\(a,BC=\sqrt{AB^2+AC^2}=15\left(cm\right)\left(pytago\right)\)
Áp dụng HTL:
\(\left\{{}\begin{matrix}AB^2=BH\cdot BC\\AC^2=CH\cdot BC\\AH^2=BH\cdot CH\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}BH=\dfrac{AB^2}{BC}=5,4\left(cm\right)\\CH=\dfrac{AC^2}{BC}=9,6\left(cm\right)\\AH=\sqrt{5,4\cdot9,6}=51,84\left(cm\right)\end{matrix}\right.\)
\(b,\sin B=\cos C=\dfrac{AC}{BC}=\dfrac{4}{5}\\ \cos B=\sin C=\dfrac{AB}{BC}=\dfrac{3}{5}\\ \tan B=\cot C=\dfrac{AC}{AB}=\dfrac{4}{3}\\ \cot B=\tan C=\dfrac{AB}{AC}=\dfrac{3}{4}\)
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