\documentclass[a4paper,11pt]{article}

\usepackage[utf8]{inputenc}
\usepackage[T1]{fontenc}
\usepackage{lmodern}
\usepackage[french]{babel}
\usepackage[a4paper,top=3.2cm,bottom=2cm,left=2cm,right=2cm]{geometry}
\usepackage{graphicx}
\usepackage[export]{adjustbox}
\usepackage{fancyhdr}
\usepackage{qrcode}
\usepackage[table]{xcolor}
\usepackage{amsmath,amssymb}
\usepackage{enumitem}
% Puces nettes (babel-french remplace sinon la puce par un tiret « – » peu pro).
% Vaut pour TOUTES les fiches qui importent ce module. AtBeginDocument pour
% passer après la personnalisation des listes faite par babel-french.
\AtBeginDocument{%
  \setlist[itemize]{leftmargin=1.8em,itemsep=0.5ex,topsep=0.4ex,parsep=0pt}%
  \setlist[itemize,1]{label=\textcolor{red!75!black}{\small\textbullet}}%
  \setlist[itemize,2]{label=\textcolor{red!75!black}{\scriptsize\textbullet}}%
}
% Titres compacts (gain de place vertical, fiches plus denses) — vaut pour tous.
\usepackage{titlesec}
\titlespacing*{\section}{0pt}{1.4ex}{0.6ex}
\titlespacing*{\subsection}{0pt}{1.0ex}{0.3ex}
% multicol : listes denses (calcul mental) sur 2 colonnes — neutre tant qu'il
% n'est pas utilisé, dispo pour tous les générateurs.
\usepackage{multicol}
\setlength{\columnsep}{1.5em}
% needspace : empêche un titre d'exercice de rester orphelin en bas de page
% (réserve quelques lignes ; sinon le titre passe à la page suivante avec sa
% liste). Neutre tant qu'il n'est pas appelé — dispo pour tous les générateurs.
\usepackage{needspace}
\usepackage{tikz}
\usetikzlibrary{decorations.text}

\usepackage{tgheros}
\pagestyle{fancy}
\fancyhf{}
\setlength{\headheight}{75pt}
\setlength{\headsep}{10pt}

\newcommand{\logowouf}{%
\begin{tikzpicture}[baseline={(current bounding box.center)},scale=0.55,
                    line cap=round,line join=round]
    \path[use as bounding box] (-2.20,-1.45) rectangle (2.20,2.20);
    \path[
        postaction={decorate},
        decoration={
            text along path,
            text={|\fontfamily{qhv}\bfseries\fontsize{9}{9}\selectfont|site2wouf.fr},
            text align={center},
            raise=0.03ex
        }
    ] (-1.29,0.48) arc[start angle=160,end angle=20,radius=1.29];
    \begin{scope}[xshift=-0.06cm]
        \fill[black] (-0.66,0.14) ellipse (0.21 and 0.31);
        \fill[black] ( 0.66,0.14) ellipse (0.21 and 0.31);
        \fill[black] (-0.23,0.49) ellipse (0.22 and 0.35);
        \fill[black] ( 0.23,0.49) ellipse (0.22 and 0.35);
        \fill[black]
            (0,-0.10)
            .. controls (-0.36,-0.02) and (-0.63,-0.24) .. (-0.58,-0.58)
            .. controls (-0.52,-0.93) and (-0.16,-1.12) .. (0,-0.90)
            .. controls (0.16,-1.12) and (0.52,-0.93) .. (0.58,-0.58)
            .. controls (0.63,-0.24) and (0.36,-0.02) .. (0,-0.10)
            -- cycle;
    \end{scope}
\end{tikzpicture}%
}

\newcommand{\titre}{Trigonométrie}
\newcommand{\soustitre}{Cycle 4 --- Fiche n\textdegree 235}

