\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 78}

\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=78}\\[2pt]
    {\small\ttfamily\bfseries\color{red} TRIG0078}
\end{minipage}%
}
}

\begin{document}

\sloppy

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

 Dans le triangle VGS rectangle en V, on sait que :
\begin{itemize}
\item VG=9,3 cm
\item \(\widehat{GSV}\) =  32\textdegree{}
\end{itemize}

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

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

 Dans le triangle BJM rectangle en B, on sait que :
\begin{itemize}
\item JM=9,3 cm
\item \(\widehat{BJM}\) = 69\textdegree{}
\end{itemize}

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

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

 Dans le triangle CNF rectangle en C, on sait que :
\begin{itemize}
\item CF=7,8 cm
\item \(\widehat{NFC}\) =  34\textdegree{}
\end{itemize}

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

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

 Dans le triangle KZM rectangle en K, on sait que :
\begin{itemize}
\item KZ=1,1 cm
\item KM=4,6 cm
\end{itemize}

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

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

 Dans le triangle VAN rectangle en V, on sait que :
\begin{itemize}
\item VA=2,1 cm
\item VN=4,7 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{VAN}\). \newpage
\section*{\textcolor{red}{Correction}}

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

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.260,3.400) -- (1.252,3.047) -- (1.263,2.703) -- (1.252,2.352) -- (1.260,2.002) -- (1.256,1.650) -- (1.254,1.298) -- (1.256,0.949) -- (1.260,0.600);
  \draw[black] (1.260,0.600) -- (1.603,0.595) -- (1.950,0.589) -- (2.294,0.594) -- (2.639,0.610) -- (2.985,0.607) -- (3.326,0.602) -- (3.670,0.590) -- (4.016,0.598) -- (4.362,0.592) -- (4.707,0.610) -- (5.052,0.592) -- (5.393,0.608) -- (5.740,0.600);
  \draw[black] (5.740,0.600) -- (5.432,0.777) -- (5.143,0.972) -- (4.840,1.151) -- (4.545,1.345) -- (4.246,1.531) -- (3.940,1.711) -- (3.654,1.914) -- (3.358,2.100) -- (3.051,2.272) -- (2.749,2.464) -- (2.458,2.663) -- (2.160,2.851) -- (1.855,3.019) -- (1.553,3.206) -- (1.260,3.400);
  \draw[black] (1.260,0.880) -- (1.266,0.787) -- (1.254,0.693) -- (1.260,0.600);
  \draw[black] (1.260,0.600) -- (1.353,0.602) -- (1.446,0.602) -- (1.540,0.600);
  \draw[black] (1.260,0.880) -- (1.352,0.880) -- (1.446,0.883) -- (1.540,0.880);
  \draw[black] (1.540,0.880) -- (1.546,0.786) -- (1.541,0.693) -- (1.540,0.600);
  \draw[black] (5.398,0.814) -- (5.398,0.792) -- (5.385,0.778) -- (5.375,0.763) -- (5.376,0.742) -- (5.368,0.726) -- (5.353,0.711) -- (5.360,0.690) -- (5.346,0.674) -- (5.344,0.656) -- (5.346,0.637) -- (5.342,0.619) -- (5.344,0.600);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.993,0.467) {V};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.993,3.467) {G};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.840,0.467) {S};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.760,1.933) {9,3 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 (5.100,0.717) {32°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{VSG}\)) = \(\dfrac{VG}{GS}\)
\par\addvspace{.5em}
 D'où sin(32\textdegree{}) = \(\dfrac{9{,}3}{GS}\)
\par\addvspace{.5em}
On a donc GS =\(\dfrac{9{,}3}{\sin(32^{\circ})}\) \(\approx\) 17,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,2.940) -- (1.047,2.704) -- (1.052,2.469) -- (1.049,2.237) -- (1.040,2.001) -- (1.040,1.764) -- (1.055,1.527) -- (1.063,1.292) -- (1.050,1.060);
  \draw[black] (1.050,1.060) -- (1.399,1.047) -- (1.752,1.061) -- (2.103,1.072) -- (2.451,1.058) -- (2.803,1.049) -- (3.151,1.066) -- (3.502,1.053) -- (3.853,1.061) -- (4.200,1.069) -- (4.550,1.072) -- (4.897,1.070) -- (5.252,1.071) -- (5.599,1.057) -- (5.950,1.060);
  \draw[black] (5.950,1.060) -- (5.624,1.187) -- (5.292,1.308) -- (4.967,1.436) -- (4.646,1.568) -- (4.314,1.688) -- (3.990,1.809) -- (3.659,1.926) -- (3.338,2.064) -- (3.014,2.197) -- (2.683,2.309) -- (2.356,2.442) -- (2.030,2.564) -- (1.706,2.700) -- (1.377,2.817) -- (1.050,2.940);
  \draw[black] (1.050,1.248) -- (1.045,1.184) -- (1.045,1.124) -- (1.050,1.060);
  \draw[black] (1.050,1.060) -- (1.114,1.065) -- (1.175,1.056) -- (1.238,1.060);
  \draw[black] (1.050,1.248) -- (1.112,1.244) -- (1.175,1.248) -- (1.238,1.248);
  \draw[black] (1.238,1.248) -- (1.242,1.184) -- (1.238,1.122) -- (1.238,1.060);
  \draw[black] (1.421,2.798) -- (1.408,2.760) -- (1.386,2.726) -- (1.367,2.690) -- (1.335,2.665) -- (1.302,2.642) -- (1.279,2.607) -- (1.244,2.586) -- (1.208,2.568) -- (1.171,2.550) -- (1.130,2.547) -- (1.089,2.551) -- (1.050,2.533);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,0.926) {B};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,3.007) {J};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,0.926) {M};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.726) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {9,3 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.428,2.324) {69°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{BJM}\)) =  \(\dfrac{BM}{JM}\)
\par\addvspace{.5em}
 d'où sin(69\textdegree{}) = \(\dfrac{BM}{9{,}3}\)

