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。

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