\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 60}

\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=60}\\[2pt]
    {\small\ttfamily\bfseries\color{red} TRIG0060}
\end{minipage}%
}
}

\begin{document}

\sloppy

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

 Dans le triangle MVN rectangle en M, on sait que :
\begin{itemize}
\item MV=2,9 cm
\item MN=4,4 cm
\end{itemize}

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

 Dans le triangle KVZ rectangle en K, on sait que :
\begin{itemize}
\item KZ=4,2 cm
\item VZ=6,9 cm
\end{itemize}

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

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

 Dans le triangle SVM rectangle en S, on sait que :
\begin{itemize}
\item VM=1,7 cm
\item \(\widehat{VMS}\) =  23\textdegree{}
\end{itemize}

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

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

 Dans le triangle ACJ rectangle en A, on sait que :
\begin{itemize}
\item CJ=5 cm
\item \(\widehat{ACJ}\) = 80\textdegree{}
\end{itemize}

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

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

 Dans le triangle GDJ rectangle en G, on sait que :
\begin{itemize}
\item GJ=8,5 cm
\item \(\widehat{GDJ}\) = 56\textdegree{}
\end{itemize}

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

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

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

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.376,3.400) -- (1.374,3.049) -- (1.373,2.703) -- (1.381,2.353) -- (1.374,2.001) -- (1.383,1.652) -- (1.364,1.302) -- (1.365,0.947) -- (1.376,0.600);
  \draw[black] (1.376,0.600) -- (1.732,0.608) -- (2.085,0.608) -- (2.437,0.610) -- (2.791,0.600) -- (3.147,0.591) -- (3.500,0.612) -- (3.852,0.611) -- (4.207,0.603) -- (4.565,0.599) -- (4.916,0.593) -- (5.268,0.591) -- (5.624,0.600);
  \draw[black] (5.624,0.600) -- (5.341,0.793) -- (5.064,0.983) -- (4.776,1.162) -- (4.488,1.347) -- (4.201,1.526) -- (3.920,1.714) -- (3.645,1.906) -- (3.361,2.091) -- (3.079,2.291) -- (2.792,2.469) -- (2.514,2.657) -- (2.220,2.830) -- (1.941,3.025) -- (1.660,3.218) -- (1.376,3.400);
  \draw[black] (1.376,0.880) -- (1.375,0.788) -- (1.378,0.693) -- (1.376,0.600);
  \draw[black] (1.376,0.600) -- (1.468,0.595) -- (1.562,0.597) -- (1.656,0.600);
  \draw[black] (1.376,0.880) -- (1.470,0.882) -- (1.562,0.883) -- (1.656,0.880);
  \draw[black] (1.656,0.880) -- (1.655,0.786) -- (1.660,0.693) -- (1.656,0.600);
  \draw[black] (1.706,3.182) -- (1.695,3.150) -- (1.663,3.134) -- (1.651,3.100) -- (1.621,3.084) -- (1.591,3.070) -- (1.564,3.051) -- (1.538,3.029) -- (1.505,3.021) -- (1.472,3.020) -- (1.440,3.012) -- (1.408,3.010) -- (1.376,2.999);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.109,0.467) {M};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.109,3.467) {V};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.724,0.467) {N};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.876,1.933) {2,9 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.267) {4,4 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.692,2.746) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{MVN}\)) = \(\dfrac{MN}{MV}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{MVN}\)) = \(\dfrac{4{,}4}{2{,}9}\)
\par\addvspace{.5em}
 On a donc MVN = ArcTan(\(\dfrac{4{,}4}{2{,}9}\)) \(\approx\) 57\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (2.426,3.400) -- (2.423,3.050) -- (2.433,2.699) -- (2.426,2.348) -- (2.417,1.998) -- (2.430,1.653) -- (2.435,1.299) -- (2.436,0.952) -- (2.426,0.600);
