题目内容
如图,在?ABCD中,AB:AD=3:2,∠ADB=60°,那么cos∠A的值等于( )
A.
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B.
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C.
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D.
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【答案】分析:作出辅助线,构造直角三角形,运用三角形面积相等,求出三角形的高,然后运用sin2α+cos2α=1,根据题中所给的条件,在直角三角形中解题,由角的余弦值与三角形边的关系求解.
解答:
解:作AF⊥DB于F,作DE⊥AB于E.
设DF=x,则AD=2x,
∵∠ADB=60°,
∴AF=
x,
又∵AB:AD=3:2,
∴AB=3x,于是BF=
x,
∴3x•DE=(
+1)x•
x,
DE=
x,sin∠A=
,
cos∠A=
=
.
故选A.
点评:考查三角函数的定义及三角形面积公式.
解答:

设DF=x,则AD=2x,
∵∠ADB=60°,
∴AF=

又∵AB:AD=3:2,
∴AB=3x,于是BF=

∴3x•DE=(


DE=


cos∠A=


故选A.
点评:考查三角函数的定义及三角形面积公式.

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