\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 119}

\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=119}\\[2pt]
    {\small\ttfamily\bfseries\color{red} TRIG0119}
\end{minipage}%
}
}

\begin{document}

\sloppy

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

 Dans le triangle WSP rectangle en W, on sait que :
\begin{itemize}
\item WP=4,3 cm
\item SP=9,6 cm
\end{itemize}

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

 Dans le triangle DRV rectangle en D, on sait que :
\begin{itemize}
\item DR=7,5 cm
\item \(\widehat{DRV}\) = 58\textdegree{}
\end{itemize}

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

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

 Dans le triangle ZGR rectangle en Z, on sait que :
\begin{itemize}
\item ZG=2,6 cm
\item ZR=5,4 cm
\end{itemize}

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

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

 Dans le triangle JFS rectangle en J, on sait que :
\begin{itemize}
\item JF=8,7 cm
\item \(\widehat{JFS}\) = 79\textdegree{}
\end{itemize}

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

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

 Dans le triangle HPD rectangle en H, on sait que :
\begin{itemize}
\item HD=6,6 cm
\item \(\widehat{HPD}\) = 74\textdegree{}
\end{itemize}

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

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

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

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (2.799,3.400) -- (2.790,3.050) -- (2.793,2.697) -- (2.788,2.353) -- (2.806,2.000) -- (2.799,1.647) -- (2.796,1.300) -- (2.791,0.953) -- (2.799,0.600);
  \draw[black] (2.799,0.600) -- (2.977,0.594) -- (3.152,0.597) -- (3.324,0.593) -- (3.498,0.597) -- (3.677,0.599) -- (3.850,0.587) -- (4.025,0.609) -- (4.201,0.600);
  \draw[black] (4.201,0.600) -- (4.050,0.910) -- (3.889,1.221) -- (3.722,1.528) -- (3.586,1.849) -- (3.432,2.162) -- (3.260,2.464) -- (3.107,2.775) -- (2.946,3.087) -- (2.799,3.400);
  \draw[black] (2.799,0.740) -- (2.803,0.692) -- (2.797,0.648) -- (2.799,0.600);
  \draw[black] (2.799,0.600) -- (2.847,0.593) -- (2.893,0.605) -- (2.939,0.600);
  \draw[black] (2.799,0.740) -- (2.846,0.744) -- (2.893,0.740) -- (2.939,0.740);
  \draw[black] (2.939,0.740) -- (2.934,0.693) -- (2.940,0.648) -- (2.939,0.600);
  \draw[black] (2.975,3.049) -- (2.968,3.027) -- (2.953,3.021) -- (2.935,3.023) -- (2.918,3.027) -- (2.904,3.022) -- (2.890,3.016) -- (2.876,3.007) -- (2.860,3.009) -- (2.844,3.012) -- (2.830,2.997) -- (2.814,2.996) -- (2.799,2.992);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.532,0.467) {W};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.532,3.467) {S};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (4.301,0.467) {P};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.267) {4,3 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {9,6 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.952,2.685) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{WSP}\)) =  \(\dfrac{WP}{SP}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{WSP}\)) = \(\dfrac{4{,}3}{9{,}6}\)
\par\addvspace{.5em}
On a donc \(\widehat{WSP}\) =  ArcSin(\(\dfrac{4{,}3}{9{,}6}\)) \(\approx\) 27\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.260,3.400) -- (1.252,3.053) -- (1.268,2.698) -- (1.260,2.349) -- (1.249,1.997) -- (1.271,1.653) -- (1.253,1.300) -- (1.264,0.951) -- (1.260,0.600);
  \draw[black] (1.260,0.600) -- (1.604,0.610) -- (1.952,0.605) -- (2.295,0.612) -- (2.635,0.605) -- (2.984,0.592) -- (3.328,0.598) -- (3.672,0.610) -- (4.014,0.604) -- (4.362,0.611) -- (4.708,0.587) -- (5.053,0.590) -- (5.398,0.597) -- (5.740,0.600);
  \draw[black] (5.740,0.600) -- (5.440,0.784) -- (5.143,0.969) -- (4.848,1.162) -- (4.548,1.347) -- (4.252,1.537) -- (3.954,1.727) -- (3.657,1.914) -- (3.353,2.100) -- (3.056,2.293) -- (2.759,2.470) -- (2.452,2.646) -- (2.152,2.836) -- (1.852,3.017) -- (1.555,3.206) -- (1.260,3.400);
  \draw[black] (1.260,0.880) -- (1.258,0.788) -- (1.259,0.693) -- (1.260,0.600);
  \draw[black] (1.260,0.600) -- (1.353,0.593) -- (1.448,0.595) -- (1.540,0.600);
  \draw[black] (1.260,0.880) -- (1.352,0.878) -- (1.445,0.883) -- (1.540,0.880);
  \draw[black] (1.540,0.880) -- (1.538,0.788) -- (1.542,0.692) -- (1.540,0.600);
