\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 75}

\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=75}\\[2pt]
    {\small\ttfamily\bfseries\color{red} TRIG0075}
\end{minipage}%
}
}

\begin{document}

\sloppy

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

 Dans le triangle FDL rectangle en F, on sait que :
\begin{itemize}
\item FL=0,7 cm
\item \(\widehat{FDL}\) = 79\textdegree{}
\end{itemize}

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

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

 Dans le triangle GPJ rectangle en G, on sait que :
\begin{itemize}
\item GP=4,6 cm
\item \(\widehat{PJG}\) =  16\textdegree{}
\end{itemize}

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

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

 Dans le triangle HDZ rectangle en H, on sait que :
\begin{itemize}
\item HD=2,1 cm
\item DZ=7,9 cm
\end{itemize}

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

 Dans le triangle DFL rectangle en D, on sait que :
\begin{itemize}
\item DL=8 cm
\item \(\widehat{FLD}\) =  38\textdegree{}
\end{itemize}

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

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

 Dans le triangle WBJ rectangle en W, on sait que :
\begin{itemize}
\item WB=3,1 cm
\item BJ=9,1 cm
\end{itemize}

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

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

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

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,2.476) -- (1.048,2.357) -- (1.045,2.241) -- (1.040,2.120) -- (1.038,2.002) -- (1.051,1.882) -- (1.047,1.765) -- (1.062,1.644) -- (1.050,1.524);
  \draw[black] (1.050,1.524) -- (1.402,1.536) -- (1.749,1.534) -- (2.097,1.523) -- (2.449,1.512) -- (2.798,1.522) -- (3.151,1.531) -- (3.502,1.518) -- (3.853,1.512) -- (4.197,1.525) -- (4.553,1.533) -- (4.900,1.511) -- (5.248,1.537) -- (5.598,1.517) -- (5.950,1.524);
  \draw[black] (5.950,1.524) -- (5.599,1.585) -- (5.249,1.666) -- (4.900,1.725) -- (4.550,1.796) -- (4.205,1.876) -- (3.851,1.929) -- (3.498,2.001) -- (3.151,2.076) -- (2.799,2.133) -- (2.451,2.212) -- (2.105,2.282) -- (1.747,2.341) -- (1.398,2.410) -- (1.050,2.476);
  \draw[black] (1.050,1.619) -- (1.052,1.586) -- (1.054,1.555) -- (1.050,1.524);
  \draw[black] (1.050,1.524) -- (1.083,1.520) -- (1.114,1.525) -- (1.145,1.524);
  \draw[black] (1.050,1.619) -- (1.083,1.617) -- (1.113,1.614) -- (1.145,1.619);
  \draw[black] (1.145,1.619) -- (1.147,1.586) -- (1.147,1.557) -- (1.145,1.524);
  \draw[black] (1.437,2.401) -- (1.431,2.355) -- (1.415,2.312) -- (1.388,2.275) -- (1.375,2.228) -- (1.333,2.204) -- (1.308,2.163) -- (1.266,2.143) -- (1.231,2.111) -- (1.188,2.093) -- (1.140,2.092) -- (1.095,2.089) -- (1.050,2.076);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,1.390) {F};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,2.543) {D};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,1.390) {L};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,1.190) {0,7 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 (1.474,1.895) {79°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{FDL}\)) =  \(\dfrac{FL}{DL}\)
\par\addvspace{.5em}
 d'où sin(79\textdegree{}) = \(\dfrac{0{,}7}{DL}\)
\par\addvspace{.5em}
 On a donc DL = \(\dfrac{0{,}7}{\sin(79^{\circ})}\) \(\approx\) 0,7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,2.703) -- (1.062,2.528) -- (1.056,2.351) -- (1.058,2.173) -- (1.040,2.003) -- (1.042,1.825) -- (1.052,1.648) -- (1.059,1.476) -- (1.050,1.297);
  \draw[black] (1.050,1.297) -- (1.399,1.307) -- (1.748,1.307) -- (2.098,1.292) -- (2.450,1.294) -- (2.803,1.288) -- (3.147,1.300) -- (3.497,1.287) -- (3.847,1.309) -- (4.203,1.306) -- (4.551,1.289) -- (4.897,1.302) -- (5.249,1.298) -- (5.597,1.289) -- (5.950,1.297);
  \draw[black] (5.950,1.297) -- (5.621,1.390) -- (5.297,1.483) -- (4.969,1.577) -- (4.645,1.672) -- (4.318,1.765) -- (3.984,1.849) -- (3.665,1.954) -- (3.337,2.039) -- (3.007,2.132) -- (2.683,2.235) -- (2.360,2.330) -- (2.032,2.417) -- (1.699,2.510) -- (1.371,2.600) -- (1.050,2.703);
  \draw[black] (1.050,1.438) -- (1.052,1.391) -- (1.043,1.344) -- (1.050,1.297);
  \draw[black] (1.050,1.297) -- (1.098,1.296) -- (1.145,1.292) -- (1.191,1.297);
  \draw[black] (1.050,1.438) -- (1.096,1.443) -- (1.143,1.434) -- (1.191,1.438);
  \draw[black] (1.191,1.438) -- (1.191,1.392) -- (1.186,1.345) -- (1.191,1.297);