  \draw[black] (2.426,0.600) -- (2.693,0.591) -- (2.964,0.596) -- (3.230,0.598) -- (3.501,0.594) -- (3.772,0.599) -- (4.036,0.602) -- (4.302,0.593) -- (4.574,0.600);
  \draw[black] (4.574,0.600) -- (4.353,0.878) -- (4.145,1.162) -- (3.929,1.437) -- (3.718,1.720) -- (3.503,2.000) -- (3.282,2.274) -- (3.073,2.565) -- (2.860,2.841) -- (2.649,3.123) -- (2.426,3.400);
  \draw[black] (2.426,0.815) -- (2.426,0.745) -- (2.422,0.670) -- (2.426,0.600);
  \draw[black] (2.426,0.600) -- (2.497,0.595) -- (2.568,0.594) -- (2.641,0.600);
  \draw[black] (2.426,0.815) -- (2.496,0.818) -- (2.571,0.819) -- (2.641,0.815);
  \draw[black] (2.641,0.815) -- (2.646,0.743) -- (2.641,0.671) -- (2.641,0.600);
  \draw[black] (4.331,0.917) -- (4.306,0.899) -- (4.281,0.880) -- (4.267,0.852) -- (4.244,0.831) -- (4.227,0.806) -- (4.220,0.774) -- (4.200,0.750) -- (4.201,0.718) -- (4.178,0.692) -- (4.181,0.661) -- (4.185,0.630) -- (4.171,0.600);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.159,0.467) {K};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.159,3.467) {V};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (4.674,0.467) {Z};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.267) {4,2 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {6,9 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.976,0.828) {?};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{KZV}\)) = \(\dfrac{KZ}{VZ}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{KZV}\)) =  \(\dfrac{4{,}2}{6{,}9}\)
\par\addvspace{.5em}
On a donc \(\widehat{KZV}\) = Arccos (\(\dfrac{4{,}2}{6{,}9}\)) \(\approx\) 53\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,3.040) -- (1.044,2.782) -- (1.041,2.523) -- (1.042,2.260) -- (1.059,2.001) -- (1.056,1.740) -- (1.060,1.482) -- (1.056,1.218) -- (1.050,0.960);
  \draw[black] (1.050,0.960) -- (1.400,0.958) -- (1.751,0.971) -- (2.101,0.952) -- (2.447,0.959) -- (2.800,0.948) -- (3.150,0.949) -- (3.502,0.956) -- (3.847,0.973) -- (4.200,0.970) -- (4.549,0.965) -- (4.900,0.950) -- (5.250,0.957) -- (5.601,0.971) -- (5.950,0.960);
  \draw[black] (5.950,0.960) -- (5.617,1.087) -- (5.301,1.244) -- (4.970,1.378) -- (4.647,1.518) -- (4.313,1.651) -- (3.995,1.798) -- (3.664,1.932) -- (3.336,2.060) -- (3.005,2.199) -- (2.684,2.350) -- (2.355,2.479) -- (2.028,2.624) -- (1.700,2.754) -- (1.380,2.905) -- (1.050,3.040);
  \draw[black] (1.050,1.168) -- (1.046,1.097) -- (1.052,1.030) -- (1.050,0.960);
  \draw[black] (1.050,0.960) -- (1.121,0.962) -- (1.189,0.958) -- (1.258,0.960);
  \draw[black] (1.050,1.168) -- (1.121,1.171) -- (1.188,1.164) -- (1.258,1.168);
  \draw[black] (1.258,1.168) -- (1.260,1.100) -- (1.257,1.030) -- (1.258,0.960);
  \draw[black] (5.590,1.113) -- (5.586,1.100) -- (5.579,1.089) -- (5.561,1.081) -- (5.556,1.068) -- (5.553,1.055) -- (5.557,1.040) -- (5.562,1.026) -- (5.553,1.014) -- (5.544,1.001) -- (5.542,0.987) -- (5.540,0.974) -- (5.543,0.960);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,0.827) {S};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,3.107) {V};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,0.827) {M};
  \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) {1,7 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.297,1.026) {23°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{SMV}\)) = \(\dfrac{SV}{VM}\)
\par\addvspace{.5em}
 D'où sin(23\textdegree{}) = \(\dfrac{SV}{1{,}7}\)