\fancyhead[C]{%
\raisebox{4pt}{%
\begin{minipage}[c]{0.15\textwidth}
    \centering
    \logowouf
\end{minipage}
\begin{minipage}[c]{0.62\textwidth}
    \centering
    {\Large\bfseries \titre\par}
    {\small\color{gray} \soustitre}
\end{minipage}
\begin{minipage}[c]{0.21\textwidth}
    \centering
    \qrcode[height=1.5cm]{https://site2wouf.fr/exercices_trigonometrie.php?x=235}\\[2pt]
    {\small\ttfamily\bfseries\color{red} TRIG0235}
\end{minipage}%
}
}

\begin{document}

\sloppy

\needspace{4\baselineskip}
\section*{
Exercice 1}

 Dans le triangle TMG rectangle en T, on sait que :
\begin{itemize}
\item TM=9,5 cm
\item \(\widehat{MGT}\) =  23\textdegree{}
\end{itemize}

Après avoir fait un schéma, calcule la longueur du segment [GM]. (Arrondir au dixième)

\needspace{4\baselineskip}
\section*{
Exercice 2}

 Dans le triangle CBT rectangle en C, on sait que :
\begin{itemize}
\item CT=5 cm
\item BT=8,6 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{CBT}\).
\needspace{4\baselineskip}
\section*{
Exercice 3}

 Dans le triangle MAS rectangle en M, on sait que :
\begin{itemize}
\item MA=2,2 cm
\item AS=7,4 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{MSA}\).

\needspace{4\baselineskip}
\section*{
Exercice 4}

 Dans le triangle FSC rectangle en F, on sait que :
\begin{itemize}
\item FC=5,7 cm
\item \(\widehat{FSC}\) = 66\textdegree{}
\end{itemize}

Après avoir fait un schéma, calcule la longueur du segment [FS]. (Arrondir au dixième)

\needspace{4\baselineskip}
\section*{
Exercice 5}

 Dans le triangle ARG rectangle en A, on sait que :
\begin{itemize}
\item AR=3,6 cm
\item \(\widehat{RGA}\) =  38\textdegree{}
\end{itemize}

Après avoir fait un schéma, calcule la longueur du segment [AG]. (Arrondir au dixième)

\newpage
\section*{\textcolor{red}{Correction}}

\needspace{4\baselineskip}
\subsection*{
Exercice 1}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,3.040) -- (1.041,2.780) -- (1.043,2.520) -- (1.042,2.260) -- (1.061,2.003) -- (1.041,1.741) -- (1.053,1.481) -- (1.054,1.222) -- (1.050,0.960);
  \draw[black] (1.050,0.960) -- (1.397,0.973) -- (1.753,0.952) -- (2.101,0.954) -- (2.451,0.950) -- (2.802,0.950) -- (3.148,0.963) -- (3.501,0.958) -- (3.849,0.955) -- (4.202,0.968) -- (4.548,0.965) -- (4.900,0.963) -- (5.250,0.972) -- (5.600,0.952) -- (5.950,0.960);
  \draw[black] (5.950,0.960) -- (5.621,1.093) -- (5.293,1.237) -- (4.972,1.386) -- (4.642,1.511) -- (4.311,1.647) -- (3.985,1.782) -- (3.661,1.930) -- (3.334,2.068) -- (3.011,2.210) -- (2.684,2.341) -- (2.358,2.484) -- (2.027,2.615) -- (1.703,2.768) -- (1.379,2.901) -- (1.050,3.040);
  \draw[black] (1.050,1.168) -- (1.045,1.099) -- (1.053,1.028) -- (1.050,0.960);
  \draw[black] (1.050,0.960) -- (1.120,0.965) -- (1.187,0.961) -- (1.258,0.960);
  \draw[black] (1.050,1.168) -- (1.118,1.172) -- (1.188,1.170) -- (1.258,1.168);
  \draw[black] (1.258,1.168) -- (1.260,1.100) -- (1.260,1.030) -- (1.258,0.960);
  \draw[black] (5.576,1.119) -- (5.580,1.103) -- (5.572,1.091) -- (5.562,1.081) -- (5.562,1.066) -- (5.565,1.052) -- (5.552,1.041) -- (5.548,1.028) -- (5.560,1.012) -- (5.547,1.001) -- (5.554,0.987) -- (5.548,0.974) -- (5.548,0.960);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,0.827) {T};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,3.107) {M};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,0.827) {G};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {9,5 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.297,1.026) {23°};
\end{tikzpicture}\end{center}


