题目内容
如图,点O是△ABC的两条角平分线的交点,若∠BOC=118°,则∠A的大小是 .
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56°.
试题分析: ∵△BOC中,∠BOC=118°,∴∠1+∠2=180°﹣118°=62°,∵BO和CO是△ABC的角平分线,∴∠ABC+∠ACB=2(∠1+∠2)=2×62°=124°,在△ABC中,∵∠ABC+∠ACB=124°,∴∠A=180°﹣(∠ABC+∠ACB)=180°﹣124°=56°.故答案为:56°.
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