\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 14}

\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=14}\\[2pt]
    {\small\ttfamily\bfseries\color{red} TRIG0014}
\end{minipage}%
}
}

\begin{document}

\sloppy

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

 Dans le triangle CTM rectangle en C, on sait que :
\begin{itemize}
\item CT=8,4 cm
\item \(\widehat{TMC}\) =  30\textdegree{}
\end{itemize}

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

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

 Dans le triangle RMN rectangle en R, on sait que :
\begin{itemize}
\item RN=2,2 cm
\item \(\widehat{MNR}\) =  38\textdegree{}
\end{itemize}

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

\needspace{4\baselineskip}
\section*{
Exercice 3}

 Dans le triangle MRV rectangle en M, on sait que :
\begin{itemize}
\item MV=5,4 cm
\item RV=7 cm
\end{itemize}

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

 Dans le triangle GSJ rectangle en G, on sait que :
\begin{itemize}
\item GS=1,4 cm
\item SJ=9,2 cm
\end{itemize}

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

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

 Dans le triangle KCF rectangle en K, on sait que :
\begin{itemize}
\item KC=0,6 cm
\item \(\widehat{KCF}\) = 79\textdegree{}
\end{itemize}

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

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

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

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.075,3.400) -- (1.074,3.049) -- (1.084,2.699) -- (1.085,2.351) -- (1.086,2.002) -- (1.062,1.651) -- (1.062,1.300) -- (1.072,0.948) -- (1.075,0.600);
  \draw[black] (1.075,0.600) -- (1.419,0.595) -- (1.769,0.598) -- (2.112,0.609) -- (2.461,0.590) -- (2.807,0.608) -- (3.156,0.608) -- (3.497,0.600) -- (3.844,0.592) -- (4.190,0.606) -- (4.537,0.594) -- (4.883,0.599) -- (5.233,0.601) -- (5.578,0.599) -- (5.925,0.600);
  \draw[black] (5.925,0.600) -- (5.622,0.774) -- (5.322,0.951) -- (5.017,1.124) -- (4.715,1.310) -- (4.406,1.464) -- (4.109,1.655) -- (3.796,1.815) -- (3.506,2.010) -- (3.194,2.172) -- (2.901,2.359) -- (2.590,2.519) -- (2.289,2.703) -- (1.982,2.869) -- (1.676,3.039) -- (1.384,3.236) -- (1.075,3.400);
  \draw[black] (1.075,0.880) -- (1.078,0.785) -- (1.077,0.694) -- (1.075,0.600);
  \draw[black] (1.075,0.600) -- (1.167,0.606) -- (1.261,0.604) -- (1.355,0.600);
  \draw[black] (1.075,0.880) -- (1.168,0.875) -- (1.262,0.883) -- (1.355,0.880);
  \draw[black] (1.355,0.880) -- (1.350,0.787) -- (1.352,0.694) -- (1.355,0.600);
  \draw[black] (5.577,0.801) -- (5.568,0.786) -- (5.554,0.773) -- (5.551,0.755) -- (5.549,0.737) -- (5.537,0.722) -- (5.541,0.703) -- (5.537,0.686) -- (5.522,0.671) -- (5.527,0.652) -- (5.521,0.635) -- (5.529,0.617) -- (5.517,0.600);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.808,0.467) {C};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.808,3.467) {T};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.025,0.467) {M};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.575,1.933) {8,4 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 (5.281,0.706) {30°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{CMT}\)) = \(\dfrac{CT}{TM}\)
\par\addvspace{.5em}
 D'où sin(30\textdegree{}) = \(\dfrac{8{,}4}{TM}\)
\par\addvspace{.5em}
On a donc TM =\(\dfrac{8{,}4}{\sin(30^{\circ})}\) \(\approx\) 16,8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.708,3.400) -- (1.695,3.048) -- (1.700,2.697) -- (1.706,2.348) -- (1.703,1.999) -- (1.709,1.647) -- (1.713,1.303) -- (1.712,0.947) -- (1.708,0.600);
