\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 13}

\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=13}\\[2pt]
    {\small\ttfamily\bfseries\color{red} TRIG0013}
\end{minipage}%
}
}

\begin{document}

\sloppy

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

 Dans le triangle ZHF rectangle en Z, on sait que :
\begin{itemize}
\item ZF=6,1 cm
\item \(\widehat{HFZ}\) =  27\textdegree{}
\end{itemize}

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

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

 Dans le triangle CKV rectangle en C, on sait que :
\begin{itemize}
\item CV=2,1 cm
\item \(\widehat{KVC}\) =  16\textdegree{}
\end{itemize}

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

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

 Dans le triangle MPG rectangle en M, on sait que :
\begin{itemize}
\item MP=2,9 cm
\item MG=3,8 cm
\end{itemize}

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

 Dans le triangle CWS rectangle en C, on sait que :
\begin{itemize}
\item WS=6,5 cm
\item \(\widehat{CWS}\) = 56\textdegree{}
\end{itemize}

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

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

 Dans le triangle MKG rectangle en M, on sait que :
\begin{itemize}
\item MK=1,2 cm
\item MG=3,9 cm
\end{itemize}

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

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

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

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,3.248) -- (1.042,2.938) -- (1.041,2.624) -- (1.043,2.312) -- (1.052,2.001) -- (1.052,1.689) -- (1.060,1.374) -- (1.042,1.061) -- (1.050,0.752);
  \draw[black] (1.050,0.752) -- (1.400,0.747) -- (1.748,0.745) -- (2.100,0.747) -- (2.448,0.760) -- (2.798,0.751) -- (3.151,0.753) -- (3.498,0.754) -- (3.848,0.752) -- (4.198,0.759) -- (4.547,0.755) -- (4.903,0.748) -- (5.251,0.755) -- (5.597,0.740) -- (5.950,0.752);
  \draw[black] (5.950,0.752) -- (5.638,0.900) -- (5.341,1.072) -- (5.028,1.211) -- (4.727,1.386) -- (4.424,1.538) -- (4.118,1.696) -- (3.811,1.848) -- (3.505,2.012) -- (3.198,2.160) -- (2.888,2.315) -- (2.583,2.470) -- (2.279,2.629) -- (1.968,2.774) -- (1.667,2.939) -- (1.355,3.084) -- (1.050,3.248);
  \draw[black] (1.050,1.001) -- (1.048,0.918) -- (1.049,0.836) -- (1.050,0.752);
  \draw[black] (1.050,0.752) -- (1.133,0.752) -- (1.215,0.757) -- (1.300,0.752);
  \draw[black] (1.050,1.001) -- (1.132,0.995) -- (1.216,1.004) -- (1.300,1.001);
  \draw[black] (1.300,1.001) -- (1.298,0.920) -- (1.300,0.835) -- (1.300,0.752);
  \draw[black] (5.588,0.936) -- (5.596,0.915) -- (5.584,0.903) -- (5.575,0.890) -- (5.561,0.878) -- (5.565,0.860) -- (5.562,0.845) -- (5.554,0.830) -- (5.559,0.814) -- (5.562,0.798) -- (5.551,0.783) -- (5.542,0.768) -- (5.545,0.752);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,0.618) {Z};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,3.315) {H};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,0.618) {F};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.418) {6,1 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 (5.302,0.841) {27°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{ZFH}\)) = \(\dfrac{ZF}{HF}\)
\par\addvspace{.5em}
  d'où    cos(27\textdegree{}) =  \(\dfrac{6{,}1}{HF}\)
\par\addvspace{.5em}
 On a donc HF =  \(\dfrac{6{,}1}{\cos(27^{\circ})}\) \(\approx\) 6,8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,2.703) -- (1.060,2.525) -- (1.058,2.353) -- (1.053,2.179) -- (1.060,2.001) -- (1.052,1.822) -- (1.037,1.652) -- (1.049,1.476) -- (1.050,1.297);
  \draw[black] (1.050,1.297) -- (1.399,1.289) -- (1.750,1.287) -- (2.097,1.307) -- (2.452,1.304) -- (2.800,1.307) -- (3.149,1.301) -- (3.498,1.289) -- (3.851,1.292) -- (4.200,1.286) -- (4.550,1.286) -- (4.901,1.296) -- (5.253,1.288) -- (5.598,1.308) -- (5.950,1.297);
  \draw[black] (5.950,1.297) -- (5.622,1.393) -- (5.298,1.485) -- (4.975,1.590) -- (4.643,1.661) -- (4.312,1.754) -- (3.986,1.850) -- (3.663,1.963) -- (3.339,2.055) -- (3.009,2.144) -- (2.686,2.239) -- (2.361,2.334) -- (2.025,2.418) -- (1.705,2.514) -- (1.379,2.616) -- (1.050,2.703);
  \draw[black] (1.050,1.438) -- (1.057,1.393) -- (1.053,1.345) -- (1.050,1.297);
  \draw[black] (1.050,1.297) -- (1.096,1.292) -- (1.145,1.293) -- (1.191,1.297);
  \draw[black] (1.050,1.438) -- (1.098,1.439) -- (1.144,1.434) -- (1.191,1.438);
  \draw[black] (1.191,1.438) -- (1.194,1.391) -- (1.185,1.344) -- (1.191,1.297);
