\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 40}

\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=40}\\[2pt]
    {\small\ttfamily\bfseries\color{red} TRIG0040}
\end{minipage}%
}
}

\begin{document}

\sloppy

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

 Dans le triangle HLT rectangle en H, on sait que :
\begin{itemize}
\item HL=2,9 cm
\item LT=7,4 cm
\end{itemize}

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

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

 Dans le triangle GRK rectangle en G, on sait que :
\begin{itemize}
\item GR=2,9 cm
\item GK=5,9 cm
\end{itemize}

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

 Dans le triangle ZTN rectangle en Z, on sait que :
\begin{itemize}
\item ZT=2,2 cm
\item \(\widehat{ZTN}\) = 49\textdegree{}
\end{itemize}

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

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

 Dans le triangle DFM rectangle en D, on sait que :
\begin{itemize}
\item DF=6,8 cm
\item \(\widehat{DFM}\) = 48\textdegree{}
\end{itemize}

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

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

 Dans le triangle JLR rectangle en J, on sait que :
\begin{itemize}
\item LR=9,2 cm
\item \(\widehat{LRJ}\) =  41\textdegree{}
\end{itemize}

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

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

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

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,3.044) -- (1.038,2.785) -- (1.050,2.520) -- (1.040,2.264) -- (1.040,1.998) -- (1.055,1.737) -- (1.057,1.481) -- (1.059,1.219) -- (1.050,0.956);
  \draw[black] (1.050,0.956) -- (1.400,0.948) -- (1.753,0.943) -- (2.100,0.964) -- (2.451,0.968) -- (2.801,0.953) -- (3.151,0.967) -- (3.502,0.967) -- (3.847,0.957) -- (4.203,0.944) -- (4.553,0.961) -- (4.900,0.962) -- (5.252,0.954) -- (5.597,0.968) -- (5.950,0.956);
  \draw[black] (5.950,0.956) -- (5.620,1.087) -- (5.298,1.243) -- (4.964,1.365) -- (4.643,1.504) -- (4.314,1.653) -- (3.995,1.801) -- (3.668,1.937) -- (3.330,2.062) -- (3.014,2.212) -- (2.680,2.345) -- (2.354,2.475) -- (2.035,2.633) -- (1.703,2.763) -- (1.378,2.903) -- (1.050,3.044);
  \draw[black] (1.050,1.165) -- (1.053,1.095) -- (1.047,1.024) -- (1.050,0.956);
  \draw[black] (1.050,0.956) -- (1.118,0.962) -- (1.189,0.952) -- (1.259,0.956);
  \draw[black] (1.050,1.165) -- (1.119,1.161) -- (1.190,1.167) -- (1.259,1.165);
  \draw[black] (1.259,1.165) -- (1.260,1.095) -- (1.265,1.025) -- (1.259,0.956);
  \draw[black] (5.580,1.114) -- (5.581,1.099) -- (5.573,1.088) -- (5.565,1.076) -- (5.556,1.065) -- (5.570,1.047) -- (5.550,1.038) -- (5.554,1.023) -- (5.563,1.009) -- (5.555,0.996) -- (5.546,0.984) -- (5.558,0.970) -- (5.548,0.956);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,0.823) {H};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,3.110) {L};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,0.823) {T};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {2,9 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 (5.297,1.023) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{HTL}\)) = \(\dfrac{HL}{LT}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{HTL}\)) =\(\dfrac{2{,}9}{7{,}4}\)
\par\addvspace{.5em}
 On a donc \(\widehat{HTL}\) = ArcSin(\(\dfrac{2{,}9}{7{,}4}\)) \(\approx\) 23\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,3.204) -- (1.037,2.900) -- (1.051,2.601) -- (1.041,2.304) -- (1.052,2.001) -- (1.045,1.696) -- (1.052,1.398) -- (1.058,1.096) -- (1.050,0.796);
