\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 331}

\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=331}\\[2pt]
    {\small\ttfamily\bfseries\color{red} TRIG0331}
\end{minipage}%
}
}

\begin{document}

\sloppy

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

 Dans le triangle CBA rectangle en C, on sait que :
\begin{itemize}
\item CB=2,3 cm
\item BA=7,6 cm
\end{itemize}

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

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

 Dans le triangle NKJ rectangle en N, on sait que :
\begin{itemize}
\item NK=2,9 cm
\item NJ=6,2 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{NKJ}\).
\needspace{4\baselineskip}
\section*{
Exercice 3}

 Dans le triangle TLP rectangle en T, on sait que :
\begin{itemize}
\item TL=6,4 cm
\item \(\widehat{TLP}\) = 74\textdegree{}
\end{itemize}

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

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

 Dans le triangle AHR rectangle en A, on sait que :
\begin{itemize}
\item AR=4,8 cm
\item \(\widehat{AHR}\) = 48\textdegree{}
\end{itemize}

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

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

 Dans le triangle DJV rectangle en D, on sait que :
\begin{itemize}
\item DV=5,7 cm
\item \(\widehat{JVD}\) =  25\textdegree{}
\end{itemize}

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

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

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

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,2.778) -- (1.043,2.583) -- (1.061,2.388) -- (1.038,2.193) -- (1.049,2.000) -- (1.045,1.805) -- (1.063,1.610) -- (1.062,1.419) -- (1.050,1.222);
  \draw[black] (1.050,1.222) -- (1.400,1.222) -- (1.752,1.209) -- (2.101,1.227) -- (2.447,1.223) -- (2.801,1.234) -- (3.151,1.234) -- (3.502,1.212) -- (3.847,1.226) -- (4.202,1.230) -- (4.549,1.234) -- (4.902,1.215) -- (5.247,1.217) -- (5.597,1.214) -- (5.950,1.222);
  \draw[black] (5.950,1.222) -- (5.627,1.333) -- (5.296,1.431) -- (4.971,1.525) -- (4.641,1.641) -- (4.319,1.754) -- (3.997,1.855) -- (3.665,1.949) -- (3.337,2.052) -- (3.013,2.156) -- (2.683,2.266) -- (2.353,2.351) -- (2.030,2.467) -- (1.705,2.570) -- (1.379,2.677) -- (1.050,2.778);
  \draw[black] (1.050,1.378) -- (1.045,1.326) -- (1.055,1.272) -- (1.050,1.222);
  \draw[black] (1.050,1.222) -- (1.102,1.220) -- (1.154,1.222) -- (1.206,1.222);
  \draw[black] (1.050,1.378) -- (1.101,1.381) -- (1.154,1.375) -- (1.206,1.378);
  \draw[black] (1.206,1.378) -- (1.208,1.325) -- (1.210,1.275) -- (1.206,1.222);
  \draw[black] (5.569,1.343) -- (5.564,1.334) -- (5.566,1.323) -- (5.567,1.312) -- (5.568,1.302) -- (5.551,1.294) -- (5.555,1.283) -- (5.563,1.272) -- (5.558,1.262) -- (5.551,1.253) -- (5.547,1.243) -- (5.551,1.232) -- (5.544,1.222);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,1.089) {C};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,2.845) {B};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,1.089) {A};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {2,3 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {7,6 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.291,1.257) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{CAB}\)) = \(\dfrac{CB}{BA}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{CAB}\)) =\(\dfrac{2{,}3}{7{,}6}\)
\par\addvspace{.5em}
 On a donc \(\widehat{CAB}\) = ArcSin(\(\dfrac{2{,}3}{7{,}6}\)) \(\approx\) 18\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,3.146) -- (1.040,2.863) -- (1.061,2.572) -- (1.044,2.286) -- (1.060,1.998) -- (1.061,1.717) -- (1.037,1.427) -- (1.048,1.138) -- (1.050,0.854);
  \draw[black] (1.050,0.854) -- (1.398,0.847) -- (1.750,0.845) -- (2.100,0.853) -- (2.451,0.841) -- (2.800,0.851) -- (3.150,0.854) -- (3.501,0.843) -- (3.847,0.851) -- (4.202,0.852) -- (4.548,0.864) -- (4.901,0.850) -- (5.248,0.860) -- (5.603,0.847) -- (5.950,0.854);
  \draw[black] (5.950,0.854) -- (5.642,0.998) -- (5.336,1.138) -- (5.035,1.291) -- (4.726,1.430) -- (4.419,1.573) -- (4.117,1.720) -- (3.806,1.849) -- (3.505,2.011) -- (3.191,2.145) -- (2.886,2.280) -- (2.580,2.424) -- (2.270,2.562) -- (1.976,2.724) -- (1.663,2.855) -- (1.363,3.011) -- (1.050,3.146);
  \draw[black] (1.050,1.083) -- (1.053,1.007) -- (1.045,0.929) -- (1.050,0.854);
  \draw[black] (1.050,0.854) -- (1.128,0.860) -- (1.203,0.854) -- (1.279,0.854);
  \draw[black] (1.050,1.083) -- (1.127,1.085) -- (1.201,1.087) -- (1.279,1.083);
  \draw[black] (1.279,1.083) -- (1.277,1.006) -- (1.284,0.931) -- (1.279,0.854);
  \draw[black] (1.420,2.973) -- (1.402,2.939) -- (1.375,2.911) -- (1.345,2.887) -- (1.322,2.857) -- (1.293,2.834) -- (1.270,2.801) -- (1.233,2.788) -- (1.200,2.768) -- (1.160,2.767) -- (1.126,2.751) -- (1.088,2.745) -- (1.050,2.748);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,0.721) {N};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,3.213) {K};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,0.721) {J};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {2,9 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.521) {6,2 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.408,2.517) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{NKJ}\)) = \(\dfrac{NJ}{NK}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{NKJ}\)) = \(\dfrac{6{,}2}{2{,}9}\)
\par\addvspace{.5em}
 On a donc NKJ = ArcTan(\(\dfrac{6{,}2}{2{,}9}\)) \(\approx\) 65\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,2.703) -- (1.053,2.529) -- (1.063,2.350) -- (1.046,2.175) -- (1.042,1.999) -- (1.038,1.826) -- (1.055,1.649) -- (1.040,1.474) -- (1.050,1.297);
  \draw[black] (1.050,1.297) -- (1.397,1.307) -- (1.747,1.285) -- (2.097,1.301) -- (2.452,1.293) -- (2.800,1.308) -- (3.148,1.285) -- (3.501,1.296) -- (3.851,1.295) -- (4.198,1.310) -- (4.551,1.306) -- (4.900,1.296) -- (5.248,1.289) -- (5.602,1.295) -- (5.950,1.297);
  \draw[black] (5.950,1.297) -- (5.624,1.388) -- (5.298,1.486) -- (4.973,1.586) -- (4.639,1.660) -- (4.311,1.756) -- (3.988,1.855) -- (3.660,1.947) -- (3.335,2.040) -- (3.009,2.147) -- (2.685,2.238) -- (2.353,2.323) -- (2.031,2.431) -- (1.703,2.505) -- (1.380,2.617) -- (1.050,2.703);
  \draw[black] (1.050,1.438) -- (1.054,1.392) -- (1.048,1.343) -- (1.050,1.297);
  \draw[black] (1.050,1.297) -- (1.097,1.292) -- (1.143,1.300) -- (1.191,1.297);
  \draw[black] (1.050,1.438) -- (1.096,1.436) -- (1.143,1.444) -- (1.191,1.438);
  \draw[black] (1.191,1.438) -- (1.196,1.392) -- (1.191,1.345) -- (1.191,1.297);
  \draw[black] (1.435,2.592) -- (1.418,2.553) -- (1.410,2.508) -- (1.373,2.480) -- (1.359,2.437) -- (1.320,2.414) -- (1.288,2.386) -- (1.253,2.363) -- (1.214,2.344) -- (1.174,2.332) -- (1.136,2.311) -- (1.092,2.312) -- (1.050,2.293);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,1.164) {T};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,2.769) {L};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,1.164) {P};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {6,4 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.964) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.451,2.103) {74°};
\end{tikzpicture}\end{center}