 Dans le triangle TMG rectangle en T, on cherche une relation entre l'angle aigu   \(\widehat{TGM}\) son coté opposé et l'hypoténuse du triangle.

\par\addvspace{.5em}
sin(\(\widehat{TGM}\)) = \(\dfrac{TM}{MG}\)
\par\addvspace{.5em}
 D'où sin(23\textdegree{}) = \(\dfrac{9{,}5}{MG}\)
\par\addvspace{.5em}
On a donc MG =\(\dfrac{9{,}5}{\sin(23^{\circ})}\) \(\approx\) 24,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (2.500,3.400) -- (2.490,3.051) -- (2.491,2.699) -- (2.487,2.351) -- (2.501,2.000) -- (2.504,1.653) -- (2.504,1.300) -- (2.500,0.948) -- (2.500,0.600);
  \draw[black] (2.500,0.600) -- (2.751,0.593) -- (3.003,0.604) -- (3.250,0.598) -- (3.500,0.589) -- (3.753,0.593) -- (3.998,0.589) -- (4.253,0.609) -- (4.500,0.600);
  \draw[black] (4.500,0.600) -- (4.307,0.887) -- (4.095,1.158) -- (3.892,1.430) -- (3.698,1.717) -- (3.511,2.005) -- (3.300,2.280) -- (3.102,2.561) -- (2.893,2.838) -- (2.696,3.115) -- (2.500,3.400);
  \draw[black] (2.500,0.800) -- (2.497,0.735) -- (2.506,0.666) -- (2.500,0.600);
  \draw[black] (2.500,0.600) -- (2.566,0.600) -- (2.632,0.595) -- (2.700,0.600);
  \draw[black] (2.500,0.800) -- (2.567,0.805) -- (2.633,0.797) -- (2.700,0.800);
  \draw[black] (2.700,0.800) -- (2.703,0.733) -- (2.697,0.667) -- (2.700,0.600);
  \draw[black] (2.727,3.082) -- (2.714,3.065) -- (2.696,3.054) -- (2.683,3.036) -- (2.656,3.043) -- (2.639,3.031) -- (2.623,3.015) -- (2.604,3.004) -- (2.584,3.000) -- (2.562,3.003) -- (2.541,3.000) -- (2.520,2.999) -- (2.500,3.001);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.233,0.467) {C};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.233,3.467) {B};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (4.600,0.467) {T};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.267) {5 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {8,6 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.703,2.698) {?};
\end{tikzpicture}\end{center}


 Dans le triangle CBT rectangle en C, on cherche une relation entre l'angle aigu   \(\widehat{CBT}\) son coté opposé et l'hypoténuse du triangle.

