△ABC中,∠A=70°,△ABC的角平分线BD,CE交于点O,则∠BOC=_度.

问题描述:

△ABC中,∠A=70°,△ABC的角平分线BD,CE交于点O,则∠BOC=______度.

∵BD平分∠ABC,则∠4=

1
2
∠ABC,
CE平分∠ACB,则∠2=
1
2
∠ACB,
∴∠2+∠4=
1
2
(∠ABC+∠ACB)=
1
2
(180°-∠A)=
1
2
(180°-70°)=55°,
∴在△BOC中,∠BOC=180°-55°=125°.