a, Ta có \(HC=BC-BH=6\left(cm\right)\)
Áp dụng HTL: \(\left\{{}\begin{matrix}AB^2=BH\cdot BC=16\\AC^2=CH\cdot BC=48\\AH^2=BH\cdot CH=12\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}AB=4\left(cm\right)\\AC=4\sqrt{3}\left(cm\right)\\AH=2\sqrt{3}\left(cm\right)\end{matrix}\right.\)
b, \(\tan\widehat{BAH}=\dfrac{BH}{AH}=\dfrac{2}{2\sqrt{3}}=\dfrac{\sqrt{3}}{3}=\tan30^0\Leftrightarrow\widehat{BAH}=30^0\)
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