如图,在△ABC中,∠A=60°.(1)若∠ABC和∠ACB的角平分线交于点O(如图1),求∠BOC的度数;(2)若∠ABC和∠ACB外角的平分线交于点D(如图2),求∠BDC的度数.
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![如图,在△ABC中,∠A=60°.(1)若∠ABC和∠ACB的角平分线交于点O(如图1),求∠BOC的度数;(2)若∠ABC和∠ACB外角的平分线交于点D(如图2),求∠BDC的度数.](/uploads/image/z/1254438-54-8.jpg?t=%E5%A6%82%E5%9B%BE%2C%E5%9C%A8%E2%96%B3ABC%E4%B8%AD%2C%E2%88%A0A%3D60%C2%B0.%281%29%E8%8B%A5%E2%88%A0ABC%E5%92%8C%E2%88%A0ACB%E7%9A%84%E8%A7%92%E5%B9%B3%E5%88%86%E7%BA%BF%E4%BA%A4%E4%BA%8E%E7%82%B9O%28%E5%A6%82%E5%9B%BE1%29%2C%E6%B1%82%E2%88%A0BOC%E7%9A%84%E5%BA%A6%E6%95%B0%EF%BC%9B%EF%BC%882%EF%BC%89%E8%8B%A5%E2%88%A0ABC%E5%92%8C%E2%88%A0ACB%E5%A4%96%E8%A7%92%E7%9A%84%E5%B9%B3%E5%88%86%E7%BA%BF%E4%BA%A4%E4%BA%8E%E7%82%B9D%EF%BC%88%E5%A6%82%E5%9B%BE2%EF%BC%89%2C%E6%B1%82%E2%88%A0BDC%E7%9A%84%E5%BA%A6%E6%95%B0.)
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如图,在△ABC中,∠A=60°.(1)若∠ABC和∠ACB的角平分线交于点O(如图1),求∠BOC的度数;(2)若∠ABC和∠ACB外角的平分线交于点D(如图2),求∠BDC的度数.
如图,在△ABC中,∠A=60°.(1)若∠ABC和∠ACB的角平分线交于点O(如图1),求∠BOC的度数;
(2)若∠ABC和∠ACB外角的平分线交于点D(如图2),求∠BDC的度数.
如图,在△ABC中,∠A=60°.(1)若∠ABC和∠ACB的角平分线交于点O(如图1),求∠BOC的度数;(2)若∠ABC和∠ACB外角的平分线交于点D(如图2),求∠BDC的度数.
∵∠A=60°
∴∠ABC+∠ACB=120°
∴∠BOC=180°- ½(∠ABC+∠ACB)=120°.
∵∠ABC和∠ACB的平分线BD、CE相交于点O,
∴∠1=∠2,∠3=∠4,
∴∠2+∠4= ½(180°-∠A)= ½(180°-60°)=60°,
故∠BOC=180°-(∠2+∠4)=180°-60°=120°