如图一在△abc中,∠abc,∠acb角平分线交于点o,则∠boc=90+½∠a=½×180°+½∠a;如图二,在△abc中,∠abc,∠acb的三等分线交于o1,o2,则∠bo1c=2/3×180°+1/3∠a;∠bo2c=1/3×180°2/3∠a,更具以上信息,
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![如图一在△abc中,∠abc,∠acb角平分线交于点o,则∠boc=90+½∠a=½×180°+½∠a;如图二,在△abc中,∠abc,∠acb的三等分线交于o1,o2,则∠bo1c=2/3×180°+1/3∠a;∠bo2c=1/3×180°2/3∠a,更具以上信息,](/uploads/image/z/754849-1-9.jpg?t=%E5%A6%82%E5%9B%BE%E4%B8%80%E5%9C%A8%E2%96%B3abc%E4%B8%AD%2C%E2%88%A0abc%2C%E2%88%A0acb%E8%A7%92%E5%B9%B3%E5%88%86%E7%BA%BF%E4%BA%A4%E4%BA%8E%E7%82%B9o%2C%E5%88%99%E2%88%A0boc%3D90%2B%26%23189%3B%E2%88%A0a%3D%26%23189%3B%C3%97180%C2%B0%2B%26%23189%3B%E2%88%A0a%EF%BC%9B%E5%A6%82%E5%9B%BE%E4%BA%8C%2C%E5%9C%A8%E2%96%B3abc%E4%B8%AD%2C%E2%88%A0abc%2C%E2%88%A0acb%E7%9A%84%E4%B8%89%E7%AD%89%E5%88%86%E7%BA%BF%E4%BA%A4%E4%BA%8Eo1%2Co2%2C%E5%88%99%E2%88%A0bo1c%3D2%2F3%C3%97180%C2%B0%2B1%2F3%E2%88%A0a%EF%BC%9B%E2%88%A0bo2c%3D1%2F3%C3%97180%C2%B02%2F3%E2%88%A0a%2C%E6%9B%B4%E5%85%B7%E4%BB%A5%E4%B8%8A%E4%BF%A1%E6%81%AF%2C)
如图一在△abc中,∠abc,∠acb角平分线交于点o,则∠boc=90+½∠a=½×180°+½∠a;如图二,在△abc中,∠abc,∠acb的三等分线交于o1,o2,则∠bo1c=2/3×180°+1/3∠a;∠bo2c=1/3×180°2/3∠a,更具以上信息,
如图一在△abc中,∠abc,∠acb角平分线交于点o,则∠boc=90+½∠a=½×180°+½∠a;
如图二,在△abc中,∠abc,∠acb的三等分线交于o1,o2,则∠bo1c=2/3×180°+1/3∠a;∠bo2c=1/3×180°2/3∠a,更具以上信息,回答下列问题:
(1)你能猜出它的规律嘛?(n等分时,内部n-1个点),∠bo1c=( ),∠bon-1c=( ),用n的代数式表示.
(2)根据你的猜想,取n=4时,说明∠bo3c的表达示成立.
我先说:注意哦,我要详细的讲解,而不是只有答案,我们可能要考这个,大师们!
如图一在△abc中,∠abc,∠acb角平分线交于点o,则∠boc=90+½∠a=½×180°+½∠a;如图二,在△abc中,∠abc,∠acb的三等分线交于o1,o2,则∠bo1c=2/3×180°+1/3∠a;∠bo2c=1/3×180°2/3∠a,更具以上信息,
(1)在图2 中,n=3时,∠BO1C=180°-1/3(∠B﹢∠C)
=180°-1/3(180°-∠A)
=180°-1/3*180°+1/3∠A
=2/3*180°+1/3∠A
∠BO2C=180°-2/3(∠B﹢∠C)
=180°-2/3(180°-∠A)
=180°-2/3*180°+2/3∠A
=1/3*180°+1/3∠A
n等分时 ∠BO1C=180°-1/n*(∠B﹢∠C)
=180°-1/n*(180°-∠A)
=180°-1/n*180°+1/n*∠A
=(n-1)/n*180°+1/n*∠A
∠BOn-1C=180°-(n-1)/n*(∠B﹢∠C)
=180°-(n-1)/n(180°-∠A)
=180°-(n-1)/n*180°+(n-1)/n*∠A
=1/n*180°+(n-1)/n∠A
(2) 当n=4时,∠BO3C=180°-3/4(∠B﹢∠C)
=180°-3/4(180°-∠A)
=180°-3/4*180°+3/4∠A
=1/4*180°+3/4∠A
根据三角形内角和180°,作一点推算即可。
∠boc=180 - 1/2(∠B+∠C)=180 -1/2(180-∠a)=½×180°+½∠a;
∠bo1c=180 - 2/3(∠B+∠C)=180 -2/3(180-∠a)=1/3×180°+2/3 ∠a;