如图,在△ABC中,∠ABC与∠ACB的角平分线交于点O,且∠A=α,则∠BOC的度数是(  ) A. 180°-

2025-03-23 05:06:22
推荐回答(1个)
回答1:

∵∠A=α,
∴∠ABC+∠ACB=180°-α,
∵BO、CO分别是∠ABC与∠ACB的角平分线,
∴∠OBC+∠OCB=
1
2
∠ABC+
1
2
∠ACB
=
1
2
(∠ABC+∠ACB)
=
1
2
(180°-α)
=90°-
1
2
α,
∴∠BOC=180°-(∠OBC+∠OCB)=90°+
1
2
α.
故选B.