+1/2∠A
∠BOC=3∠A
∴∠BOC=108° ∠A=36°