如图,已知在△ABC中,∠ABC和∠ACB的平分线BD、CE相交于点O,且∠A=60°,求∠BOC

2025-04-24 09:51:24
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回答1:

∵∠A=60°,∴∠ABC+∠ACB=120°,∴∠OBC+∠OCB=1/2∠ABC+1/2∠ACB=120÷2=60°
∴∠BOC=180-60=120°