  \draw[black] (5.571,1.406) -- (5.565,1.398) -- (5.561,1.390) -- (5.565,1.379) -- (5.555,1.372) -- (5.564,1.361) -- (5.548,1.354) -- (5.559,1.343) -- (5.560,1.334) -- (5.549,1.326) -- (5.554,1.316) -- (5.542,1.307) -- (5.550,1.297);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,1.164) {C};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,2.769) {K};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,1.164) {V};
  \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) {2,1 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.290,1.324) {16°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{CVK}\)) = \(\dfrac{CK}{CV}\)
\par\addvspace{.5em}
d'où  tan(16\textdegree{}) =  \(\dfrac{CK}{2{,}1}\)
\par\addvspace{.5em}
On a donc CK =  2,1 \(\times\) tan(16\textdegree{})  \(\approx\) 0,6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.666,3.400) -- (1.652,3.052) -- (1.665,2.703) -- (1.669,2.352) -- (1.666,2.001) -- (1.675,1.651) -- (1.673,1.301) -- (1.676,0.949) -- (1.666,0.600);
  \draw[black] (1.666,0.600) -- (1.997,0.610) -- (2.333,0.611) -- (2.668,0.610) -- (2.999,0.613) -- (3.334,0.604) -- (3.669,0.598) -- (3.997,0.600) -- (4.334,0.590) -- (4.667,0.606) -- (4.999,0.610) -- (5.334,0.600);
  \draw[black] (5.334,0.600) -- (5.049,0.812) -- (4.772,1.038) -- (4.482,1.244) -- (4.197,1.452) -- (3.922,1.675) -- (3.650,1.899) -- (3.357,2.100) -- (3.075,2.326) -- (2.800,2.546) -- (2.511,2.753) -- (2.221,2.959) -- (1.953,3.194) -- (1.666,3.400);
  \draw[black] (1.666,0.880) -- (1.663,0.785) -- (1.665,0.693) -- (1.666,0.600);
  \draw[black] (1.666,0.600) -- (1.760,0.602) -- (1.851,0.605) -- (1.946,0.600);
  \draw[black] (1.666,0.880) -- (1.759,0.886) -- (1.851,0.874) -- (1.946,0.880);
  \draw[black] (1.946,0.880) -- (1.942,0.787) -- (1.951,0.693) -- (1.946,0.600);
  \draw[black] (1.984,3.157) -- (1.958,3.139) -- (1.941,3.114) -- (1.914,3.098) -- (1.899,3.068) -- (1.873,3.051) -- (1.845,3.037) -- (1.812,3.037) -- (1.788,3.013) -- (1.756,3.014) -- (1.726,3.009) -- (1.696,3.006) -- (1.666,2.994);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.399,0.467) {M};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.399,3.467) {P};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.434,0.467) {G};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.166,1.933) {2,9 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.267) {3,8 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.961,2.736) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{MPG}\)) = \(\dfrac{MG}{MP}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{MPG}\)) = \(\dfrac{3{,}8}{2{,}9}\)
\par\addvspace{.5em}
 On a donc MPG = ArcTan(\(\dfrac{3{,}8}{2{,}9}\)) \(\approx\) 53\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.424,3.400) -- (1.434,3.049) -- (1.425,2.699) -- (1.417,2.348) -- (1.431,2.003) -- (1.413,1.649) -- (1.421,1.302) -- (1.428,0.950) -- (1.424,0.600);
  \draw[black] (1.424,0.600) -- (1.769,0.602) -- (2.115,0.605) -- (2.460,0.608) -- (2.806,0.609) -- (3.151,0.602) -- (3.500,0.588) -- (3.845,0.610) -- (4.190,0.608) -- (4.540,0.588) -- (4.883,0.603) -- (5.229,0.589) -- (5.576,0.600);
  \draw[black] (5.576,0.600) -- (5.294,0.776) -- (5.015,0.968) -- (4.752,1.164) -- (4.470,1.344) -- (4.197,1.541) -- (3.916,1.718) -- (3.645,1.911) -- (3.368,2.096) -- (3.094,2.289) -- (2.805,2.461) -- (2.537,2.658) -- (2.261,2.849) -- (1.980,3.029) -- (1.695,3.208) -- (1.424,3.400);
  \draw[black] (1.424,0.880) -- (1.426,0.787) -- (1.428,0.693) -- (1.424,0.600);
  \draw[black] (1.424,0.600) -- (1.519,0.604) -- (1.610,0.604) -- (1.704,0.600);
  \draw[black] (1.424,0.880) -- (1.518,0.886) -- (1.611,0.878) -- (1.704,0.880);
  \draw[black] (1.704,0.880) -- (1.699,0.787) -- (1.700,0.693) -- (1.704,0.600);
  \draw[black] (1.756,3.176) -- (1.737,3.150) -- (1.717,3.124) -- (1.686,3.109) -- (1.667,3.081) -- (1.642,3.060) -- (1.608,3.055) -- (1.585,3.029) -- (1.552,3.021) -- (1.520,3.018) -- (1.490,3.001) -- (1.456,3.011) -- (1.424,2.996);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.158,0.467) {C};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.158,3.467) {W};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.676,0.467) {S};
  \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 (3.667,2.000) {6,5 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.737,2.745) {56°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{CWS}\)) =  \(\dfrac{CS}{WS}\)
\par\addvspace{.5em}
 d'où sin(56\textdegree{}) = \(\dfrac{CS}{6{,}5}\)

