\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 8}

\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=8}\\[2pt]
    {\small\ttfamily\bfseries\color{red} TRIG0008}
\end{minipage}%
}
}

\begin{document}

\sloppy

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

 Dans le triangle TMR rectangle en T, on sait que :
\begin{itemize}
\item TR=4 cm
\item MR=9,8 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{TMR}\).
\needspace{4\baselineskip}
\section*{
Exercice 2}

 Dans le triangle DKJ rectangle en D, on sait que :
\begin{itemize}
\item DJ=7,8 cm
\item \(\widehat{KJD}\) =  32\textdegree{}
\end{itemize}

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

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

 Dans le triangle NSZ rectangle en N, on sait que :
\begin{itemize}
\item NS=4,4 cm
\item \(\widehat{SZN}\) =  20\textdegree{}
\end{itemize}

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

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

 Dans le triangle JLH rectangle en J, on sait que :
\begin{itemize}
\item JL=4,2 cm
\item \(\widehat{LHJ}\) =  44\textdegree{}
\end{itemize}

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

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

 Dans le triangle DAN rectangle en D, on sait que :
\begin{itemize}
\item DA=2,5 cm
\item DN=4,1 cm
\end{itemize}

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

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

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

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (2.874,3.400) -- (2.867,3.049) -- (2.874,2.700) -- (2.863,2.348) -- (2.870,2.003) -- (2.866,1.651) -- (2.865,1.303) -- (2.886,0.952) -- (2.874,0.600);
  \draw[black] (2.874,0.600) -- (3.033,0.611) -- (3.185,0.600) -- (3.344,0.592) -- (3.497,0.599) -- (3.654,0.595) -- (3.813,0.590) -- (3.969,0.598) -- (4.126,0.600);
  \draw[black] (4.126,0.600) -- (3.987,0.914) -- (3.847,1.219) -- (3.711,1.535) -- (3.565,1.842) -- (3.434,2.157) -- (3.295,2.467) -- (3.149,2.775) -- (3.016,3.092) -- (2.874,3.400);
  \draw[black] (2.874,0.725) -- (2.869,0.682) -- (2.873,0.642) -- (2.874,0.600);
  \draw[black] (2.874,0.600) -- (2.915,0.595) -- (2.956,0.604) -- (2.999,0.600);
  \draw[black] (2.874,0.725) -- (2.915,0.728) -- (2.957,0.722) -- (2.999,0.725);
  \draw[black] (2.999,0.725) -- (2.997,0.685) -- (2.993,0.641) -- (2.999,0.600);
  \draw[black] (3.033,3.043) -- (3.028,3.021) -- (3.009,3.031) -- (2.996,3.027) -- (2.986,3.010) -- (2.971,3.014) -- (2.959,3.002) -- (2.944,3.005) -- (2.931,2.996) -- (2.915,3.009) -- (2.902,3.000) -- (2.888,3.002) -- (2.874,3.008);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.607,0.467) {T};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.607,3.467) {M};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (4.226,0.467) {R};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.267) {4 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {9,8 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.013,2.681) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{TMR}\)) =  \(\dfrac{TR}{MR}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{TMR}\)) = \(\dfrac{4}{9{,}8}\)
\par\addvspace{.5em}
On a donc \(\widehat{TMR}\) =  ArcSin(\(\dfrac{4}{9{,}8}\)) \(\approx\) 24\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.260,3.400) -- (1.264,3.050) -- (1.270,2.702) -- (1.267,2.348) -- (1.252,2.001) -- (1.270,1.649) -- (1.256,1.299) -- (1.259,0.953) -- (1.260,0.600);
  \draw[black] (1.260,0.600) -- (1.605,0.589) -- (1.949,0.592) -- (2.291,0.612) -- (2.636,0.610) -- (2.984,0.613) -- (3.329,0.589) -- (3.673,0.591) -- (4.019,0.587) -- (4.364,0.597) -- (4.706,0.598) -- (5.050,0.605) -- (5.394,0.599) -- (5.740,0.600);
