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

\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}

\begin{document}\everymath{\displaystyle}
	
\thispagestyle{empty}
\boxit{4pt}{Minitest 10}\rightarrowfill\quad\boxit{3pt}{Corrigé}\doublespacing

\bigskip

Dans le triangle $CAN$ rectangle en $A$, on a : $NA=3$ et $CA=4$, calculer $NC$

\begin{minipage}{4cm}
\begin{tikzpicture}[scale=0.8]
	
	% Points
	\coordinate (A) at (0,0);
	\coordinate (N) at (3,0);
	\coordinate (C) at (0,4);
	
	% Triangle
	\draw (A) -- (N) -- (C) -- cycle;
	
	% Angle droit en A
	\draw (0,0) rectangle (0.4,0.4);
	
	% Labels des points
	\node[below left] at (A) {$A$};
	\node[below right] at (N) {$N$};
	\node[above left] at (C) {$C$};
	
	% Longueurs
	\node[below] at (1.5,0) {$3$};
	\node[left] at (0,2) {$4$};
	
\end{tikzpicture}
\end{minipage}
\begin{minipage}{12cm}\red
	Dans le triangle $CAN$ rectangle $A$ d'après le théorème de Pythagore, on a: $CN^2  =AN^2+ AC^2 = 3^2+4^2 = 9+16 = 25$ 
	
	donc $AN = \sqrt{25} = 5$
\end{minipage}

Dans le triangle $ZUT$ rectangle en $U$, on a : $ZT=10$ et $\widehat{UZT} = 30^\circ$, calculer $UT$

\begin{minipage}{4cm}
	\begin{tikzpicture}[scale=0.5]
		
		% Points
		\coordinate (U) at (0,0);
		\coordinate (T) at (5,0);
		\coordinate (Z) at (0,{5*sqrt(3)});
		
		% Triangle
		\draw (Z) -- (U) -- (T) -- cycle;
		
		% Angle droit en U (bien aligné)
		\draw (0.4,0) -- (0.4,0.4) -- (0,0.4);
		
		% Angle 30° en Z
		\draw ($(Z)+(0,-1.6)$) arc (-90:-60:1.6);
		\node at ($(Z)+(0.6,-2.2)$) {$30^\circ$};
		
		% Labels
		\node[left] at (Z) {$Z$};
		\node[below left] at (U) {$U$};
		\node[below] at (T) {$T$};
		
		% Longueurs
		\node[right] at ($(Z)!0.5!(T)$) {$10$};
		
	\end{tikzpicture}
	
	
\end{minipage}
\begin{minipage}{12cm}\red
	Dans le triangle $ZUT$, rectangle $U$, on a : $\sin \widehat{TZU} =\dfrac{UT}{ZT}$ 
	
	donc $\sin 30^\circ = \dfrac{UT}{10}$ donc $UT = 10\times \sin{30^\circ} = 10\times \dfrac12 = 5$
\end{minipage}

\'Ecrire le tableau des angles remarquables du cours avec leurs sinus et leurs cosinus

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

\newpage

\thispagestyle{empty}
\boxit{4pt}{Minitest 10}\rightarrowfill\quad\boxit{3pt}{Corrigé}\doublespacing

\bigskip

Dans le triangle $BEL$ rectangle en $B$, on a : $BL=4$ et $BE=3$, calculer $EL$

\begin{minipage}{4cm}
	\begin{tikzpicture}[scale=0.8]
		
		% Points
		\coordinate (B) at (0,0);
		\coordinate (E) at (3,0);
		\coordinate (L) at (0,4);
		
		% Triangle
		\draw (B) -- (E) -- (L) -- cycle;
		
		% Angle droit en A
		\draw (0,0) rectangle (0.4,0.4);
		
		% Labels des points
		\node[below left] at (B) {$B$};
		\node[below right] at (E) {$E$};
		\node[above left] at (L) {$L$};
		
		% Longueurs
		\node[below] at (1.5,0) {$3$};
		\node[left] at (0,2) {$4$};
		
	\end{tikzpicture}
\end{minipage}
\begin{minipage}{12cm}\red
	Dans le triangle $BEL$ rectangle $B$ d'après le théorème de Pythagore, on a: $EL^2  =BL^2+ BE^2 = 3^2+4^2 = 9+16 = 25$ 
	
	donc $EL = \sqrt{25} = 5$
\end{minipage}


Dans le triangle $GRO$ rectangle en $O$, on a : $OR=12$ et $\widehat{OGR} = 60^\circ$, calculer $GR$

\begin{minipage}{6.5cm}
	\begin{tikzpicture}[scale=0.5]
		
		% Points
		\coordinate (O) at (0,0);
		\coordinate (R) at (12,0);
		\coordinate (G) at (0,{12/sqrt(3)});
		
		% Triangle
		\draw (G) -- (R) -- (O) -- cycle;
		
		% Angle droit en U (bien aligné)
		\draw (0.4,0) -- (0.4,0.4) -- (0,0.4);
		
		% Angle 60° en Z
		\draw ($(G)+(0,-1.8)$) arc (-90:-30:1.8);
		\node at ($(G)+(1.2,-2.0)$) {$60^\circ$};
		
		% Labels
		\node[left] at (G) {$G$};
		\node[below right] at (R) {$R$};
		\node[below] at (O) {$O$};
		
		% Longueurs
		\node[below] at (6,0) {$12$};
		
	\end{tikzpicture}
	
	
\end{minipage}
\begin{minipage}{12cm}\red
	Dans le triangle $GRO$, rectangle $O$, on a : $\sin \widehat{OGR} =\dfrac{OR}{GR}$ 
	
	donc $\sin 60^\circ = \dfrac{12}{GR}$ donc $GR\times \sin{60^\circ}= 12$ \\
	
	donc  $GR = \dfrac{12}{\sin(60^\circ)} = \dfrac{12}{\frac{\sqrt{3}}{2}} = \dfrac{24}{\sqrt{3}} = \dfrac{24\sqrt{3}}{3} = 8\sqrt{3}$
\end{minipage}

\'Ecrire le tableau des angles remarquables du cours avec leurs sinus et leurs cosinus

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

\end{document} 