On a donc CS = 6,5 \(\times\) sin(56\textdegree{}) \(\approx\) 5,4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,2.754) -- (1.045,2.567) -- (1.039,2.375) -- (1.053,2.190) -- (1.037,2.000) -- (1.042,1.813) -- (1.051,1.622) -- (1.043,1.432) -- (1.050,1.246);
  \draw[black] (1.050,1.246) -- (1.401,1.237) -- (1.751,1.234) -- (2.101,1.238) -- (2.450,1.246) -- (2.799,1.238) -- (3.152,1.248) -- (3.502,1.252) -- (3.852,1.259) -- (4.201,1.247) -- (4.548,1.238) -- (4.902,1.259) -- (5.250,1.234) -- (5.603,1.243) -- (5.950,1.246);
  \draw[black] (5.950,1.246) -- (5.622,1.353) -- (5.295,1.452) -- (4.968,1.538) -- (4.650,1.660) -- (4.318,1.745) -- (3.987,1.837) -- (3.657,1.938) -- (3.331,2.040) -- (3.005,2.141) -- (2.682,2.259) -- (2.359,2.358) -- (2.030,2.454) -- (1.699,2.540) -- (1.372,2.641) -- (1.050,2.754);
  \draw[black] (1.050,1.397) -- (1.049,1.345) -- (1.056,1.297) -- (1.050,1.246);
  \draw[black] (1.050,1.246) -- (1.102,1.247) -- (1.150,1.252) -- (1.201,1.246);
  \draw[black] (1.050,1.397) -- (1.099,1.399) -- (1.150,1.402) -- (1.201,1.397);
  \draw[black] (1.201,1.397) -- (1.204,1.347) -- (1.198,1.296) -- (1.201,1.246);
  \draw[black] (5.562,1.366) -- (5.565,1.354) -- (5.570,1.343) -- (5.559,1.335) -- (5.562,1.324) -- (5.564,1.314) -- (5.555,1.305) -- (5.559,1.295) -- (5.562,1.285) -- (5.560,1.275) -- (5.550,1.266) -- (5.551,1.256) -- (5.554,1.246);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,1.113) {M};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,2.821) {K};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,1.113) {G};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {1,2 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.913) {3,9 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.291,1.279) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{MGK}\)) = \(\dfrac{MK}{MG}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{MGK}\)) =  \(\dfrac{1{,}2}{3{,}9}\)
\par\addvspace{.5em}
 On a donc \(\widehat{MGK}\) =  ArcTan(\(\dfrac{1{,}2}{3{,}9}\)) \(\approx\) 17\textdegree{}\end{document}