  \draw[black] (1.050,0.796) -- (1.398,0.809) -- (1.750,0.789) -- (2.098,0.800) -- (2.450,0.783) -- (2.797,0.788) -- (3.153,0.784) -- (3.498,0.789) -- (3.850,0.799) -- (4.198,0.790) -- (4.547,0.796) -- (4.899,0.807) -- (5.249,0.803) -- (5.598,0.806) -- (5.950,0.796);
  \draw[black] (5.950,0.796) -- (5.649,0.955) -- (5.330,1.088) -- (5.032,1.250) -- (4.725,1.397) -- (4.427,1.559) -- (4.113,1.702) -- (3.805,1.842) -- (3.498,1.995) -- (3.190,2.148) -- (2.891,2.302) -- (2.584,2.451) -- (2.271,2.595) -- (1.966,2.754) -- (1.665,2.908) -- (1.353,3.046) -- (1.050,3.204);
  \draw[black] (1.050,1.037) -- (1.050,0.955) -- (1.056,0.878) -- (1.050,0.796);
  \draw[black] (1.050,0.796) -- (1.132,0.801) -- (1.210,0.792) -- (1.291,0.796);
  \draw[black] (1.050,1.037) -- (1.131,1.037) -- (1.210,1.030) -- (1.291,1.037);
  \draw[black] (1.291,1.037) -- (1.290,0.955) -- (1.285,0.877) -- (1.291,0.796);
  \draw[black] (1.417,3.024) -- (1.390,2.996) -- (1.376,2.960) -- (1.340,2.942) -- (1.324,2.906) -- (1.290,2.888) -- (1.258,2.869) -- (1.230,2.845) -- (1.198,2.824) -- (1.162,2.812) -- (1.122,2.819) -- (1.088,2.801) -- (1.050,2.805);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,0.662) {G};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,3.271) {R};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,0.662) {K};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {2,9 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.462) {5,9 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.402,2.572) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{GRK}\)) = \(\dfrac{GK}{GR}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{GRK}\)) = \(\dfrac{5{,}9}{2{,}9}\)
\par\addvspace{.5em}
 On a donc GRK = ArcTan(\(\dfrac{5{,}9}{2{,}9}\)) \(\approx\) 64\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.889,3.400) -- (1.902,3.048) -- (1.902,2.702) -- (1.898,2.351) -- (1.888,2.003) -- (1.894,1.653) -- (1.877,1.299) -- (1.883,0.950) -- (1.889,0.600);
  \draw[black] (1.889,0.600) -- (2.246,0.598) -- (2.603,0.598) -- (2.963,0.604) -- (3.321,0.591) -- (3.677,0.594) -- (4.035,0.590) -- (4.398,0.588) -- (4.753,0.590) -- (5.111,0.600);
  \draw[black] (5.111,0.600) -- (4.837,0.827) -- (4.580,1.077) -- (4.305,1.298) -- (4.027,1.525) -- (3.769,1.764) -- (3.496,1.992) -- (3.233,2.232) -- (2.960,2.467) -- (2.695,2.701) -- (2.435,2.941) -- (2.151,3.157) -- (1.889,3.400);
  \draw[black] (1.889,0.880) -- (1.892,0.787) -- (1.887,0.695) -- (1.889,0.600);
  \draw[black] (1.889,0.600) -- (1.983,0.606) -- (2.075,0.599) -- (2.169,0.600);
  \draw[black] (1.889,0.880) -- (1.983,0.876) -- (2.075,0.881) -- (2.169,0.880);
  \draw[black] (2.169,0.880) -- (2.164,0.786) -- (2.175,0.693) -- (2.169,0.600);
  \draw[black] (2.199,3.131) -- (2.169,3.120) -- (2.150,3.099) -- (2.127,3.083) -- (2.109,3.058) -- (2.082,3.046) -- (2.055,3.036) -- (2.031,3.020) -- (2.003,3.013) -- (1.974,3.010) -- (1.945,3.012) -- (1.918,2.996) -- (1.889,3.001);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.623,0.467) {Z};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.623,3.467) {T};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.211,0.467) {N};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.389,1.933) {2,2 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 (2.166,2.727) {49°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{ZTN}\)) = \(\dfrac{ZN}{ZT}\)
\par\addvspace{.5em}
d'où tan(49\textdegree{}) =\(\dfrac{ZN}{2{,}2}\)
\par\addvspace{.5em}
