若O为△ABC的外心,I为三角形的内心,且∠BIC=110°,则∠BOC=( )
A.70°
B.80°
C.90°
D.100°
答案:B
∵I是△ABC的内心,
∴∠IBC=∠ABC,∠ICB=
∠ACB.
∵∠ABC+∠ACB=180°-∠A,
∠IBC+∠ICB+∠BIC=180°,
∴∠BIC=180°-(∠IBC+∠ICB)=180°-(∠ABC+
∠ACB)
=180°-(180°-∠A)=90°+
∠A.
∴∠BIC=90°+∠A=90°+
×
∠BOC=90°+
∠BOC
∵∠BIC=110°
∴∠BOC=80°
故选B