\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 378}

\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=378}\\[2pt]
    {\small\ttfamily\bfseries\color{red} TRIG0378}
\end{minipage}%
}
}

\begin{document}

\sloppy

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

 Dans le triangle BWV rectangle en B, on sait que :
\begin{itemize}
\item BW=3,2 cm
\item WV=9,9 cm
\end{itemize}

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

 Dans le triangle MKL rectangle en M, on sait que :
\begin{itemize}
\item MK=8,6 cm
\item \(\widehat{MKL}\) = 71\textdegree{}
\end{itemize}

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

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

 Dans le triangle VSZ rectangle en V, on sait que :
\begin{itemize}
\item SZ=5,8 cm
\item \(\widehat{SZV}\) =  25\textdegree{}
\end{itemize}

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

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

 Dans le triangle BAJ rectangle en B, on sait que :
\begin{itemize}
\item BA=2,7 cm
\item AJ=8,2 cm
\end{itemize}

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

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

 Dans le triangle GMW rectangle en G, on sait que :
\begin{itemize}
\item GW=0,4 cm
\item \(\widehat{MWG}\) =  35\textdegree{}
\end{itemize}

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

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

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

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,2.837) -- (1.046,2.630) -- (1.057,2.418) -- (1.050,2.209) -- (1.046,2.000) -- (1.039,1.792) -- (1.050,1.579) -- (1.043,1.369) -- (1.050,1.163);
  \draw[black] (1.050,1.163) -- (1.399,1.152) -- (1.747,1.176) -- (2.099,1.160) -- (2.451,1.159) -- (2.801,1.165) -- (3.152,1.156) -- (3.501,1.160) -- (3.852,1.154) -- (4.203,1.172) -- (4.549,1.158) -- (4.900,1.166) -- (5.251,1.168) -- (5.602,1.171) -- (5.950,1.163);
  \draw[black] (5.950,1.163) -- (5.628,1.282) -- (5.294,1.384) -- (4.966,1.490) -- (4.649,1.617) -- (4.312,1.713) -- (3.990,1.837) -- (3.667,1.945) -- (3.335,2.053) -- (3.010,2.160) -- (2.687,2.282) -- (2.357,2.385) -- (2.028,2.503) -- (1.704,2.618) -- (1.372,2.719) -- (1.050,2.837);
  \draw[black] (1.050,1.331) -- (1.045,1.276) -- (1.046,1.220) -- (1.050,1.163);
  \draw[black] (1.050,1.163) -- (1.107,1.167) -- (1.163,1.167) -- (1.217,1.163);
  \draw[black] (1.050,1.331) -- (1.105,1.325) -- (1.160,1.329) -- (1.217,1.331);
  \draw[black] (1.217,1.331) -- (1.212,1.274) -- (1.214,1.218) -- (1.217,1.163);
  \draw[black] (1.434,2.706) -- (1.405,2.673) -- (1.399,2.630) -- (1.372,2.597) -- (1.351,2.560) -- (1.316,2.537) -- (1.283,2.511) -- (1.244,2.497) -- (1.208,2.478) -- (1.171,2.461) -- (1.134,2.436) -- (1.090,2.448) -- (1.050,2.444);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,1.030) {B};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,2.904) {W};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,1.030) {V};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {3,2 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {9,9 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.438,2.228) {?};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{BWV}\)) = \(\dfrac{BW}{WV}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{BWV}\)) = \(\dfrac{3{,}2}{9{,}9}\)
\par\addvspace{.5em}
On a donc \(\widehat{BWV}\) =ArcCos(\(\dfrac{3{,}2}{9{,}9}\)) \(\approx\)71\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,2.844) -- (1.042,2.633) -- (1.037,2.420) -- (1.054,2.214) -- (1.044,2.003) -- (1.055,1.791) -- (1.038,1.578) -- (1.039,1.368) -- (1.050,1.156);