On a donc ZN = 2,2 \(\times\) tan(49\textdegree{}) \(\approx\) 2,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.945,3.400) -- (1.939,3.050) -- (1.949,2.701) -- (1.952,2.352) -- (1.933,1.997) -- (1.950,1.652) -- (1.944,1.300) -- (1.942,0.948) -- (1.945,0.600);
  \draw[black] (1.945,0.600) -- (2.291,0.607) -- (2.636,0.587) -- (2.982,0.602) -- (3.328,0.594) -- (3.675,0.609) -- (4.020,0.589) -- (4.366,0.600) -- (4.707,0.592) -- (5.055,0.600);
  \draw[black] (5.055,0.600) -- (4.791,0.824) -- (4.542,1.072) -- (4.279,1.298) -- (4.021,1.534) -- (3.750,1.760) -- (3.508,2.004) -- (3.242,2.229) -- (2.989,2.470) -- (2.732,2.706) -- (2.463,2.930) -- (2.204,3.169) -- (1.945,3.400);
  \draw[black] (1.945,0.880) -- (1.941,0.785) -- (1.940,0.693) -- (1.945,0.600);
  \draw[black] (1.945,0.600) -- (2.038,0.605) -- (2.132,0.597) -- (2.225,0.600);
  \draw[black] (1.945,0.880) -- (2.040,0.873) -- (2.130,0.874) -- (2.225,0.880);
  \draw[black] (2.225,0.880) -- (2.220,0.788) -- (2.224,0.693) -- (2.225,0.600);
  \draw[black] (2.236,3.138) -- (2.229,3.106) -- (2.205,3.090) -- (2.176,3.082) -- (2.152,3.068) -- (2.135,3.043) -- (2.109,3.033) -- (2.081,3.026) -- (2.056,3.014) -- (2.030,3.002) -- (2.001,3.005) -- (1.973,3.005) -- (1.945,3.003);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.678,0.467) {D};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.678,3.467) {F};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.155,0.467) {M};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.445,1.933) {6,8 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 (2.216,2.724) {48°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{DFM}\)) = \(\dfrac{DF}{FM}\)
\par\addvspace{.5em}
 d'où cos(48\textdegree{}) = \(\dfrac{6{,}8}{FM}\)
\par\addvspace{.5em}
On a donc FM = \(\dfrac{6{,}8}{\cos(48^{\circ})}\)\(\approx\) 10,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.889,3.400) -- (1.893,3.051) -- (1.883,2.702) -- (1.881,2.348) -- (1.900,1.998) -- (1.892,1.647) -- (1.885,1.301) -- (1.893,0.948) -- (1.889,0.600);
  \draw[black] (1.889,0.600) -- (2.250,0.592) -- (2.605,0.588) -- (2.962,0.594) -- (3.322,0.605) -- (3.682,0.600) -- (4.040,0.595) -- (4.392,0.587) -- (4.755,0.601) -- (5.111,0.600);
  \draw[black] (5.111,0.600) -- (4.846,0.835) -- (4.564,1.060) -- (4.303,1.293) -- (4.040,1.533) -- (3.766,1.766) -- (3.492,1.996) -- (3.235,2.239) -- (2.970,2.470) -- (2.690,2.693) -- (2.431,2.939) -- (2.150,3.160) -- (1.889,3.400);
  \draw[black] (1.889,0.880) -- (1.885,0.787) -- (1.894,0.692) -- (1.889,0.600);
  \draw[black] (1.889,0.600) -- (1.984,0.602) -- (2.077,0.600) -- (2.169,0.600);
  \draw[black] (1.889,0.880) -- (1.984,0.886) -- (2.077,0.880) -- (2.169,0.880);
  \draw[black] (2.169,0.880) -- (2.169,0.787) -- (2.168,0.693) -- (2.169,0.600);
  \draw[black] (4.802,0.869) -- (4.788,0.848) -- (4.787,0.820) -- (4.761,0.808) -- (4.752,0.785) -- (4.753,0.759) -- (4.730,0.742) -- (4.728,0.717) -- (4.716,0.696) -- (4.715,0.672) -- (4.710,0.648) -- (4.720,0.623) -- (4.720,0.600);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.623,0.467) {J};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.623,3.467) {L};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.211,0.467) {R};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.389,1.933) {?};
  \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 (4.486,0.767) {41°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{JRL}\)) = \(\dfrac{JL}{LR}\)
\par\addvspace{.5em}
 D'où sin(41\textdegree{}) = \(\dfrac{JL}{9{,}2}\)

On a donc JL = 9,2 \(\times\) sin(41\textdegree{}) \(\approx\) 6 cm\end{document}
