解:(1)∵∠ABC和∠ACB的平分線BD,CE相交于點O,
∴∠DBC=
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∠ABC,∠ECB=
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∠ACB,又∠ABC=32°,∠ACB=58°,
∴∠DBC=16°,∠ECB=29°,
∴∠BOC=180°-∠DBC-∠ECB=180°-16°-29°=135°,
故答案為:135°;
(2)∵∠A=76°,
∴∠ABC+∠ACB=180°-76°=104°,
又∠ABC和∠ACB的平分線BD,CE相交于點O,
∴∠DBC=
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∠ABC,∠ECB=
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∠ACB,
∴∠DBC+∠ECB=
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(∠ABC+∠ACB)=52°,
則∠BOC=180°-(∠DBC+∠ECB)=180°-52°=128°;
(3)β=90+
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α,
理由如下:∵∠ABC、∠ACB的平分線相交于點O,
∴∠OBC=
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∠ABC、∠0CB=
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∠ACB,
∴∠OBC+∠0CB=
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∠ABC+
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∠ACB=
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(180°-α)=90°-
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α,
∴β=180°-(∠OBC+∠0CB)=180°-(90°-
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α)=90°+
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α.
分析:(1)先利用角平分線的定義求出∠DBC和∠ECB的度數(shù),再運用△BOC的內(nèi)角和是180°,求解∠BOC的度數(shù),
(2)利用互補的性質(zhì)計算,
(3)利用互余和角平分線的性質(zhì)計算.
點評:本題主要考查了角平分線定義以及三角形內(nèi)角和定理,難度適中.