\par\addvspace{.5em}
sin(\(\widehat{CBT}\)) =  \(\dfrac{CT}{BT}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{CBT}\)) = \(\dfrac{5}{8{,}6}\)
\par\addvspace{.5em}
On a donc \(\widehat{CBT}\) =  ArcSin(\(\dfrac{5}{8{,}6}\)) \(\approx\) 36\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,2.763) -- (1.037,2.573) -- (1.043,2.383) -- (1.056,2.191) -- (1.039,2.000) -- (1.045,1.811) -- (1.053,1.619) -- (1.044,1.430) -- (1.050,1.237);
  \draw[black] (1.050,1.237) -- (1.401,1.231) -- (1.750,1.225) -- (2.100,1.242) -- (2.452,1.233) -- (2.799,1.247) -- (3.153,1.242) -- (3.500,1.248) -- (3.853,1.242) -- (4.199,1.239) -- (4.552,1.226) -- (4.903,1.225) -- (5.251,1.224) -- (5.597,1.231) -- (5.950,1.237);
  \draw[black] (5.950,1.237) -- (5.626,1.350) -- (5.294,1.435) -- (4.976,1.550) -- (4.645,1.645) -- (4.315,1.738) -- (3.995,1.854) -- (3.663,1.947) -- (3.333,2.048) -- (3.015,2.163) -- (2.680,2.247) -- (2.353,2.346) -- (2.035,2.468) -- (1.703,2.554) -- (1.377,2.657) -- (1.050,2.763);
  \draw[black] (1.050,1.390) -- (1.055,1.340) -- (1.054,1.288) -- (1.050,1.237);
  \draw[black] (1.050,1.237) -- (1.102,1.244) -- (1.153,1.232) -- (1.203,1.237);
  \draw[black] (1.050,1.390) -- (1.100,1.392) -- (1.153,1.394) -- (1.203,1.390);
  \draw[black] (1.203,1.390) -- (1.200,1.339) -- (1.197,1.290) -- (1.203,1.237);
  \draw[black] (5.578,1.353) -- (5.565,1.346) -- (5.561,1.337) -- (5.555,1.328) -- (5.549,1.319) -- (5.564,1.306) -- (5.562,1.296) -- (5.562,1.286) -- (5.553,1.277) -- (5.546,1.268) -- (5.559,1.257) -- (5.550,1.247) -- (5.546,1.237);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,1.104) {M};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,2.830) {A};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,1.104) {S};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {2,2 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {7,4 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.291,1.271) {?};
\end{tikzpicture}\end{center}


 Dans le triangle MAS rectangle en M, on cherche une relation entre l'angle aigu   \(\widehat{MSA}\) son coté opposé et l'hypoténuse du triangle.

\par\addvspace{.5em}
sin(\(\widehat{MSA}\)) = \(\dfrac{MA}{AS}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{MSA}\)) =\(\dfrac{2{,}2}{7{,}4}\)
\par\addvspace{.5em}
 On a donc \(\widehat{MSA}\) = ArcSin(\(\dfrac{2{,}2}{7{,}4}\)) \(\approx\) 17\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,3.091) -- (1.050,2.821) -- (1.052,2.548) -- (1.041,2.275) -- (1.061,2.002) -- (1.051,1.727) -- (1.058,1.453) -- (1.055,1.180) -- (1.050,0.909);
  \draw[black] (1.050,0.909) -- (1.397,0.906) -- (1.751,0.906) -- (2.098,0.905) -- (2.450,0.896) -- (2.799,0.904) -- (3.149,0.916) -- (3.498,0.922) -- (3.851,0.922) -- (4.201,0.898) -- (4.547,0.920) -- (4.899,0.910) -- (5.251,0.920) -- (5.603,0.920) -- (5.950,0.909);
  \draw[black] (5.950,0.909) -- (5.648,1.054) -- (5.342,1.189) -- (5.034,1.330) -- (4.721,1.442) -- (4.420,1.589) -- (4.108,1.719) -- (3.804,1.853) -- (3.504,2.011) -- (3.193,2.142) -- (2.888,2.278) -- (2.577,2.401) -- (2.275,2.538) -- (1.971,2.685) -- (1.668,2.828) -- (1.358,2.955) -- (1.050,3.091);
  \draw[black] (1.050,1.127) -- (1.051,1.053) -- (1.049,0.984) -- (1.050,0.909);
  \draw[black] (1.050,0.909) -- (1.124,0.915) -- (1.197,0.909) -- (1.268,0.909);
  \draw[black] (1.050,1.127) -- (1.124,1.125) -- (1.194,1.122) -- (1.268,1.127);
  \draw[black] (1.268,1.127) -- (1.266,1.054) -- (1.274,0.982) -- (1.268,0.909);
  \draw[black] (1.410,2.931) -- (1.404,2.890) -- (1.370,2.867) -- (1.350,2.835) -- (1.324,2.807) -- (1.294,2.784) -- (1.270,2.753) -- (1.237,2.733) -- (1.202,2.714) -- (1.163,2.709) -- (1.127,2.693) -- (1.089,2.684) -- (1.050,2.685);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,0.776) {F};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,3.157) {S};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,0.776) {C};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.576) {5,7 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.413,2.465) {66°};
\end{tikzpicture}\end{center}


