a) \(BC^2=AB^2+AC^2\left(Pitago\right)\)
\(\Rightarrow AC^2=BC^2-AB^2=100+36=136\)
\(\Rightarrow AC=2\sqrt[]{34}\left(cm\right)\)
\(tanB=\dfrac{AB}{AC}=\dfrac{6}{2\sqrt[]{34}}=\dfrac{3}{\sqrt[]{34}}\)
\(tanC=tan\left(90^o-B\right)=cotB=\dfrac{1}{tanB}=\dfrac{\sqrt[]{34}}{3}\)
\(tanB=\dfrac{AH}{BH}=\dfrac{3}{\sqrt[]{34}}\)
\(tanC=\dfrac{AH}{HC}=\dfrac{\sqrt[]{34}}{3}\)
\(\Rightarrow\dfrac{tanB}{tanC}=\dfrac{HC}{BH}=\dfrac{9}{34}\)
\(\Rightarrow\dfrac{HC}{9}=\dfrac{BH}{34}=\dfrac{HC+BH}{9+34}=\dfrac{10}{43}\)
\(\Rightarrow\left\{{}\begin{matrix}HC=9.\dfrac{10}{43}=\dfrac{90}{43}\\BH=34.\dfrac{10}{43}=\dfrac{340}{43}\end{matrix}\right.\)
\(AH^2=HB.HC=\dfrac{340}{43}.\dfrac{90}{43}\)
\(\Rightarrow AH=\dfrac{30\sqrt[]{34}}{43}\)
\(sin\widehat{HAC}=\dfrac{HC}{AC}=\dfrac{90}{43.2\sqrt[]{34}}=\dfrac{45}{43\sqrt[]{34}}\)
\(cos\widehat{HAC}=\dfrac{AH}{AC}=\dfrac{30\sqrt[]{34}}{43.2\sqrt[]{34}}=\dfrac{15}{43}\)