a) $AC = \sqrt{BC^{2} - AB^{2}} = \sqrt{16^{2} - 12^{2}} = 4\sqrt{7} \, cm$
$\begin{cases}sin\widehat{B} = \dfrac{AC}{BC} = \dfrac{4\sqrt{7}}{16} = \dfrac{\sqrt{7}}{4}\\cos\widehat{B} = \dfrac{AB}{BC} = \dfrac{12}{16}=\dfrac{3}{4}\\tan\widehat{B} = \dfrac{AC}{AB} = \dfrac{\dfrac{4\sqrt{7}}{16}}{\dfrac{12}{16}} = \dfrac{\sqrt{7}}{3}\\cot\widehat{B} = \dfrac{1}{tan\widehat{B}} = \dfrac{1}{\dfrac{\sqrt{7}}{3}}=\dfrac{3}{\sqrt{7}}=\dfrac{3\sqrt{7}}{7}\end{cases}$
⇒ $\begin{cases}sin\widehat{C} = cos\widehat{B} = \dfrac{3}{4}\\cos\widehat{C} = sin\widehat{B} = \dfrac{\sqrt{7}}{4}\\tan\widehat{C} = cot\widehat{B} = \dfrac{3\sqrt{7}}{7} \\cot\widehat{C} =tan\widehat{B} = \dfrac{\sqrt{7}}{3}\end{cases}$
b) $BC = BH + CH = 16 + 9 = 25 \, cm$
$AB = \sqrt{BH.BC} = 20 \, cm$
$AC = \sqrt{CH.BC} = 15 \, cm$
$\begin{cases}sin\widehat{B} = \dfrac{AC}{BC} = \dfrac{15}{25} = \dfrac{3}{5}\\cos\widehat{B} = \dfrac{AB}{BC} = \dfrac{20}{25}=\dfrac{4}{5}\\tan\widehat{B} = \dfrac{AC}{AB} = \dfrac{15}{20} = \dfrac{3}{4}\\cot\widehat{B} = \dfrac{1}{tan\widehat{B}} = \dfrac{1}{\dfrac{3}{4}}=\dfrac{4}{3}\end{cases}$
⇒ $\begin{cases}sin\widehat{C} = cos\widehat{B} = \dfrac{4}{5}\\cos\widehat{C} = sin\widehat{B} = \dfrac{\sqrt{3}}{5}\\tan\widehat{C} = cot\widehat{B} = \dfrac{4}{3} \\cot\widehat{C} =tan\widehat{B} = \dfrac{3}{4}\end{cases}$
c) $BC = \sqrt{AB^{2} + AC^{2}} = \sqrt{6^{2} + 8^{2}} = 10 \, cm$
$\begin{cases}sin\widehat{B} = \dfrac{AC}{BC} = \dfrac{8}{10} = \dfrac{4}{5}\\cos\widehat{B} = \dfrac{AB}{BC} = \dfrac{6}{10}=\dfrac{3}{5}\\tan\widehat{B} = \dfrac{AC}{AB} = \dfrac{8}{6} = \dfrac{4}{3}\\cot\widehat{B} = \dfrac{1}{tan\widehat{B}} = \dfrac{1}{\dfrac{4}{3}}=\dfrac{3}{4}\end{cases}$
⇒ $\begin{cases}sin\widehat{C} = cos\widehat{B} = \dfrac{3}{5}\\cos\widehat{C} = sin\widehat{B} = \dfrac{\sqrt{4}}{5}\\tan\widehat{C} = cot\widehat{B} = \dfrac{3}{4} \\cot\widehat{C} =tan\widehat{B} = \dfrac{4}{3}\end{cases}$