  \draw[black] (5.740,0.600) -- (5.440,0.781) -- (5.138,0.963) -- (4.851,1.169) -- (4.550,1.348) -- (4.244,1.532) -- (3.951,1.728) -- (3.648,1.902) -- (3.356,2.100) -- (3.057,2.289) -- (2.748,2.458) -- (2.452,2.651) -- (2.163,2.848) -- (1.856,3.024) -- (1.564,3.223) -- (1.260,3.400);
  \draw[black] (1.260,0.880) -- (1.257,0.787) -- (1.257,0.693) -- (1.260,0.600);
  \draw[black] (1.260,0.600) -- (1.352,0.598) -- (1.446,0.596) -- (1.540,0.600);
  \draw[black] (1.260,0.880) -- (1.352,0.886) -- (1.447,0.880) -- (1.540,0.880);
  \draw[black] (1.540,0.880) -- (1.533,0.786) -- (1.538,0.692) -- (1.540,0.600);
  \draw[black] (5.393,0.817) -- (5.396,0.794) -- (5.377,0.782) -- (5.379,0.761) -- (5.359,0.749) -- (5.359,0.729) -- (5.361,0.709) -- (5.356,0.691) -- (5.352,0.673) -- (5.344,0.656) -- (5.343,0.637) -- (5.346,0.618) -- (5.334,0.600);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.993,0.467) {D};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.993,3.467) {K};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.840,0.467) {J};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.760,1.933) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.267) {7,8 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.100,0.717) {32°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{DJK}\)) = \(\dfrac{DK}{DJ}\)
\par\addvspace{.5em}
d'où  tan(32\textdegree{}) =  \(\dfrac{DK}{7{,}8}\)
\par\addvspace{.5em}
On a donc DK =  7,8 \(\times\) tan(32\textdegree{})  \(\approx\) 4,9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,2.892) -- (1.040,2.668) -- (1.054,2.444) -- (1.042,2.221) -- (1.047,2.001) -- (1.055,1.778) -- (1.055,1.554) -- (1.060,1.330) -- (1.050,1.108);
  \draw[black] (1.050,1.108) -- (1.398,1.112) -- (1.749,1.115) -- (2.102,1.110) -- (2.448,1.119) -- (2.798,1.114) -- (3.147,1.102) -- (3.500,1.120) -- (3.852,1.102) -- (4.199,1.107) -- (4.552,1.098) -- (4.900,1.109) -- (5.253,1.097) -- (5.598,1.097) -- (5.950,1.108);
  \draw[black] (5.950,1.108) -- (5.622,1.223) -- (5.292,1.338) -- (4.964,1.455) -- (4.643,1.586) -- (4.313,1.702) -- (3.988,1.825) -- (3.666,1.952) -- (3.341,2.072) -- (3.007,2.176) -- (2.685,2.298) -- (2.358,2.421) -- (2.024,2.526) -- (1.700,2.655) -- (1.374,2.775) -- (1.050,2.892);
  \draw[black] (1.050,1.287) -- (1.053,1.228) -- (1.050,1.167) -- (1.050,1.108);
  \draw[black] (1.050,1.108) -- (1.110,1.104) -- (1.169,1.113) -- (1.228,1.108);
  \draw[black] (1.050,1.287) -- (1.109,1.293) -- (1.168,1.281) -- (1.228,1.287);
  \draw[black] (1.228,1.287) -- (1.226,1.227) -- (1.233,1.167) -- (1.228,1.108);
  \draw[black] (5.582,1.242) -- (5.576,1.232) -- (5.565,1.223) -- (5.571,1.210) -- (5.552,1.203) -- (5.560,1.189) -- (5.554,1.178) -- (5.545,1.168) -- (5.560,1.154) -- (5.542,1.144) -- (5.560,1.131) -- (5.548,1.120) -- (5.547,1.108);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,0.975) {N};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,2.958) {S};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,0.975) {Z};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {4,4 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.775) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.293,1.157) {20°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{NZS}\)) = \(\dfrac{NS}{NZ}\)
\par\addvspace{.5em}
d'où  tan(20\textdegree{}) =  \(\dfrac{4{,}4}{NZ}\)
\par\addvspace{.5em}
On a donc NZ = \(\dfrac{4{,}4}{\tan(20^{\circ})}\) \(\approx\) 12,1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (2.050,3.400) -- (2.043,3.048) -- (2.044,2.699) -- (2.057,2.348) -- (2.056,1.998) -- (2.052,1.649) -- (2.050,1.299) -- (2.054,0.949) -- (2.050,0.600);
