Đáp án:
$\begin{array}{l}
a)\widehat B + \widehat C = {90^0}\\
\Leftrightarrow \cot \widehat C = \tan \widehat B = \sqrt 2 \\
\Leftrightarrow \tan \widehat C = \dfrac{1}{{\cot \widehat C}} = \dfrac{1}{{\sqrt 2 }} = \dfrac{{\sqrt 2 }}{2}\\
Do:\dfrac{1}{{{{\sin }^2}\widehat C}} = {\cot ^2}\widehat C + 1 = 3\\
\Leftrightarrow \sin \widehat C = \dfrac{1}{{\sqrt 3 }} = \dfrac{{\sqrt 3 }}{3}\\
\Leftrightarrow \cos \widehat C = \dfrac{{\sin \widehat C}}{{\tan \widehat C}} = \dfrac{{\dfrac{{\sqrt 3 }}{3}}}{{\dfrac{{\sqrt 2 }}{2}}} = \dfrac{{\sqrt 6 }}{3}\\
Vậy\,\sin \widehat C = \dfrac{{\sqrt 3 }}{3};\cos \widehat C = \dfrac{{\sqrt 6 }}{3};\tan \widehat C = \dfrac{{\sqrt 2 }}{2};\cot \widehat C = \sqrt 2 \\
b)Trong:\Delta ABH:\widehat H = {90^0};\\
\tan \widehat B = \dfrac{{AH}}{{BH}} = \sqrt 2 \\
\Leftrightarrow BH = \dfrac{{2\sqrt 3 }}{{\sqrt 2 }} = \sqrt 6 \left( {cm} \right)\\
Theo\,Pytago:A{B^2} = A{H^2} + B{H^2}\\
\Leftrightarrow AB = \sqrt {{{\left( {2\sqrt 3 } \right)}^2} + 6} = 3\sqrt 2 \left( {cm} \right)\\
Trong:\Delta ACH:\widehat H = {90^0}\\
\tan \widehat C = \dfrac{{\sqrt 2 }}{2} = \dfrac{{AH}}{{CH}}\\
\Leftrightarrow CH = AH.\sqrt 2 = 2\sqrt 6 \left( {cm} \right)\\
\Leftrightarrow \left\{ \begin{array}{l}
BC = BH + CH = \sqrt 6 + 2\sqrt 6 = 3\sqrt 6 \left( {cm} \right)\\
AC = \sqrt {{{\left( {2\sqrt 3 } \right)}^2} + {{\left( {2\sqrt 6 } \right)}^2}} = 6\left( {cm} \right)
\end{array} \right.\\
Vậy\,AB = 3\sqrt 2 cm;AC = 6cm;BC = 3\sqrt 6 cm
\end{array}$
