来自戚浩峰的问题
如图,在△ABC中,∠ABC和∠ACB的外角平分线交于点O,且∠BOC=40°,则∠A=()A.10°B.70°C.100°D.160°
如图,在△ABC中,∠ABC和∠ACB的外角平分线交于点O,且∠BOC=40°,则∠A=()
A.10°
B.70°
C.100°
D.160°
1回答
2020-03-04 16:58
如图,在△ABC中,∠ABC和∠ACB的外角平分线交于点O,且∠BOC=40°,则∠A=()A.10°B.70°C.100°D.160°
如图,在△ABC中,∠ABC和∠ACB的外角平分线交于点O,且∠BOC=40°,则∠A=()
A.10°
B.70°
C.100°
D.160°
∠BOC=40°
∴∠OBC+∠OCB=140°
∴∠ABC+∠ACB=180°×2-140°×2=80°
∴∠A=180°-(∠ABC+∠ACB)=180°-80°=100°.
故选C.