  \draw[black] (1.708,0.600) -- (2.067,0.596) -- (2.428,0.591) -- (2.785,0.612) -- (3.139,0.610) -- (3.498,0.596) -- (3.860,0.613) -- (4.220,0.612) -- (4.573,0.590) -- (4.933,0.590) -- (5.292,0.600);
  \draw[black] (5.292,0.600) -- (5.021,0.824) -- (4.735,1.019) -- (4.462,1.240) -- (4.191,1.464) -- (3.913,1.679) -- (3.644,1.895) -- (3.362,2.103) -- (3.081,2.315) -- (2.806,2.535) -- (2.534,2.749) -- (2.256,2.965) -- (1.989,3.189) -- (1.708,3.400);
  \draw[black] (1.708,0.880) -- (1.712,0.786) -- (1.703,0.695) -- (1.708,0.600);
  \draw[black] (1.708,0.600) -- (1.803,0.598) -- (1.894,0.602) -- (1.988,0.600);
  \draw[black] (1.708,0.880) -- (1.801,0.879) -- (1.896,0.876) -- (1.988,0.880);
  \draw[black] (1.988,0.880) -- (1.982,0.786) -- (1.990,0.693) -- (1.988,0.600);
  \draw[black] (4.981,0.843) -- (4.958,0.832) -- (4.943,0.815) -- (4.936,0.793) -- (4.924,0.774) -- (4.927,0.749) -- (4.907,0.733) -- (4.915,0.707) -- (4.896,0.689) -- (4.898,0.666) -- (4.891,0.644) -- (4.902,0.622) -- (4.898,0.600);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.441,0.467) {R};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.441,3.467) {M};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.392,0.467) {N};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.208,1.933) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.267) {2,2 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (4.662,0.750) {38°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{RNM}\)) = \(\dfrac{RM}{RN}\)
\par\addvspace{.5em}
d'où  tan(38\textdegree{}) =  \(\dfrac{RM}{2{,}2}\)
\par\addvspace{.5em}
On a donc RM =  2,2 \(\times\) tan(38\textdegree{})  \(\approx\) 1,7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.803,3.400) -- (1.815,3.053) -- (1.808,2.701) -- (1.808,2.349) -- (1.796,2.002) -- (1.789,1.652) -- (1.796,1.297) -- (1.793,0.952) -- (1.803,0.600);
  \draw[black] (1.803,0.600) -- (2.145,0.590) -- (2.484,0.598) -- (2.819,0.592) -- (3.163,0.607) -- (3.499,0.596) -- (3.843,0.599) -- (4.176,0.610) -- (4.517,0.590) -- (4.860,0.605) -- (5.197,0.600);
  \draw[black] (5.197,0.600) -- (4.931,0.809) -- (4.677,1.038) -- (4.411,1.241) -- (4.153,1.458) -- (3.899,1.687) -- (3.635,1.900) -- (3.368,2.110) -- (3.101,2.314) -- (2.845,2.541) -- (2.589,2.754) -- (2.319,2.960) -- (2.072,3.195) -- (1.803,3.400);
  \draw[black] (1.803,0.880) -- (1.808,0.786) -- (1.797,0.692) -- (1.803,0.600);
  \draw[black] (1.803,0.600) -- (1.895,0.599) -- (1.988,0.594) -- (2.083,0.600);
  \draw[black] (1.803,0.880) -- (1.895,0.879) -- (1.990,0.880) -- (2.083,0.880);
  \draw[black] (2.083,0.880) -- (2.078,0.785) -- (2.088,0.693) -- (2.083,0.600);
  \draw[black] (2.117,3.141) -- (2.087,3.128) -- (2.071,3.103) -- (2.051,3.081) -- (2.020,3.074) -- (1.998,3.054) -- (1.974,3.038) -- (1.945,3.030) -- (1.920,3.012) -- (1.890,3.009) -- (1.860,3.010) -- (1.832,3.000) -- (1.803,3.003);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.536,0.467) {M};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.536,3.467) {R};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.297,0.467) {V};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.267) {5,4 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {7 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.087,2.730) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{MRV}\)) =  \(\dfrac{MV}{RV}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{MRV}\)) = \(\dfrac{5{,}4}{7}\)
\par\addvspace{.5em}
On a donc \(\widehat{MRV}\) =  ArcSin(\(\dfrac{5{,}4}{7}\)) \(\approx\) 50\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,2.377) -- (1.060,2.285) -- (1.058,2.187) -- (1.043,2.097) -- (1.056,2.002) -- (1.054,1.908) -- (1.062,1.813) -- (1.059,1.717) -- (1.050,1.623);
