Theo định lý cosin ta có
\(AD^2=AM^2+MD^2-2.MA.MD.cos\widehat{ÀMD}\)
Xé \(\Delta ABM\)có \(BM=\frac{a}{2}\)
\(AM=\sqrt{AB^2+BM^2}=\sqrt{a^2+\left(\frac{a}{2}\right)^2}=\frac{\sqrt{5}a}{2}\)
Xét \(\Delta DCM\)có \(CM=\frac{a}{2}\)
\(\Rightarrow DM=\sqrt{DC^2+CM^2}=\sqrt{a^2+\left(\frac{a}{2}\right)^2}=\frac{\sqrt{5}a}{2}\)
\(\Rightarrow\cos\widehat{AMD}=\frac{AM^2+MD^2-AD^2}{2.MA.MD}=\frac{\frac{5a^2}{4}+\frac{5a^2}{4}-a^2}{\frac{\sqrt{5}a}{2}.\frac{\sqrt{5}a}{2}}=\frac{3}{5}\)
Vậy \(\cos\widehat{AMD}=\frac{3}{5}\)
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