Đáp án:
$\begin{array}{l}
a){\mathop{\rm sinB}\nolimits} = \dfrac{{AC}}{{BC}}\\
\sin C = \dfrac{{AB}}{{BC}}\\
\Rightarrow \dfrac{{\sin B}}{{\sin C}} = \dfrac{{AC}}{{BC}}:\dfrac{{AB}}{{BC}} = \dfrac{{AC}}{{AB}}\\
b)Trong\,\Delta ABH \bot tai\,H\\
Theo\,Pytago:\\
B{H^2} = A{B^2} - A{H^2} = 7,{5^2} - {6^2}\\
\Rightarrow BH = 4,5\left( {cm} \right)\\
\Rightarrow \cos B = \dfrac{{BH}}{{AB}} = \dfrac{{4,5}}{{7,5}} = \dfrac{3}{5}\\
\Rightarrow \sin B = \dfrac{{AH}}{{AB}} = \dfrac{4}{5} = \cos C\\
c)\cot B = \dfrac{{BH}}{{AH}}\\
\cot C = \dfrac{{CH}}{{AH}}\\
\Rightarrow \dfrac{{\cot B}}{{\cot C}} = \dfrac{{BH}}{{CH}} = 3\\
d)AM = \dfrac{1}{2}BC\\
Do:AM = AC\\
\Rightarrow AC = \dfrac{1}{2}BC\\
\Rightarrow BC = 2AC\\
\Rightarrow A{B^2} + A{C^2} = 4A{C^2}\\
\Rightarrow A{B^2} = 3A{C^2}\\
Do:\tan B = \dfrac{{AC}}{{AB}};\tan C = \dfrac{{AB}}{{AC}}\\
\Rightarrow \dfrac{{\tan B}}{{\tan C}} = \dfrac{{A{C^2}}}{{A{B^2}}} = \dfrac{1}{3}
\end{array}$