  \draw[black] (2.050,0.600) -- (2.412,0.594) -- (2.775,0.597) -- (3.137,0.601) -- (3.501,0.599) -- (3.860,0.609) -- (4.222,0.591) -- (4.589,0.592) -- (4.950,0.600);
  \draw[black] (4.950,0.600) -- (4.702,0.829) -- (4.460,1.064) -- (4.221,1.292) -- (3.974,1.525) -- (3.737,1.760) -- (3.508,2.010) -- (3.262,2.238) -- (3.022,2.467) -- (2.782,2.705) -- (2.540,2.940) -- (2.302,3.172) -- (2.050,3.400);
  \draw[black] (2.050,0.880) -- (2.044,0.788) -- (2.050,0.693) -- (2.050,0.600);
  \draw[black] (2.050,0.600) -- (2.144,0.594) -- (2.238,0.594) -- (2.330,0.600);
  \draw[black] (2.050,0.880) -- (2.142,0.881) -- (2.238,0.885) -- (2.330,0.880);
  \draw[black] (2.330,0.880) -- (2.324,0.786) -- (2.332,0.693) -- (2.330,0.600);
  \draw[black] (4.660,0.880) -- (4.649,0.856) -- (4.626,0.841) -- (4.619,0.815) -- (4.609,0.791) -- (4.597,0.769) -- (4.579,0.750) -- (4.566,0.727) -- (4.568,0.700) -- (4.559,0.676) -- (4.549,0.652) -- (4.555,0.625) -- (4.549,0.600);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.784,0.467) {J};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.784,3.467) {L};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.050,0.467) {H};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.550,1.933) {4,2 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 (4.332,0.783) {44°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{JHL}\)) = \(\dfrac{JL}{LH}\)
\par\addvspace{.5em}
 D'où sin(44\textdegree{}) = \(\dfrac{4{,}2}{LH}\)
\par\addvspace{.5em}
On a donc LH =\(\dfrac{4{,}2}{\sin(44^{\circ})}\) \(\approx\) 6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.204,3.400) -- (1.196,3.049) -- (1.213,2.702) -- (1.211,2.349) -- (1.201,2.002) -- (1.199,1.650) -- (1.210,1.300) -- (1.191,0.952) -- (1.204,0.600);
  \draw[black] (1.204,0.600) -- (1.558,0.607) -- (1.913,0.591) -- (2.265,0.603) -- (2.616,0.610) -- (2.971,0.590) -- (3.324,0.601) -- (3.677,0.600) -- (4.033,0.590) -- (4.385,0.608) -- (4.737,0.588) -- (5.089,0.600) -- (5.443,0.600) -- (5.796,0.600);
  \draw[black] (5.796,0.600) -- (5.513,0.776) -- (5.230,0.961) -- (4.937,1.125) -- (4.652,1.302) -- (4.359,1.471) -- (4.071,1.646) -- (3.782,1.817) -- (3.503,2.007) -- (3.217,2.187) -- (2.926,2.349) -- (2.640,2.532) -- (2.352,2.702) -- (2.068,2.879) -- (1.779,3.050) -- (1.490,3.224) -- (1.204,3.400);
  \draw[black] (1.204,0.880) -- (1.200,0.788) -- (1.209,0.692) -- (1.204,0.600);
  \draw[black] (1.204,0.600) -- (1.298,0.600) -- (1.391,0.594) -- (1.484,0.600);
  \draw[black] (1.204,0.880) -- (1.299,0.879) -- (1.390,0.878) -- (1.484,0.880);
  \draw[black] (1.484,0.880) -- (1.488,0.788) -- (1.480,0.692) -- (1.484,0.600);
  \draw[black] (5.457,0.807) -- (5.445,0.793) -- (5.434,0.778) -- (5.427,0.761) -- (5.427,0.741) -- (5.408,0.728) -- (5.410,0.708) -- (5.402,0.691) -- (5.399,0.673) -- (5.401,0.654) -- (5.394,0.637) -- (5.403,0.618) -- (5.395,0.600);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.937,0.467) {D};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.937,3.467) {A};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.896,0.467) {N};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.704,1.933) {2,5 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.267) {4,1 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.154,0.714) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{DNA}\)) = \(\dfrac{DA}{DN}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{DNA}\)) =  \(\dfrac{2{,}5}{4{,}1}\)
\par\addvspace{.5em}
 On a donc \(\widehat{DNA}\) =  ArcTan(\(\dfrac{2{,}5}{4{,}1}\)) \(\approx\) 31\textdegree{}\end{document}
