Đáp án: 19.D 20.C
19 Do góc A lớn hơn 90 độ nên trực tâm H nằm ngoài tam giác ABC
Ta có:
$\begin{array}{l}
\widehat {AHB} + \widehat {BHC} + \widehat {CHA} = 2\widehat {BHC} = 2\left( {\widehat {AHB} + \widehat {CHA}} \right)\\
Do:\left\{ \begin{array}{l}
\widehat {AHB} = {90^0} - \widehat {{A_1}}\\
\widehat {CHA} = {90^0} - \widehat {{A_2}}
\end{array} \right.\\
\Rightarrow \widehat {AHB} + \widehat {CHA} = {180^0} - \left( {\widehat {{A_2}} + \widehat {{A_1}}} \right) = {180^0} - \widehat A = {75^0}\\
\Rightarrow \widehat {AHB} + \widehat {BHC} + \widehat {CHA} = {2.75^0} = {150^0}\\
\Rightarrow \alpha = {150^0}\\
\Rightarrow cos\alpha = - \frac{{\sqrt 3 }}{2}
\end{array}$
20)
$\begin{array}{l}
\left[ \begin{array}{l}
- 3 < m + 2 < - 1\\
- 1 < m - 1 < 2\\
2 < m - 1 < 4
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
- 5 < m < - 3\\
0 < m < 3\\
3 < m < 5
\end{array} \right.\\
\Rightarrow \left| m \right| < 5;m \ne 0
\end{array}$