  \draw[black] (5.559,1.410) -- (5.563,1.399) -- (5.559,1.390) -- (5.550,1.382) -- (5.547,1.373) -- (5.563,1.361) -- (5.554,1.353) -- (5.550,1.344) -- (5.561,1.334) -- (5.544,1.326) -- (5.557,1.316) -- (5.559,1.307) -- (5.543,1.297);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,1.164) {G};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,2.769) {P};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,1.164) {J};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {4,6 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.964) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.290,1.324) {16°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{GJP}\)) = \(\dfrac{GP}{GJ}\)
\par\addvspace{.5em}
d'où  tan(16\textdegree{}) =  \(\dfrac{4{,}6}{GJ}\)
\par\addvspace{.5em}
On a donc GJ = \(\dfrac{4{,}6}{\tan(16^{\circ})}\) \(\approx\) 16 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,2.676) -- (1.037,2.505) -- (1.054,2.338) -- (1.040,2.171) -- (1.063,2.003) -- (1.060,1.831) -- (1.063,1.662) -- (1.052,1.492) -- (1.050,1.324);
  \draw[black] (1.050,1.324) -- (1.401,1.317) -- (1.748,1.317) -- (2.098,1.326) -- (2.453,1.331) -- (2.800,1.312) -- (3.153,1.334) -- (3.503,1.316) -- (3.851,1.322) -- (4.201,1.324) -- (4.548,1.319) -- (4.898,1.318) -- (5.251,1.318) -- (5.597,1.314) -- (5.950,1.324);
  \draw[black] (5.950,1.324) -- (5.620,1.412) -- (5.302,1.514) -- (4.970,1.588) -- (4.641,1.679) -- (4.321,1.778) -- (3.989,1.868) -- (3.664,1.950) -- (3.334,2.034) -- (3.014,2.142) -- (2.679,2.215) -- (2.357,2.312) -- (2.028,2.397) -- (1.698,2.484) -- (1.383,2.597) -- (1.050,2.676);
  \draw[black] (1.050,1.460) -- (1.047,1.415) -- (1.046,1.370) -- (1.050,1.324);
  \draw[black] (1.050,1.324) -- (1.097,1.329) -- (1.139,1.322) -- (1.185,1.324);
  \draw[black] (1.050,1.460) -- (1.095,1.460) -- (1.140,1.460) -- (1.185,1.460);
  \draw[black] (1.185,1.460) -- (1.190,1.413) -- (1.188,1.368) -- (1.185,1.324);
  \draw[black] (1.436,2.569) -- (1.422,2.528) -- (1.395,2.493) -- (1.383,2.451) -- (1.354,2.418) -- (1.321,2.390) -- (1.296,2.353) -- (1.254,2.337) -- (1.216,2.318) -- (1.177,2.300) -- (1.134,2.295) -- (1.093,2.281) -- (1.050,2.276);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,1.191) {H};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,2.742) {D};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,1.191) {Z};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {2,1 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {7,9 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.454,2.079) {?};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{HDZ}\)) = \(\dfrac{HD}{DZ}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{HDZ}\)) = \(\dfrac{2{,}1}{7{,}9}\)
\par\addvspace{.5em}
On a donc \(\widehat{HDZ}\) =ArcCos(\(\dfrac{2{,}1}{7{,}9}\)) \(\approx\)75\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.708,3.400) -- (1.711,3.052) -- (1.718,2.697) -- (1.709,2.347) -- (1.705,1.998) -- (1.699,1.649) -- (1.705,1.299) -- (1.712,0.953) -- (1.708,0.600);
  \draw[black] (1.708,0.600) -- (2.066,0.610) -- (2.422,0.587) -- (2.786,0.611) -- (3.141,0.611) -- (3.498,0.589) -- (3.860,0.595) -- (4.213,0.612) -- (4.572,0.611) -- (4.935,0.602) -- (5.292,0.600);
  \draw[black] (5.292,0.600) -- (5.020,0.825) -- (4.737,1.023) -- (4.470,1.255) -- (4.190,1.463) -- (3.919,1.684) -- (3.630,1.882) -- (3.363,2.104) -- (3.091,2.332) -- (2.811,2.540) -- (2.538,2.752) -- (2.251,2.962) -- (1.980,3.182) -- (1.708,3.400);
  \draw[black] (1.708,0.880) -- (1.707,0.786) -- (1.706,0.694) -- (1.708,0.600);
  \draw[black] (1.708,0.600) -- (1.802,0.605) -- (1.893,0.598) -- (1.988,0.600);
  \draw[black] (1.708,0.880) -- (1.800,0.877) -- (1.896,0.881) -- (1.988,0.880);
  \draw[black] (1.988,0.880) -- (1.988,0.788) -- (1.985,0.693) -- (1.988,0.600);
  \draw[black] (4.984,0.840) -- (4.965,0.827) -- (4.951,0.810) -- (4.946,0.788) -- (4.935,0.769) -- (4.914,0.754) -- (4.917,0.729) -- (4.910,0.708) -- (4.909,0.686) -- (4.903,0.665) -- (4.898,0.644) -- (4.883,0.623) -- (4.884,0.600);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.441,0.467) {D};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.441,3.467) {F};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.392,0.467) {L};
  \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) {8 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 DFL rectangle en D, on cherche une relation entre l'angle aigu   \(\widehat{DLF}\) son coté opposé et son coté adjacent.