  \draw[black] (1.606,3.183) -- (1.582,3.159) -- (1.560,3.133) -- (1.530,3.115) -- (1.512,3.084) -- (1.478,3.075) -- (1.458,3.042) -- (1.420,3.042) -- (1.392,3.024) -- (1.358,3.021) -- (1.328,2.996) -- (1.294,2.996) -- (1.260,2.999);
  \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) {R};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.840,0.467) {V};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.760,1.933) {7,5 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 (1.583,2.750) {58°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{DRV}\)) = \(\dfrac{DR}{RV}\)
\par\addvspace{.5em}
 d'où cos(58\textdegree{}) = \(\dfrac{7{,}5}{RV}\)
\par\addvspace{.5em}
On a donc RV = \(\dfrac{7{,}5}{\cos(58^{\circ})}\)\(\approx\) 14,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,3.180) -- (1.038,2.883) -- (1.056,2.591) -- (1.049,2.297) -- (1.040,2.003) -- (1.053,1.706) -- (1.047,1.410) -- (1.051,1.113) -- (1.050,0.820);
  \draw[black] (1.050,0.820) -- (1.397,0.809) -- (1.747,0.816) -- (2.102,0.825) -- (2.453,0.832) -- (2.801,0.812) -- (3.148,0.812) -- (3.503,0.826) -- (3.848,0.816) -- (4.200,0.812) -- (4.552,0.820) -- (4.899,0.833) -- (5.250,0.822) -- (5.598,0.820) -- (5.950,0.820);
  \draw[black] (5.950,0.820) -- (5.640,0.961) -- (5.340,1.127) -- (5.035,1.269) -- (4.726,1.419) -- (4.427,1.568) -- (4.117,1.716) -- (3.814,1.864) -- (3.499,1.991) -- (3.195,2.147) -- (2.888,2.293) -- (2.575,2.434) -- (2.277,2.598) -- (1.970,2.738) -- (1.659,2.874) -- (1.354,3.026) -- (1.050,3.180);
  \draw[black] (1.050,1.056) -- (1.049,0.978) -- (1.044,0.900) -- (1.050,0.820);
  \draw[black] (1.050,0.820) -- (1.129,0.818) -- (1.209,0.817) -- (1.286,0.820);
  \draw[black] (1.050,1.056) -- (1.127,1.050) -- (1.206,1.056) -- (1.286,1.056);
  \draw[black] (1.286,1.056) -- (1.289,0.978) -- (1.280,0.899) -- (1.286,0.820);
  \draw[black] (5.582,0.997) -- (5.587,0.979) -- (5.572,0.969) -- (5.574,0.952) -- (5.566,0.939) -- (5.562,0.924) -- (5.553,0.911) -- (5.558,0.895) -- (5.556,0.880) -- (5.544,0.866) -- (5.560,0.850) -- (5.549,0.835) -- (5.549,0.820);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,0.687) {Z};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,3.246) {G};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,0.687) {R};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {2,6 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.487) {5,4 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.300,0.902) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{ZRG}\)) = \(\dfrac{ZG}{ZR}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{ZRG}\)) =  \(\dfrac{2{,}6}{5{,}4}\)
\par\addvspace{.5em}
 On a donc \(\widehat{ZRG}\) =  ArcTan(\(\dfrac{2{,}6}{5{,}4}\)) \(\approx\) 26\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,2.476) -- (1.056,2.360) -- (1.048,2.241) -- (1.055,2.118) -- (1.056,2.000) -- (1.060,1.882) -- (1.050,1.761) -- (1.056,1.641) -- (1.050,1.524);
  \draw[black] (1.050,1.524) -- (1.400,1.511) -- (1.752,1.518) -- (2.102,1.535) -- (2.449,1.534) -- (2.800,1.519) -- (3.150,1.522) -- (3.502,1.536) -- (3.852,1.524) -- (4.203,1.523) -- (4.548,1.525) -- (4.903,1.531) -- (5.251,1.518) -- (5.599,1.536) -- (5.950,1.524);
  \draw[black] (5.950,1.524) -- (5.600,1.597) -- (5.254,1.666) -- (4.897,1.718) -- (4.548,1.789) -- (4.205,1.872) -- (3.849,1.922) -- (3.496,1.992) -- (3.150,2.055) -- (2.805,2.143) -- (2.449,2.196) -- (2.098,2.267) -- (1.752,2.343) -- (1.401,2.412) -- (1.050,2.476);
  \draw[black] (1.050,1.619) -- (1.056,1.586) -- (1.055,1.555) -- (1.050,1.524);
  \draw[black] (1.050,1.524) -- (1.082,1.526) -- (1.115,1.529) -- (1.145,1.524);
  \draw[black] (1.050,1.619) -- (1.082,1.623) -- (1.114,1.615) -- (1.145,1.619);
  \draw[black] (1.145,1.619) -- (1.141,1.587) -- (1.140,1.555) -- (1.145,1.524);
  \draw[black] (1.439,2.401) -- (1.428,2.356) -- (1.418,2.311) -- (1.395,2.271) -- (1.365,2.236) -- (1.344,2.193) -- (1.302,2.171) -- (1.263,2.147) -- (1.225,2.123) -- (1.185,2.101) -- (1.142,2.085) -- (1.095,2.086) -- (1.050,2.076);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,1.390) {J};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,2.543) {F};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,1.390) {S};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {8,7 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 JFS rectangle en J, on cherche une relation entre l'angle aigu   \(\widehat{JFS}\) son coté adjacent et son coté opposé.