  \draw[black] (1.050,1.156) -- (1.403,1.154) -- (1.749,1.146) -- (2.100,1.150) -- (2.447,1.165) -- (2.801,1.167) -- (3.152,1.159) -- (3.500,1.158) -- (3.848,1.145) -- (4.197,1.169) -- (4.549,1.168) -- (4.903,1.154) -- (5.251,1.149) -- (5.599,1.145) -- (5.950,1.156);
  \draw[black] (5.950,1.156) -- (5.624,1.261) -- (5.296,1.380) -- (4.975,1.503) -- (4.649,1.615) -- (4.322,1.724) -- (3.995,1.843) -- (3.668,1.949) -- (3.332,2.050) -- (3.011,2.176) -- (2.681,2.268) -- (2.359,2.395) -- (2.030,2.508) -- (1.702,2.606) -- (1.380,2.736) -- (1.050,2.844);
  \draw[black] (1.050,1.325) -- (1.047,1.270) -- (1.055,1.213) -- (1.050,1.156);
  \draw[black] (1.050,1.156) -- (1.108,1.162) -- (1.161,1.151) -- (1.219,1.156);
  \draw[black] (1.050,1.325) -- (1.107,1.320) -- (1.164,1.319) -- (1.219,1.325);
  \draw[black] (1.219,1.325) -- (1.218,1.269) -- (1.225,1.212) -- (1.219,1.156);
  \draw[black] (1.423,2.715) -- (1.419,2.672) -- (1.396,2.637) -- (1.368,2.606) -- (1.347,2.570) -- (1.317,2.541) -- (1.283,2.516) -- (1.245,2.501) -- (1.212,2.473) -- (1.171,2.464) -- (1.134,2.444) -- (1.091,2.453) -- (1.050,2.440);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,1.023) {M};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,2.910) {K};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,1.023) {L};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {8,6 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 (1.437,2.234) {71°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{MKL}\)) = \(\dfrac{MK}{KL}\)
\par\addvspace{.5em}
 d'où cos(71\textdegree{}) = \(\dfrac{8{,}6}{KL}\)
\par\addvspace{.5em}
On a donc KL = \(\dfrac{8{,}6}{\cos(71^{\circ})}\)\(\approx\) 26,4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,3.142) -- (1.044,2.858) -- (1.063,2.570) -- (1.042,2.287) -- (1.037,2.003) -- (1.040,1.714) -- (1.061,1.430) -- (1.045,1.144) -- (1.050,0.858);
  \draw[black] (1.050,0.858) -- (1.403,0.860) -- (1.750,0.856) -- (2.099,0.844) -- (2.451,0.857) -- (2.801,0.849) -- (3.153,0.851) -- (3.503,0.858) -- (3.850,0.870) -- (4.199,0.852) -- (4.547,0.863) -- (4.901,0.867) -- (5.252,0.867) -- (5.599,0.867) -- (5.950,0.858);
  \draw[black] (5.950,0.858) -- (5.649,1.004) -- (5.337,1.148) -- (5.028,1.285) -- (4.726,1.426) -- (4.417,1.560) -- (4.111,1.705) -- (3.804,1.846) -- (3.494,1.994) -- (3.190,2.139) -- (2.895,2.294) -- (2.577,2.421) -- (2.279,2.577) -- (1.970,2.723) -- (1.667,2.865) -- (1.356,2.992) -- (1.050,3.142);
  \draw[black] (1.050,1.086) -- (1.056,1.011) -- (1.048,0.935) -- (1.050,0.858);
  \draw[black] (1.050,0.858) -- (1.125,0.859) -- (1.201,0.864) -- (1.278,0.858);
  \draw[black] (1.050,1.086) -- (1.126,1.091) -- (1.203,1.090) -- (1.278,1.086);
  \draw[black] (1.278,1.086) -- (1.275,1.011) -- (1.278,0.935) -- (1.278,0.858);
  \draw[black] (5.591,1.025) -- (5.583,1.013) -- (5.584,0.997) -- (5.580,0.983) -- (5.576,0.970) -- (5.564,0.958) -- (5.554,0.945) -- (5.566,0.928) -- (5.563,0.914) -- (5.561,0.900) -- (5.551,0.887) -- (5.544,0.872) -- (5.551,0.858);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,0.724) {V};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,3.209) {S};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,0.724) {Z};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.524) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {5,8 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.299,0.935) {25°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{VZS}\)) = \(\dfrac{VZ}{SZ}\)
\par\addvspace{.5em}
  d'où    cos(25\textdegree{}) =  \(\dfrac{VZ}{5{,}8}\)

