如图,△ABC中,角平分线BD,CE交于点O,求∠BOC的度数
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/08/23 08:29:22
如图,△ABC中,角平分线BD,CE交于点O,求∠BOC的度数
![](http://img.wesiedu.com/upload/9/9f/99f4573033aed4cc712b432b92692467.jpg)
![](http://img.wesiedu.com/upload/9/9f/99f4573033aed4cc712b432b92692467.jpg)
∠BOC=180°-(∠OBC+∠OCB) =180°-(1/2∠ABC+1/2∠ACB)
=180°-(1/2)(∠ABC+∠ACB)
=180°-(1/2)(180-∠A)
=180°-90°+1/2∠A
=90°+1/2∠A
=90°+1/2*50°
=90°+25°
=115°∠BOC=180°-(∠OBC+∠OCB)
=180°-(1/2∠ABC+1/2∠ACB)
=180°-(1/2)(∠ABC+∠ACB)
=180°-(1/2)(180-∠A)
=180°-90°+1/2∠A
=90°+1/2∠A
=180°-(1/2)(∠ABC+∠ACB)
=180°-(1/2)(180-∠A)
=180°-90°+1/2∠A
=90°+1/2∠A
=90°+1/2*50°
=90°+25°
=115°∠BOC=180°-(∠OBC+∠OCB)
=180°-(1/2∠ABC+1/2∠ACB)
=180°-(1/2)(∠ABC+∠ACB)
=180°-(1/2)(180-∠A)
=180°-90°+1/2∠A
=90°+1/2∠A
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