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

(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