\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 9}

\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=9}\\[2pt]
    {\small\ttfamily\bfseries\color{red} TRIG0009}
\end{minipage}%
}
}

\begin{document}

\sloppy

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

 Dans le triangle DLH rectangle en D, on sait que :
\begin{itemize}
\item DL=2,4 cm
\item \(\widehat{LHD}\) =  40\textdegree{}
\end{itemize}

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

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

 Dans le triangle DHR rectangle en D, on sait que :
\begin{itemize}
\item HR=5,3 cm
\item \(\widehat{HRD}\) =  11\textdegree{}
\end{itemize}

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

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

 Dans le triangle NRP rectangle en N, on sait que :
\begin{itemize}
\item NR=3,1 cm
\item RP=7,5 cm
\end{itemize}

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

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

 Dans le triangle LCK rectangle en L, on sait que :
\begin{itemize}
\item LK=4,6 cm
\item CK=10 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{LCK}\).
\needspace{4\baselineskip}
\section*{
Exercice 5}

 Dans le triangle JDL rectangle en J, on sait que :
\begin{itemize}
\item JL=4 cm
\item \(\widehat{JDL}\) = 78\textdegree{}
\end{itemize}

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

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

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

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.832,3.400) -- (1.844,3.052) -- (1.822,2.697) -- (1.843,2.349) -- (1.824,1.997) -- (1.820,1.652) -- (1.839,1.298) -- (1.821,0.948) -- (1.832,0.600);
  \draw[black] (1.832,0.600) -- (2.165,0.589) -- (2.496,0.604) -- (2.829,0.594) -- (3.169,0.602) -- (3.498,0.591) -- (3.835,0.607) -- (4.170,0.591) -- (4.503,0.613) -- (4.837,0.608) -- (5.168,0.600);
  \draw[black] (5.168,0.600) -- (4.907,0.814) -- (4.647,1.025) -- (4.398,1.246) -- (4.144,1.461) -- (3.879,1.667) -- (3.632,1.899) -- (3.363,2.102) -- (3.121,2.326) -- (2.853,2.535) -- (2.606,2.762) -- (2.349,2.974) -- (2.078,3.177) -- (1.832,3.400);
  \draw[black] (1.832,0.880) -- (1.833,0.785) -- (1.836,0.693) -- (1.832,0.600);
  \draw[black] (1.832,0.600) -- (1.926,0.597) -- (2.018,0.600) -- (2.112,0.600);
  \draw[black] (1.832,0.880) -- (1.926,0.877) -- (2.019,0.885) -- (2.112,0.880);
  \draw[black] (2.112,0.880) -- (2.113,0.788) -- (2.116,0.692) -- (2.112,0.600);
  \draw[black] (4.858,0.860) -- (4.850,0.837) -- (4.832,0.821) -- (4.831,0.795) -- (4.818,0.776) -- (4.806,0.756) -- (4.797,0.735) -- (4.779,0.717) -- (4.783,0.691) -- (4.779,0.669) -- (4.771,0.646) -- (4.778,0.623) -- (4.771,0.600);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.565,0.467) {D};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.565,3.467) {L};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.268,0.467) {H};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.332,1.933) {2,4 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 (4.542,0.761) {40°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{DHL}\)) = \(\dfrac{DL}{DH}\)
\par\addvspace{.5em}
d'où  tan(40\textdegree{}) =  \(\dfrac{2{,}4}{DH}\)
\par\addvspace{.5em}
On a donc DH = \(\dfrac{2{,}4}{\tan(40^{\circ})}\) \(\approx\) 2,9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,2.476) -- (1.038,2.358) -- (1.060,2.238) -- (1.060,2.122) -- (1.049,1.999) -- (1.062,1.881) -- (1.045,1.760) -- (1.053,1.640) -- (1.050,1.524);
  \draw[black] (1.050,1.524) -- (1.401,1.533) -- (1.753,1.517) -- (2.097,1.524) -- (2.450,1.536) -- (2.801,1.532) -- (3.152,1.535) -- (3.497,1.531) -- (3.850,1.532) -- (4.199,1.522) -- (4.550,1.532) -- (4.900,1.512) -- (5.247,1.512) -- (5.603,1.511) -- (5.950,1.524);
  \draw[black] (5.950,1.524) -- (5.600,1.593) -- (5.248,1.661) -- (4.897,1.717) -- (4.552,1.797) -- (4.200,1.870) -- (3.847,1.922) -- (3.500,1.999) -- (3.154,2.071) -- (2.801,2.139) -- (2.454,2.214) -- (2.100,2.281) -- (1.748,2.344) -- (1.397,2.396) -- (1.050,2.476);
  \draw[black] (1.050,1.619) -- (1.048,1.587) -- (1.051,1.557) -- (1.050,1.524);
  \draw[black] (1.050,1.524) -- (1.082,1.528) -- (1.112,1.528) -- (1.145,1.524);
  \draw[black] (1.050,1.619) -- (1.082,1.621) -- (1.112,1.619) -- (1.145,1.619);
  \draw[black] (1.145,1.619) -- (1.142,1.588) -- (1.141,1.557) -- (1.145,1.524);
  \draw[black] (5.548,1.602) -- (5.554,1.594) -- (5.546,1.589) -- (5.545,1.583) -- (5.547,1.576) -- (5.560,1.568) -- (5.555,1.562) -- (5.557,1.555) -- (5.545,1.550) -- (5.557,1.543) -- (5.556,1.536) -- (5.558,1.530) -- (5.558,1.524);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,1.390) {D};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,2.543) {H};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,1.390) {R};
  \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 (3.667,2.000) {5,3 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.286,1.521) {11°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{DRH}\)) = \(\dfrac{DH}{HR}\)
\par\addvspace{.5em}
 D'où sin(11\textdegree{}) = \(\dfrac{DH}{5{,}3}\)