  \draw[black] (1.050,1.623) -- (1.401,1.614) -- (1.750,1.629) -- (2.102,1.616) -- (2.448,1.623) -- (2.802,1.631) -- (3.153,1.629) -- (3.498,1.610) -- (3.851,1.621) -- (4.200,1.612) -- (4.550,1.625) -- (4.903,1.622) -- (5.250,1.628) -- (5.597,1.614) -- (5.950,1.623);
  \draw[black] (5.950,1.623) -- (5.597,1.681) -- (5.250,1.719) -- (4.900,1.776) -- (4.549,1.838) -- (4.204,1.899) -- (3.848,1.948) -- (3.501,1.988) -- (3.152,2.056) -- (2.801,2.112) -- (2.450,2.174) -- (2.102,2.226) -- (1.752,2.277) -- (1.399,2.320) -- (1.050,2.377);
  \draw[black] (1.050,1.698) -- (1.047,1.674) -- (1.053,1.649) -- (1.050,1.623);
  \draw[black] (1.050,1.623) -- (1.077,1.628) -- (1.101,1.628) -- (1.125,1.623);
  \draw[black] (1.050,1.698) -- (1.076,1.693) -- (1.099,1.698) -- (1.125,1.698);
  \draw[black] (1.125,1.698) -- (1.126,1.675) -- (1.130,1.647) -- (1.125,1.623);
  \draw[black] (5.550,1.684) -- (5.560,1.678) -- (5.559,1.673) -- (5.549,1.669) -- (5.554,1.663) -- (5.557,1.658) -- (5.559,1.653) -- (5.544,1.649) -- (5.554,1.643) -- (5.559,1.638) -- (5.541,1.633) -- (5.542,1.628) -- (5.560,1.623);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,1.489) {G};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,2.444) {S};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,1.489) {J};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {1,4 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {9,2 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.285,1.607) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{GJS}\)) = \(\dfrac{GS}{SJ}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{GJS}\)) =\(\dfrac{1{,}4}{9{,}2}\)
\par\addvspace{.5em}
 On a donc \(\widehat{GJS}\) = ArcSin(\(\dfrac{1{,}4}{9{,}2}\)) \(\approx\) 9\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,2.476) -- (1.044,2.355) -- (1.043,2.239) -- (1.057,2.116) -- (1.042,1.997) -- (1.052,1.878) -- (1.045,1.760) -- (1.050,1.641) -- (1.050,1.524);
  \draw[black] (1.050,1.524) -- (1.398,1.524) -- (1.747,1.524) -- (2.099,1.519) -- (2.452,1.532) -- (2.799,1.523) -- (3.153,1.512) -- (3.500,1.526) -- (3.849,1.537) -- (4.200,1.511) -- (4.547,1.511) -- (4.897,1.537) -- (5.252,1.535) -- (5.602,1.511) -- (5.950,1.524);
  \draw[black] (5.950,1.524) -- (5.601,1.594) -- (5.252,1.656) -- (4.902,1.725) -- (4.551,1.804) -- (4.197,1.861) -- (3.847,1.927) -- (3.500,2.006) -- (3.150,2.071) -- (2.799,2.147) -- (2.447,2.206) -- (2.096,2.266) -- (1.751,2.350) -- (1.395,2.398) -- (1.050,2.476);
  \draw[black] (1.050,1.619) -- (1.052,1.588) -- (1.044,1.557) -- (1.050,1.524);
  \draw[black] (1.050,1.524) -- (1.081,1.522) -- (1.113,1.519) -- (1.145,1.524);
  \draw[black] (1.050,1.619) -- (1.082,1.624) -- (1.114,1.620) -- (1.145,1.619);
  \draw[black] (1.145,1.619) -- (1.148,1.588) -- (1.148,1.555) -- (1.145,1.524);
  \draw[black] (1.435,2.401) -- (1.432,2.355) -- (1.421,2.310) -- (1.397,2.270) -- (1.364,2.237) -- (1.338,2.199) -- (1.299,2.174) -- (1.267,2.141) -- (1.230,2.113) -- (1.182,2.109) -- (1.142,2.084) -- (1.097,2.070) -- (1.050,2.081);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,1.390) {K};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,2.543) {C};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,1.390) {F};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {0,6 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,1.190) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.474,1.895) {79°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{KCF}\)) = \(\dfrac{KF}{KC}\)
\par\addvspace{.5em}
d'où tan(79\textdegree{}) =\(\dfrac{KF}{0{,}6}\)
\par\addvspace{.5em}
On a donc KF = 0,6 \(\times\) tan(79\textdegree{}) \(\approx\) 3,1 cm\end{document}
