\(BC=HB+HC=35\)
\(AM=\frac{1}{2}BC=\frac{35}{2}\) \(\Rightarrow HM=HB-BM=HB-\frac{1}{2}BC=\frac{3}{2}\)
\(AH=\sqrt{HB.HC}=4\sqrt{19}\)
\(sin\widehat{HAM}=\frac{HM}{AM}=\frac{3}{35}\)
\(cos\widehat{HAM}=\frac{AH}{AM}=\frac{8\sqrt{19}}{35}\)
\(tan\widehat{HAM}=\frac{HM}{AH}=\frac{3}{8\sqrt{19}}\)