\par\addvspace{.5em}
tan(\(\widehat{JFS}\)) = \(\dfrac{JS}{JF}\)
\par\addvspace{.5em}
d'où tan(79\textdegree{}) =\(\dfrac{JS}{8{,}7}\)
\par\addvspace{.5em}
On a donc JS = 8,7 \(\times\) tan(79\textdegree{}) \(\approx\) 44,8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,2.703) -- (1.044,2.524) -- (1.039,2.354) -- (1.056,2.174) -- (1.046,1.997) -- (1.048,1.824) -- (1.059,1.650) -- (1.041,1.474) -- (1.050,1.297);
  \draw[black] (1.050,1.297) -- (1.401,1.288) -- (1.748,1.303) -- (2.100,1.289) -- (2.453,1.294) -- (2.799,1.287) -- (3.150,1.310) -- (3.498,1.301) -- (3.847,1.289) -- (4.197,1.303) -- (4.552,1.297) -- (4.902,1.296) -- (5.249,1.293) -- (5.602,1.294) -- (5.950,1.297);
  \draw[black] (5.950,1.297) -- (5.627,1.403) -- (5.301,1.494) -- (4.971,1.590) -- (4.643,1.669) -- (4.318,1.764) -- (3.990,1.866) -- (3.667,1.962) -- (3.335,2.049) -- (3.009,2.131) -- (2.685,2.233) -- (2.354,2.319) -- (2.030,2.418) -- (1.702,2.522) -- (1.381,2.613) -- (1.050,2.703);
  \draw[black] (1.050,1.438) -- (1.054,1.391) -- (1.050,1.346) -- (1.050,1.297);
  \draw[black] (1.050,1.297) -- (1.096,1.295) -- (1.143,1.296) -- (1.191,1.297);
  \draw[black] (1.050,1.438) -- (1.098,1.442) -- (1.142,1.440) -- (1.191,1.438);
  \draw[black] (1.191,1.438) -- (1.192,1.392) -- (1.192,1.345) -- (1.191,1.297);
  \draw[black] (1.425,2.595) -- (1.424,2.550) -- (1.407,2.510) -- (1.378,2.477) -- (1.350,2.444) -- (1.320,2.414) -- (1.289,2.386) -- (1.259,2.353) -- (1.219,2.334) -- (1.174,2.330) -- (1.135,2.314) -- (1.092,2.312) -- (1.050,2.309);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,1.164) {H};
  \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) {D};
  \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 (3.500,0.964) {6,6 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.451,2.103) {74°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{HPD}\)) = \(\dfrac{HD}{HP}\)
\par\addvspace{.5em}
tan(74\textdegree{}) = \(\dfrac{6{,}6}{HP}\)
\par\addvspace{.5em}
 On a donc HP = \(\dfrac{6{,}6}{\tan(74^{\circ})}\) \(\approx\) 1,9 cm\end{document}