On a donc SV = 1,7 \(\times\) sin(23\textdegree{}) \(\approx\) 0,7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,2.432) -- (1.052,2.322) -- (1.058,2.217) -- (1.038,2.108) -- (1.052,2.001) -- (1.059,1.890) -- (1.058,1.784) -- (1.043,1.678) -- (1.050,1.568);
  \draw[black] (1.050,1.568) -- (1.397,1.574) -- (1.747,1.562) -- (2.097,1.555) -- (2.450,1.571) -- (2.802,1.577) -- (3.149,1.558) -- (3.503,1.566) -- (3.852,1.574) -- (4.201,1.557) -- (4.551,1.558) -- (4.902,1.574) -- (5.251,1.557) -- (5.597,1.570) -- (5.950,1.568);
  \draw[black] (5.950,1.568) -- (5.601,1.638) -- (5.251,1.687) -- (4.899,1.751) -- (4.551,1.807) -- (4.202,1.869) -- (3.853,1.937) -- (3.498,1.991) -- (3.147,2.050) -- (2.801,2.115) -- (2.452,2.191) -- (2.101,2.237) -- (1.748,2.306) -- (1.398,2.365) -- (1.050,2.432);
  \draw[black] (1.050,1.654) -- (1.045,1.626) -- (1.055,1.596) -- (1.050,1.568);
  \draw[black] (1.050,1.568) -- (1.080,1.562) -- (1.108,1.565) -- (1.136,1.568);
  \draw[black] (1.050,1.654) -- (1.078,1.661) -- (1.108,1.661) -- (1.136,1.654);
  \draw[black] (1.136,1.654) -- (1.136,1.626) -- (1.131,1.595) -- (1.136,1.568);
  \draw[black] (1.441,2.363) -- (1.425,2.320) -- (1.416,2.274) -- (1.398,2.231) -- (1.374,2.190) -- (1.341,2.158) -- (1.308,2.125) -- (1.271,2.096) -- (1.225,2.083) -- (1.187,2.056) -- (1.142,2.045) -- (1.097,2.027) -- (1.050,2.023);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,1.435) {A};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,2.499) {C};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,1.435) {J};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,1.235) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {5 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.479,1.855) {80°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{ACJ}\)) =  \(\dfrac{AJ}{CJ}\)
\par\addvspace{.5em}
 d'où sin(80\textdegree{}) = \(\dfrac{AJ}{5}\)

On a donc AJ = 5 \(\times\) sin(80\textdegree{}) \(\approx\) 4,9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.424,3.400) -- (1.438,3.049) -- (1.411,2.701) -- (1.411,2.349) -- (1.429,1.998) -- (1.435,1.651) -- (1.423,1.297) -- (1.429,0.947) -- (1.424,0.600);
  \draw[black] (1.424,0.600) -- (1.771,0.613) -- (2.116,0.603) -- (2.465,0.603) -- (2.808,0.604) -- (3.154,0.601) -- (3.499,0.602) -- (3.843,0.591) -- (4.190,0.606) -- (4.539,0.603) -- (4.886,0.610) -- (5.228,0.588) -- (5.576,0.600);
  \draw[black] (5.576,0.600) -- (5.293,0.784) -- (5.030,0.981) -- (4.741,1.157) -- (4.475,1.353) -- (4.184,1.528) -- (3.918,1.721) -- (3.646,1.915) -- (3.357,2.090) -- (3.081,2.274) -- (2.809,2.463) -- (2.529,2.652) -- (2.247,2.833) -- (1.982,3.030) -- (1.701,3.217) -- (1.424,3.400);
  \draw[black] (1.424,0.880) -- (1.428,0.787) -- (1.420,0.693) -- (1.424,0.600);
  \draw[black] (1.424,0.600) -- (1.519,0.604) -- (1.611,0.596) -- (1.704,0.600);
  \draw[black] (1.424,0.880) -- (1.519,0.884) -- (1.612,0.886) -- (1.704,0.880);
  \draw[black] (1.704,0.880) -- (1.701,0.788) -- (1.704,0.693) -- (1.704,0.600);
  \draw[black] (1.748,3.182) -- (1.735,3.152) -- (1.714,3.127) -- (1.687,3.109) -- (1.661,3.090) -- (1.637,3.068) -- (1.615,3.041) -- (1.582,3.036) -- (1.555,3.014) -- (1.519,3.021) -- (1.489,3.006) -- (1.457,3.006) -- (1.424,2.993);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.158,0.467) {G};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.158,3.467) {D};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.676,0.467) {J};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.267) {8,5 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 (1.737,2.745) {56°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{GDJ}\)) =  \(\dfrac{GJ}{DJ}\)
\par\addvspace{.5em}
 d'où sin(56\textdegree{}) = \(\dfrac{8{,}5}{DJ}\)
\par\addvspace{.5em}
 On a donc DJ = \(\dfrac{8{,}5}{\sin(56^{\circ})}\) \(\approx\) 10,3 cm\end{document}