On a donc DH = 5,3 \(\times\) sin(11\textdegree{}) \(\approx\) 1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,3.112) -- (1.062,2.836) -- (1.054,2.553) -- (1.053,2.281) -- (1.037,2.001) -- (1.040,1.725) -- (1.052,1.445) -- (1.055,1.165) -- (1.050,0.888);
  \draw[black] (1.050,0.888) -- (1.403,0.878) -- (1.747,0.890) -- (2.100,0.893) -- (2.450,0.877) -- (2.799,0.891) -- (3.147,0.900) -- (3.497,0.893) -- (3.849,0.896) -- (4.203,0.877) -- (4.553,0.899) -- (4.901,0.876) -- (5.251,0.876) -- (5.601,0.891) -- (5.950,0.888);
  \draw[black] (5.950,0.888) -- (5.649,1.032) -- (5.340,1.170) -- (5.027,1.299) -- (4.726,1.451) -- (4.419,1.583) -- (4.104,1.711) -- (3.805,1.853) -- (3.506,2.011) -- (3.190,2.129) -- (2.890,2.283) -- (2.585,2.424) -- (2.279,2.564) -- (1.967,2.698) -- (1.663,2.834) -- (1.363,2.983) -- (1.050,3.112);
  \draw[black] (1.050,1.110) -- (1.044,1.036) -- (1.046,0.961) -- (1.050,0.888);
  \draw[black] (1.050,0.888) -- (1.123,0.892) -- (1.200,0.890) -- (1.272,0.888);
  \draw[black] (1.050,1.110) -- (1.124,1.113) -- (1.198,1.107) -- (1.272,1.110);
  \draw[black] (1.272,1.110) -- (1.266,1.038) -- (1.275,0.962) -- (1.272,0.888);
  \draw[black] (5.586,1.053) -- (5.571,1.044) -- (5.581,1.025) -- (5.577,1.011) -- (5.563,1.001) -- (5.558,0.987) -- (5.568,0.971) -- (5.561,0.958) -- (5.557,0.944) -- (5.551,0.931) -- (5.550,0.916) -- (5.541,0.902) -- (5.545,0.888);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,0.755) {N};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,3.179) {R};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,0.755) {P};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {3,1 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {7,5 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.298,0.962) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{NPR}\)) = \(\dfrac{NR}{RP}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{NPR}\)) =\(\dfrac{3{,}1}{7{,}5}\)
\par\addvspace{.5em}
 On a donc \(\widehat{NPR}\) = ArcSin(\(\dfrac{3{,}1}{7{,}5}\)) \(\approx\) 24\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (2.775,3.400) -- (2.763,3.050) -- (2.770,2.697) -- (2.783,2.353) -- (2.765,1.999) -- (2.783,1.648) -- (2.780,1.302) -- (2.780,0.953) -- (2.775,0.600);
  \draw[black] (2.775,0.600) -- (2.958,0.603) -- (3.140,0.601) -- (3.316,0.612) -- (3.500,0.609) -- (3.682,0.608) -- (3.863,0.599) -- (4.046,0.591) -- (4.225,0.600);
  \draw[black] (4.225,0.600) -- (4.075,0.914) -- (3.899,1.223) -- (3.740,1.531) -- (3.581,1.842) -- (3.427,2.161) -- (3.247,2.464) -- (3.098,2.777) -- (2.937,3.093) -- (2.775,3.400);
  \draw[black] (2.775,0.745) -- (2.777,0.698) -- (2.779,0.650) -- (2.775,0.600);
  \draw[black] (2.775,0.600) -- (2.822,0.600) -- (2.870,0.595) -- (2.920,0.600);
  \draw[black] (2.775,0.745) -- (2.823,0.752) -- (2.871,0.750) -- (2.920,0.745);