On a donc BM = 9,3 \(\times\) sin(69\textdegree{}) \(\approx\) 8,7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.424,3.400) -- (1.432,3.047) -- (1.422,2.699) -- (1.434,2.351) -- (1.436,2.000) -- (1.419,1.647) -- (1.426,1.298) -- (1.423,0.949) -- (1.424,0.600);
  \draw[black] (1.424,0.600) -- (1.768,0.589) -- (2.119,0.609) -- (2.460,0.587) -- (2.809,0.609) -- (3.154,0.591) -- (3.501,0.596) -- (3.844,0.594) -- (4.194,0.603) -- (4.539,0.609) -- (4.887,0.611) -- (5.232,0.596) -- (5.576,0.600);
  \draw[black] (5.576,0.600) -- (5.296,0.788) -- (5.018,0.967) -- (4.752,1.171) -- (4.466,1.341) -- (4.193,1.534) -- (3.913,1.720) -- (3.638,1.907) -- (3.367,2.102) -- (3.086,2.281) -- (2.800,2.457) -- (2.526,2.645) -- (2.259,2.853) -- (1.978,3.028) -- (1.707,3.223) -- (1.424,3.400);
  \draw[black] (1.424,0.880) -- (1.427,0.788) -- (1.428,0.692) -- (1.424,0.600);
  \draw[black] (1.424,0.600) -- (1.516,0.598) -- (1.612,0.599) -- (1.704,0.600);
  \draw[black] (1.424,0.880) -- (1.518,0.876) -- (1.611,0.873) -- (1.704,0.880);
  \draw[black] (1.704,0.880) -- (1.701,0.787) -- (1.706,0.693) -- (1.704,0.600);
  \draw[black] (5.245,0.823) -- (5.230,0.809) -- (5.218,0.793) -- (5.213,0.773) -- (5.215,0.750) -- (5.199,0.736) -- (5.184,0.720) -- (5.191,0.697) -- (5.180,0.679) -- (5.185,0.658) -- (5.168,0.640) -- (5.183,0.619) -- (5.181,0.600);
  \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) {N};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.676,0.467) {F};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.924,1.933) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.267) {7,8 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (4.938,0.728) {34°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{CFN}\)) = \(\dfrac{CN}{CF}\)
\par\addvspace{.5em}
d'où  tan(34\textdegree{}) =  \(\dfrac{CN}{7{,}8}\)
\par\addvspace{.5em}
On a donc CN =  7,8 \(\times\) tan(34\textdegree{})  \(\approx\) 5,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,2.586) -- (1.053,2.438) -- (1.061,2.296) -- (1.052,2.148) -- (1.057,1.999) -- (1.051,1.857) -- (1.046,1.708) -- (1.056,1.560) -- (1.050,1.414);
  \draw[black] (1.050,1.414) -- (1.401,1.421) -- (1.749,1.415) -- (2.102,1.422) -- (2.453,1.416) -- (2.803,1.407) -- (3.147,1.414) -- (3.503,1.422) -- (3.847,1.414) -- (4.201,1.421) -- (4.553,1.411) -- (4.901,1.402) -- (5.247,1.419) -- (5.601,1.409) -- (5.950,1.414);
  \draw[black] (5.950,1.414) -- (5.627,1.503) -- (5.298,1.581) -- (4.970,1.644) -- (4.647,1.739) -- (4.314,1.798) -- (3.988,1.886) -- (3.664,1.966) -- (3.336,2.044) -- (3.008,2.122) -- (2.683,2.191) -- (2.361,2.284) -- (2.034,2.360) -- (1.699,2.423) -- (1.378,2.516) -- (1.050,2.586);
  \draw[black] (1.050,1.531) -- (1.056,1.492) -- (1.047,1.455) -- (1.050,1.414);
  \draw[black] (1.050,1.414) -- (1.089,1.412) -- (1.128,1.416) -- (1.167,1.414);
  \draw[black] (1.050,1.531) -- (1.089,1.530) -- (1.127,1.526) -- (1.167,1.531);
