meifei1981
幼苗
共回答了15个问题采纳率:93.3% 举报
答:
△ABD是等腰直角三角形,三种情况:
1)AB是斜边,Ad=Bd=√10
cos∠ABC=BC/AB=2/2√5=√5/5
sin∠ABC=AC/AB=4/2√5=2√5/5
cos∠CBd=cos(∠ABC+45°)
=(√2/2)*(√5/5-2√5/5)
=-√10/10
所以:
Cd^2=BC^2+Bd^2-2BC*Bd*cos∠CBd
=4+10-4√10*(-√10/10)
=14+4
=18
所以:Cd=3√2
2)AB是直角边,A是直角顶点,AB=AD=2√5,BD=2√10
cos∠CAD
=cos(∠CAB+90°)
=-sin∠CAB
=-BC/AB
=-√5/5
所以:
CD^2=AC^2+AD^2-2AC*AD*cos∠CAD
=16+20-8*2√5*(-√5/5)
=36+16
=52
所以:CD=2√13
3)AB是直角边,B是直角顶点,AB=BD=2√5,AD‘=2√10
cos∠CBD'=-sin∠CBA=-2√5/5
CD'^2=BC^2+BD'^2-2BC*BD'*cos∠CBD'
=4+20-4*2√5*(-2√5/5)
=24+16
=40
所以:CD'=2√10
综上所述,CD=3√2或者CD=2√13或者CD=2√10
1年前
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