△ABC中,∠ABC、∠ACB的平分线相交于点O;(1)若∠ABC=40°,∠ACB=50°,则∠BOC=______;(2)若∠AB

2025-05-21 02:15:48
推荐回答(1个)
回答1:

(1)∠BOC=180°-

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(40°+50°)=135°;

(2)∠BOC=180°-
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×116°=122°;

(3)∠BOC=180°-
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×(180°-∠A)=128°;

(4)∵∠BOC=120°
∴∠OBC+∠OCB=60°
根据角平分线的定义得:∠ABC+∠ACB=2×60°=120°
∴∠A=60°;

(5)根据角平分线的定义得:∠OBC+∠OCB=
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(∠ABC+∠ACB)
∴∠BOC=180°-(∠OBC+∠OCB)=180°-
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(∠ABC+∠ACB)=180°-
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(180°-∠A)=90°+0.5x.