  \draw[black] (1.167,1.531) -- (1.172,1.493) -- (1.161,1.453) -- (1.167,1.414);
  \draw[black] (5.557,1.508) -- (5.560,1.499) -- (5.560,1.491) -- (5.560,1.483) -- (5.550,1.477) -- (5.555,1.469) -- (5.545,1.462) -- (5.544,1.454) -- (5.554,1.445) -- (5.548,1.438) -- (5.552,1.430) -- (5.554,1.422) -- (5.552,1.414);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,1.281) {K};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,2.653) {Z};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,1.281) {M};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {1,1 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,1.081) {4,6 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.288,1.426) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{KMZ}\)) = \(\dfrac{KZ}{KM}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{KMZ}\)) =  \(\dfrac{1{,}1}{4{,}6}\)
\par\addvspace{.5em}
 On a donc \(\widehat{KMZ}\) =  ArcTan(\(\dfrac{1{,}1}{4{,}6}\)) \(\approx\) 13\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,3.095) -- (1.057,2.820) -- (1.046,2.545) -- (1.050,2.272) -- (1.042,2.000) -- (1.048,1.729) -- (1.047,1.456) -- (1.053,1.179) -- (1.050,0.905);
  \draw[black] (1.050,0.905) -- (1.403,0.908) -- (1.750,0.900) -- (2.103,0.899) -- (2.447,0.900) -- (2.799,0.914) -- (3.153,0.900) -- (3.502,0.899) -- (3.847,0.894) -- (4.203,0.896) -- (4.550,0.914) -- (4.898,0.918) -- (5.253,0.909) -- (5.599,0.893) -- (5.950,0.905);
  \draw[black] (5.950,0.905) -- (5.641,1.039) -- (5.341,1.190) -- (5.035,1.327) -- (4.721,1.446) -- (4.417,1.588) -- (4.115,1.726) -- (3.805,1.854) -- (3.495,1.991) -- (3.191,2.126) -- (2.889,2.273) -- (2.578,2.404) -- (2.274,2.541) -- (1.973,2.692) -- (1.667,2.824) -- (1.357,2.963) -- (1.050,3.095);
  \draw[black] (1.050,1.124) -- (1.055,1.051) -- (1.044,0.977) -- (1.050,0.905);
  \draw[black] (1.050,0.905) -- (1.122,0.907) -- (1.197,0.905) -- (1.269,0.905);
  \draw[black] (1.050,1.124) -- (1.124,1.121) -- (1.197,1.125) -- (1.269,1.124);
  \draw[black] (1.269,1.124) -- (1.266,1.050) -- (1.266,0.977) -- (1.269,0.905);
  \draw[black] (1.412,2.933) -- (1.406,2.893) -- (1.376,2.866) -- (1.350,2.838) -- (1.329,2.805) -- (1.304,2.774) -- (1.273,2.751) -- (1.237,2.736) -- (1.199,2.724) -- (1.162,2.715) -- (1.128,2.693) -- (1.088,2.702) -- (1.050,2.695);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,0.772) {V};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,3.161) {A};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,0.772) {N};
  \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 (3.500,0.572) {4,7 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.413,2.469) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{VAN}\)) = \(\dfrac{VN}{VA}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{VAN}\)) = \(\dfrac{4{,}7}{2{,}1}\)
\par\addvspace{.5em}
 On a donc VAN = ArcTan(\(\dfrac{4{,}7}{2{,}1}\)) \(\approx\) 66\textdegree{}\end{document}
