Giải thích các bước giải:

 A)

$\begin{array}{l}
1)\cos {34^0} = \sin {56^0};\cos {29^0} = \sin {61^0}\\
\sin {56^0} < \sin {59^0} < \sin {61^0} < \sin {73^0}\\
 \Rightarrow \cos {34^0} < \sin {59^0} < \cos {29^0} < \sin {73^0}\\
2)\cot {48^0} = \tan {42^0};\cot {12^0} = \tan {78^0}\\
\tan {42^0} < \tan {64^0} < \tan {76^0} < \tan {78^0}\\
 \Rightarrow \cot {48^0} < \tan {64^0} < \tan {76^0} < \cot {12^0}\\
3)\cos {19^0} = \sin {71^0};\cos {54^0} = \sin {36^0}\\
\sin {36^0} < \sin {56^0} < \sin {65^0} < \sin {71^0}\\
 \Rightarrow \cos {54^0} < \sin {56^0} < \sin {65^0} < \cos {19^0}\\
\tan {71^0} = \dfrac{{\sin {{71}^0}}}{{\cos {{71}^0}}} > \sin {71^0}\left( {0 < \cos {{71}^0} < 1} \right)\\
 \Rightarrow \cos {54^0} < \sin {56^0} < \sin {65^0} < \cos {19^0} < \tan {71^0}\\
4)\cot {67^0} = \tan {23^0};\cot {18^0} = \tan {72^0}\\
\tan {23^0} < \tan {52^0} < \tan {62^0} < \tan {72^0}\\
 \Rightarrow \cot {67^0} < \tan {52^0} < \tan {62^0} < \cot {18^0}\\
\sin {23^0} = \tan {23^0}.\cos {23^0} < \tan {23^0}\left( {\cos {{23}^0} < 1} \right)\\
 \Rightarrow \sin {23^0} < \cot {67^0}\\
 \Rightarrow \sin {23^0} < \cot {67^0} < \tan {52^0} < \tan {62^0} < \cot {18^0}
\end{array}$

B) 

1) Ta có:

$\begin{array}{l}
\Delta ABC;\widehat A = {90^0};\widehat B = {50^0};BC = 8\\
 \Rightarrow \left\{ \begin{array}{l}
\widehat C = {90^0} - \widehat B = {40^0}\\
AC = BC.\sin \widehat B = 8.\sin {50^0} = 6,1\\
AB = BC.\sin \widehat C = 8.\sin {40^0} = 5,1
\end{array} \right.
\end{array}$

Vậy ${\widehat C = {{40}^0};AC = 6,1;AB = 5,1}$

2) 

Ta có:

$\begin{array}{l}
\Delta ABC;\widehat A = {90^0};AH \bot BC = H\\
 \Rightarrow A{H^2} = BH.HC\\
 \Rightarrow A{H^2} + B{H^2} = BH.HC + B{H^2}\\
 \Rightarrow A{H^2} + B{H^2} = BH\left( {HC + BH} \right) = BH.BC\\
 \Rightarrow A{H^2} + B{H^2} = BH.BC
\end{array}$

3) 

Ta có:

$\widehat {EAF} = \widehat {AEF} = \widehat {AFE} = {90^0}$

$\to AEHF$ là hình chữ nhật.

$ \Rightarrow \widehat {FEA} = \widehat {EAH} = \widehat {BAH} = \widehat {BCA} = {40^0}$

Vậy $\widehat {FEA} = {40^0}$

4) 

Ta có:

$\begin{array}{l}
\Delta ABC;\widehat A = {90^0}\\
 \Rightarrow {S_{\Delta ABC}} = \dfrac{1}{2}AB.AC = \dfrac{1}{2}BC.\sin C.BC.\sin B = \dfrac{{B{C^2}.\sin B.\sin C}}{2}
\end{array}$