有两种求法,一、连接AC,求出AM,CM,AC,cos(∠AMC)=(AM^2+CM^2-AC^2)/(2.AM.CM);二、求出AM、CM的直线斜率K1,K2,tan(∠AMC)=(K1-K2)/(1+K1.K2); A(0,2), B(-4,0), C(4,0), D(0,-6), M(m,m1);DE⊥AB, KDE=-1/KAB=-1/(2/4)=-2;(m1+6)/m=-2;(1)DM=AB=2√5; √((m1+6)^2+m^2)=2√5;(2)由(1), (2)解得m=-2, m1=-2;KAM=k1=(2-m1)/(-m)=2; KCM=k2=(-m1)/(4-m)=1/3;tan(∠AMC)=(k1...
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