  \draw[black] (2.920,0.745) -- (2.923,0.698) -- (2.918,0.647) -- (2.920,0.600);
  \draw[black] (2.958,3.047) -- (2.946,3.035) -- (2.928,3.035) -- (2.918,3.018) -- (2.900,3.019) -- (2.882,3.024) -- (2.871,3.007) -- (2.853,3.014) -- (2.837,3.014) -- (2.821,3.012) -- (2.807,2.994) -- (2.791,2.996) -- (2.775,3.003);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.508,0.467) {L};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.508,3.467) {C};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (4.325,0.467) {K};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.267) {4,6 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {10 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.933,2.686) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{LCK}\)) =  \(\dfrac{LK}{CK}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{LCK}\)) = \(\dfrac{4{,}6}{10}\)
\par\addvspace{.5em}
On a donc \(\widehat{LCK}\) =  ArcSin(\(\dfrac{4{,}6}{10}\)) \(\approx\) 27\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,2.521) -- (1.052,2.389) -- (1.062,2.261) -- (1.044,2.129) -- (1.052,2.001) -- (1.053,1.867) -- (1.042,1.740) -- (1.044,1.612) -- (1.050,1.479);
  \draw[black] (1.050,1.479) -- (1.401,1.489) -- (1.750,1.484) -- (2.097,1.484) -- (2.448,1.486) -- (2.803,1.477) -- (3.151,1.480) -- (3.501,1.486) -- (3.851,1.479) -- (4.200,1.492) -- (4.550,1.490) -- (4.903,1.474) -- (5.247,1.467) -- (5.599,1.479) -- (5.950,1.479);
  \draw[black] (5.950,1.479) -- (5.626,1.558) -- (5.296,1.610) -- (4.967,1.682) -- (4.645,1.765) -- (4.316,1.839) -- (3.988,1.897) -- (3.661,1.954) -- (3.336,2.041) -- (3.006,2.094) -- (2.686,2.182) -- (2.361,2.250) -- (2.032,2.324) -- (1.706,2.395) -- (1.376,2.449) -- (1.050,2.521);
  \draw[black] (1.050,1.583) -- (1.055,1.549) -- (1.047,1.514) -- (1.050,1.479);
  \draw[black] (1.050,1.479) -- (1.086,1.478) -- (1.119,1.478) -- (1.154,1.479);
  \draw[black] (1.050,1.583) -- (1.084,1.583) -- (1.119,1.579) -- (1.154,1.583);
  \draw[black] (1.154,1.583) -- (1.151,1.547) -- (1.148,1.512) -- (1.154,1.479);
  \draw[black] (1.438,2.438) -- (1.423,2.396) -- (1.416,2.350) -- (1.388,2.314) -- (1.368,2.272) -- (1.333,2.243) -- (1.296,2.216) -- (1.268,2.179) -- (1.229,2.154) -- (1.182,2.148) -- (1.138,2.139) -- (1.095,2.122) -- (1.050,2.125);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,1.346) {J};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,2.587) {D};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,1.346) {L};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,1.146) {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 (1.470,1.936) {78°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{JDL}\)) =  \(\dfrac{JL}{DL}\)
\par\addvspace{.5em}
 d'où sin(78\textdegree{}) = \(\dfrac{4}{DL}\)
\par\addvspace{.5em}
 On a donc DL = \(\dfrac{4}{\sin(78^{\circ})}\) \(\approx\) 4,1 cm\end{document}
