题目内容
已知△ABC,若对任意t∈R,
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A.锐角三角形
B.钝角三角形
C.直角三角形
D.答案不确定
【答案】分析:则根据向量的减法的几何意义,由|
-t
|≥|
|对一切实数t都成立可得|
|≥|
|,进而得到AC⊥BC,即可得到三角形为直角三角形.
解答:解:令
=
-t
,则根据向量的减法的几何意义可得M在BC上,
由|
-t
|≥|
|对一切实数t都成立可得:|
|≥|
|,
∴AC⊥BC,
则△ABC为直角三角形.
故选C
点评:本题是一道构造非常巧妙的试题,解题的关键是由|
-t
|≥|
|对一切实数t都成立可得到AC为A到BC的距离.
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解答:解:令
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由|
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∴AC⊥BC,
则△ABC为直角三角形.
故选C
点评:本题是一道构造非常巧妙的试题,解题的关键是由|

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