如图,已知圆O是三角形ABC的内切圆,且角BAC=50度,求角BOC的度数.10

∵O为ΔABC内心,
∴∠OBC+∠OCB
=1/2∠ABC+1/2∠ACB
=1/2(∠ABC+∠ACB)
=1/2(180°-∠A)
=65°,
∴∠BOC=180°-(∠OBC+∠OCB)=115°.