On a donc VZ = 5,8 \(\times\) cos(25\textdegree{}) \(\approx\) 5{,}3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,2.854) -- (1.048,2.643) -- (1.063,2.426) -- (1.052,2.214) -- (1.045,2.003) -- (1.052,1.783) -- (1.058,1.576) -- (1.054,1.361) -- (1.050,1.146);
  \draw[black] (1.050,1.146) -- (1.398,1.140) -- (1.752,1.137) -- (2.101,1.143) -- (2.449,1.143) -- (2.798,1.151) -- (3.150,1.152) -- (3.497,1.143) -- (3.849,1.151) -- (4.201,1.141) -- (4.548,1.134) -- (4.902,1.150) -- (5.247,1.138) -- (5.599,1.136) -- (5.950,1.146);
  \draw[black] (5.950,1.146) -- (5.628,1.263) -- (5.297,1.378) -- (4.966,1.475) -- (4.648,1.608) -- (4.319,1.719) -- (3.993,1.829) -- (3.657,1.934) -- (3.338,2.054) -- (3.008,2.167) -- (2.687,2.295) -- (2.353,2.396) -- (2.032,2.517) -- (1.699,2.617) -- (1.377,2.739) -- (1.050,2.854);
  \draw[black] (1.050,1.317) -- (1.045,1.258) -- (1.051,1.203) -- (1.050,1.146);
  \draw[black] (1.050,1.146) -- (1.106,1.146) -- (1.166,1.141) -- (1.221,1.146);
  \draw[black] (1.050,1.317) -- (1.105,1.313) -- (1.163,1.322) -- (1.221,1.317);
  \draw[black] (1.221,1.317) -- (1.218,1.260) -- (1.224,1.203) -- (1.221,1.146);
  \draw[black] (5.581,1.274) -- (5.574,1.265) -- (5.563,1.257) -- (5.568,1.244) -- (5.556,1.235) -- (5.559,1.223) -- (5.557,1.212) -- (5.547,1.202) -- (5.546,1.191) -- (5.555,1.179) -- (5.556,1.168) -- (5.557,1.157) -- (5.548,1.146);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,1.012) {B};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,2.921) {A};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,1.012) {J};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {2,7 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {8,2 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.293,1.190) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{BJA}\)) = \(\dfrac{BA}{AJ}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{BJA}\)) =\(\dfrac{2{,}7}{8{,}2}\)
\par\addvspace{.5em}
 On a donc \(\widehat{BJA}\) = ArcSin(\(\dfrac{2{,}7}{8{,}2}\)) \(\approx\) 19\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.501,3.400) -- (1.503,3.053) -- (1.489,2.699) -- (1.495,2.350) -- (1.493,1.997) -- (1.489,1.650) -- (1.502,1.297) -- (1.500,0.952) -- (1.501,0.600);
  \draw[black] (1.501,0.600) -- (1.867,0.610) -- (2.229,0.611) -- (2.593,0.598) -- (2.954,0.607) -- (3.316,0.591) -- (3.682,0.598) -- (4.044,0.602) -- (4.410,0.609) -- (4.771,0.611) -- (5.138,0.596) -- (5.499,0.600);
  \draw[black] (5.499,0.600) -- (5.211,0.793) -- (4.928,1.003) -- (4.643,1.198) -- (4.354,1.394) -- (4.075,1.606) -- (3.779,1.795) -- (3.504,2.008) -- (3.210,2.199) -- (2.935,2.409) -- (2.648,2.604) -- (2.359,2.800) -- (2.080,3.006) -- (1.791,3.208) -- (1.501,3.400);
  \draw[black] (1.501,0.880) -- (1.501,0.787) -- (1.507,0.693) -- (1.501,0.600);
  \draw[black] (1.501,0.600) -- (1.595,0.604) -- (1.688,0.598) -- (1.781,0.600);
  \draw[black] (1.501,0.880) -- (1.594,0.878) -- (1.689,0.877) -- (1.781,0.880);
  \draw[black] (1.781,0.880) -- (1.787,0.787) -- (1.774,0.693) -- (1.781,0.600);
  \draw[black] (5.170,0.831) -- (5.157,0.815) -- (5.144,0.799) -- (5.148,0.773) -- (5.140,0.755) -- (5.133,0.736) -- (5.110,0.723) -- (5.110,0.701) -- (5.117,0.679) -- (5.110,0.660) -- (5.093,0.642) -- (5.104,0.620) -- (5.090,0.600);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.234,0.467) {G};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.234,3.467) {M};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.599,0.467) {W};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.001,1.933) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.267) {0,4 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (4.864,0.734) {35°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{GWM}\)) = \(\dfrac{GM}{GW}\)
\par\addvspace{.5em}
d'où  tan(35\textdegree{}) =  \(\dfrac{GM}{0{,}4}\)
\par\addvspace{.5em}
On a donc GM =  0,4 \(\times\) tan(35\textdegree{})  \(\approx\) 0,3 cm\end{document}
