\documentclass[12pt,a4paper]{article}
\usepackage[latin1]{inputenc}
\usepackage[french]{babel}
\usepackage{amsmath,amssymb,makeidx}
\usepackage{amsfonts}
\usepackage{geometry}
\usepackage{array}
\usepackage{dsfont}
\usepackage{mathrsfs} 
\usepackage{setspace}
\usepackage{pstricks}
\geometry{verbose,tmargin=1cm,bmargin=1cm,lmargin=1cm,rmargin=1cm}

\def\boxit#1#2{\leavevmode\setbox1=\hbox{\kern#1{#2}\kern#1}%
\dimen1=\ht1 \advance\dimen1 by #1 \dimen2=\dp1 \advance\dimen2 by #1
\setbox1=\hbox{\vrule height\dimen1 depth\dimen2\box1\vrule}%
\setbox1=\vbox{\hrule\box1\hrule}%
\advance\dimen1 by .4pt \ht1=\dimen1
\advance\dimen2 by .4pt \dp1=\dimen2
\box1\relax}
\def\tvi{\vrule height 20pt depth  15pt width 0pt}
\def\tv{\tvi\vrule}
\def\taille{\vrule height 20pt depth 20pt width 0pt }

\newcommand{\vect}[1]{\mathchoice%
	{\overrightarrow{\displaystyle\mathstrut#1\,\,}}%
	{\overrightarrow{\textstyle\mathstrut#1\,\,}}%
	{\overrightarrow{\scriptstyle\mathstrut#1\,\,}}%
	{\overrightarrow{\scriptscriptstyle\mathstrut#1\,\,}}}
	
\newcommand{\N}{\mathds {N}}
\newcommand{\R}{\mathds {R}}

\setlength\parindent{0pt}

\newcolumntype{C}[1]{>{\centering\arraybackslash}m{#1}} % centrée et largeur
\newcolumntype{M}[1]{>{\centering\arraybackslash$}m{#1}<{$}}

\begin{document}\everymath{\displaystyle}
	
\thispagestyle{empty}
\boxit{4pt}{Minitest 11}\rightarrowfill\quad\boxit{3pt}{Corrigé}\doublespacing

\bigskip

A] \'Ecrire le tableau des angles remarquables du cours avec leurs sinus et leurs cosinus\vskip 0.5cm

\bigskip
\renewcommand{\arraystretch}{2}% augmente la hauteur des lignes
{\red
	\[
	\begin{tabular}{|*{6}{M{1.5cm}|}}
		\hline
		\text{Angle} & 0 & \frac{\pi}{6} & \frac{\pi}{4} & \frac{\pi}{3} & \frac{\pi}{2} \\
		\hline
		\sin & 0 & \frac{1}{2} & \frac{\sqrt{2}}{2} & \frac{\sqrt{3}}{2} & 1 \\
		\hline
		\cos & 1 & \frac{\sqrt{3}}{2} & \frac{\sqrt{2}}{2} & \frac{1}{2} & 0 \\
		\hline
	\end{tabular}
	\]
}

\vskip 1cm


B]
\begin{minipage}{5cm}
\psset{xunit=1cm,yunit=1cm,
	algebraic=true,dimen=middle,
	dotsize=3pt 0,linewidth=1pt,
	arrowsize=3pt 2,arrowinset=0.25}
\begin{pspicture*}(-3.5,-1)(1,4)
	\psline(0,2.7)(-2,0)(-3,0)(0,2.7)(0,0)(-0.2,0)(-0.2,0.2)(0,0.2)
	\psline[linestyle=dashed](0,0)(-2,0)
	\psdots(-3,0)(-2,0)(0,0)(0,2.7)(0,2.7)
	\rput[r](-3,-0.5){$A$}
	\rput[r](-2,-0.5){$B$}
	\rput[l](0,-0.5){$C$}
	\rput(0.4,2.8){$D$}
\end{pspicture*}
\end{minipage}
\begin{minipage}{12cm}
	Sachant que $AB= 50$m, que $\widehat{BAD} = 40^\circ$ et que $\widehat{CBD} = 50^\circ$
	
	1) Montrer que $CD = BC\times \tan 50^\circ$ et que $CD = AC\times \tan 40^\circ$
	
	{\red 
		
	Dans le triangle CBD rectangle en C, on a : $\tan \widehat{CBD} = \dfrac{CD}{BC}$ donc $CD = BC\times \tan \widehat{CBD}$ donc    $CD= BC\tan 50^\circ$\bigskip
	
	Dans le triangle CAD rectangle en C, on a : $\tan \widehat{CAD} = \dfrac{CD}{AC}$ donc $CD = AC\times \tan \widehat{CAD}$ donc    $CD= AC\tan 40^\circ$
	
	}
\end{minipage}

\vskip 1.5cm
	
	2) Sachant que $AC = 50+BC$ en déduire que : $BC\times \tan50^\circ-BC\times \tan40^\circ = 50\times \tan40^\circ$
	
	{\red  $AC = 50+BC$ et  $CD= AC\times \tan 40^\circ$ donc $CD= (50+BC)\times \tan 40^\circ = 50\times \tan40^\circ+BC \times \tan40^\circ$ 
	
	mais  $CD= BC\times \tan 50^\circ$ donc $ BC\times \tan 50^\circ = 50\times \tan40^\circ+BC\times \tan40^\circ$ 
	
	Et finalement : $BC\times \tan50^\circ-BC\times \tan40^\circ = 50\times \tan40^\circ$
	
	}

\vskip 1.5cm

	3) En déduire la longueur exacte de $BC$ puis celle de $CD$
	
	{\red 
	
	$BC\times \tan50^\circ-BC\times \tan40^\circ = 50\times \tan40^\circ$ donc $BC\times (\tan50^\circ-\tan40^\circ) = 50\times \tan40^\circ$
	
	donc $BC = \dfrac{ 50\times \tan40^\circ}{\tan50^\circ-\tan40^\circ}$
	
	Comme : $CD= BC\times \tan 50^\circ$, cela donne : \fbox{$CD = \dfrac{ 50\times \tan40^\circ\times \tan 50^\circ}{\tan50^\circ-\tan40^\circ}$}
	
	}
	


\end{document} 