\par\addvspace{.5em}
tan(\(\widehat{DLF}\)) = \(\dfrac{DF}{DL}\)
\par\addvspace{.5em}
d'où  tan(38\textdegree{}) =  \(\dfrac{DF}{8}\)
\par\addvspace{.5em}
On a donc DF =  8 \(\times\) tan(38\textdegree{})  \(\approx\) 6,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,2.888) -- (1.040,2.665) -- (1.053,2.443) -- (1.053,2.221) -- (1.049,1.999) -- (1.037,1.777) -- (1.060,1.553) -- (1.039,1.337) -- (1.050,1.112);
  \draw[black] (1.050,1.112) -- (1.402,1.107) -- (1.753,1.118) -- (2.099,1.110) -- (2.449,1.118) -- (2.798,1.117) -- (3.151,1.110) -- (3.500,1.115) -- (3.848,1.116) -- (4.199,1.103) -- (4.550,1.105) -- (4.902,1.104) -- (5.248,1.119) -- (5.602,1.117) -- (5.950,1.112);
  \draw[black] (5.950,1.112) -- (5.624,1.239) -- (5.297,1.347) -- (4.972,1.478) -- (4.646,1.591) -- (4.316,1.696) -- (3.990,1.819) -- (3.662,1.936) -- (3.338,2.068) -- (3.006,2.174) -- (2.689,2.305) -- (2.354,2.406) -- (2.028,2.523) -- (1.702,2.652) -- (1.375,2.763) -- (1.050,2.888);
  \draw[black] (1.050,1.290) -- (1.050,1.231) -- (1.046,1.172) -- (1.050,1.112);
  \draw[black] (1.050,1.112) -- (1.108,1.112) -- (1.168,1.108) -- (1.228,1.112);
  \draw[black] (1.050,1.290) -- (1.109,1.290) -- (1.168,1.289) -- (1.228,1.290);
  \draw[black] (1.228,1.290) -- (1.229,1.231) -- (1.234,1.170) -- (1.228,1.112);
  \draw[black] (5.570,1.250) -- (5.565,1.239) -- (5.565,1.227) -- (5.562,1.216) -- (5.565,1.203) -- (5.555,1.193) -- (5.548,1.183) -- (5.548,1.171) -- (5.551,1.159) -- (5.558,1.146) -- (5.554,1.135) -- (5.558,1.124) -- (5.551,1.112);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,0.979) {W};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,2.954) {B};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,0.979) {J};
  \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) {9,1 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.293,1.161) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{WJB}\)) = \(\dfrac{WB}{BJ}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{WJB}\)) =\(\dfrac{3{,}1}{9{,}1}\)
\par\addvspace{.5em}
 On a donc \(\widehat{WJB}\) = ArcSin(\(\dfrac{3{,}1}{9{,}1}\)) \(\approx\) 20\textdegree{}\end{document}
