Đáp án:
$\begin{array}{l}
11)\\
\Delta ABH \bot tai\,H:\\
\cos A = \dfrac{{AH}}{{AB}}\\
\Rightarrow \cos {60^0} = \dfrac{{AH}}{4} \Rightarrow AH = 2\left( {cm} \right)\\
\Rightarrow HC = AC - AH = 4\left( {cm} \right)\\
Do:A{H^2} + H{B^2} = A{B^2}\\
\Rightarrow H{B^2} = {4^2} - {2^2} = 12\\
\Rightarrow B{C^2} = H{B^2} + H{C^2} = 12 + 16 = 28\\
\Rightarrow BC = 2\sqrt 7 \left( {cm} \right)\\
15)\\
a)\cos B = \dfrac{{AB}}{{BC}} = \dfrac{{12}}{{16}} = \dfrac{3}{4}\\
\Rightarrow \sin B = \dfrac{{AC}}{{BC}} = \dfrac{{\sqrt 7 }}{4}\\
\Rightarrow \tan B = \dfrac{{\sin B}}{{\cos B}} = \dfrac{{\sqrt 7 }}{3}\\
\Rightarrow \cot B = \dfrac{{3\sqrt 7 }}{7}\\
\Rightarrow \sin C = {\mathop{\rm cosB}\nolimits} = \dfrac{3}{4}\\
\cos C = \sin B = \dfrac{{\sqrt 7 }}{4}\\
\tan C = \cot B;\cot C = \tan B\\
b)\cos B = \dfrac{{BH}}{{AB}} = \dfrac{5}{{13}} = \sin C\\
\Rightarrow \sin B = \cos C = \dfrac{{12}}{{13}}\\
\Rightarrow \tan B = {\mathop{\rm cotC}\nolimits} = \dfrac{{12}}{5}\\
\Rightarrow \cot B = \tan C = \dfrac{5}{{12}}\\
c)A{H^2} = BH.CH\\
\Rightarrow AH = 12\\
\Rightarrow \tan B = \cot C = \dfrac{{AH}}{{BH}} = \dfrac{{12}}{{16}} = \dfrac{3}{4}\\
\cot B = \tan C = \dfrac{4}{3}\\
d)\tan B = \cot C = \dfrac{{AC}}{{AB}} = \dfrac{4}{3}\\
\cot B = \tan C = \dfrac{3}{4}
\end{array}$