 Dans le triangle FSC rectangle en F, on cherche une relation entre l'angle aigu   \(\widehat{FSC}\) son coté adjacent et son coté opposé.

\par\addvspace{.5em}
tan(\(\widehat{FSC}\)) = \(\dfrac{FC}{FS}\)
\par\addvspace{.5em}
tan(66\textdegree{}) = \(\dfrac{5{,}7}{FS}\)
\par\addvspace{.5em}
 On a donc FS = \(\dfrac{5{,}7}{\tan(66^{\circ})}\) \(\approx\) 2,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.708,3.400) -- (1.696,3.051) -- (1.705,2.699) -- (1.712,2.351) -- (1.707,2.000) -- (1.699,1.653) -- (1.698,1.302) -- (1.708,0.951) -- (1.708,0.600);
  \draw[black] (1.708,0.600) -- (2.064,0.594) -- (2.427,0.597) -- (2.785,0.600) -- (3.138,0.589) -- (3.503,0.601) -- (3.856,0.604) -- (4.214,0.608) -- (4.577,0.604) -- (4.933,0.598) -- (5.292,0.600);
  \draw[black] (5.292,0.600) -- (5.017,0.821) -- (4.745,1.039) -- (4.468,1.247) -- (4.198,1.469) -- (3.919,1.681) -- (3.640,1.895) -- (3.372,2.114) -- (3.083,2.321) -- (2.816,2.549) -- (2.536,2.759) -- (2.260,2.970) -- (1.980,3.180) -- (1.708,3.400);
  \draw[black] (1.708,0.880) -- (1.714,0.786) -- (1.706,0.693) -- (1.708,0.600);
  \draw[black] (1.708,0.600) -- (1.803,0.598) -- (1.896,0.601) -- (1.988,0.600);
  \draw[black] (1.708,0.880) -- (1.802,0.880) -- (1.895,0.881) -- (1.988,0.880);
  \draw[black] (1.988,0.880) -- (1.986,0.788) -- (1.984,0.693) -- (1.988,0.600);
  \draw[black] (4.982,0.842) -- (4.960,0.831) -- (4.948,0.812) -- (4.935,0.794) -- (4.924,0.774) -- (4.914,0.754) -- (4.919,0.729) -- (4.909,0.709) -- (4.910,0.686) -- (4.897,0.666) -- (4.888,0.645) -- (4.898,0.622) -- (4.896,0.600);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.441,0.467) {A};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.441,3.467) {R};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.392,0.467) {G};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.208,1.933) {3,6 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.267) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (4.662,0.750) {38°};
\end{tikzpicture}\end{center}


 Dans le triangle ARG rectangle en A, on cherche une relation entre l'angle aigu   \(\widehat{AGR}\) son coté opposé et son coté adjacent.

\par\addvspace{.5em}
tan(\(\widehat{AGR}\)) = \(\dfrac{AR}{AG}\)
\par\addvspace{.5em}
d'où  tan(38\textdegree{}) =  \(\dfrac{3{,}6}{AG}\)
\par\addvspace{.5em}
On a donc AG = \(\dfrac{3{,}6}{\tan(38^{\circ})}\) \(\approx\) 4,6 cm\end{document}