 Dans le triangle TLP rectangle en T, on cherche une relation entre l'angle aigu   \(\widehat{TLP}\) son coté adjacent et son coté opposé.

\par\addvspace{.5em}
tan(\(\widehat{TLP}\)) = \(\dfrac{TP}{TL}\)
\par\addvspace{.5em}
d'où tan(74\textdegree{}) =\(\dfrac{TP}{6{,}4}\)
\par\addvspace{.5em}
On a donc TP = 6,4 \(\times\) tan(74\textdegree{}) \(\approx\) 22,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.945,3.400) -- (1.949,3.049) -- (1.951,2.697) -- (1.951,2.352) -- (1.949,1.999) -- (1.935,1.650) -- (1.938,1.303) -- (1.943,0.948) -- (1.945,0.600);
  \draw[black] (1.945,0.600) -- (2.290,0.599) -- (2.637,0.596) -- (2.984,0.606) -- (3.325,0.595) -- (3.671,0.587) -- (4.016,0.597) -- (4.366,0.613) -- (4.711,0.588) -- (5.055,0.600);
  \draw[black] (5.055,0.600) -- (4.801,0.837) -- (4.547,1.075) -- (4.279,1.304) -- (4.028,1.540) -- (3.759,1.763) -- (3.492,1.992) -- (3.238,2.234) -- (2.979,2.468) -- (2.724,2.696) -- (2.460,2.929) -- (2.207,3.166) -- (1.945,3.400);
  \draw[black] (1.945,0.880) -- (1.948,0.787) -- (1.947,0.695) -- (1.945,0.600);
  \draw[black] (1.945,0.600) -- (2.039,0.605) -- (2.131,0.602) -- (2.225,0.600);
  \draw[black] (1.945,0.880) -- (2.037,0.883) -- (2.131,0.882) -- (2.225,0.880);
  \draw[black] (2.225,0.880) -- (2.221,0.786) -- (2.219,0.694) -- (2.225,0.600);
  \draw[black] (2.239,3.135) -- (2.221,3.115) -- (2.198,3.099) -- (2.183,3.073) -- (2.153,3.068) -- (2.130,3.052) -- (2.108,3.034) -- (2.084,3.019) -- (2.053,3.023) -- (2.029,3.008) -- (2.001,3.001) -- (1.973,3.007) -- (1.945,2.992);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.678,0.467) {A};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.678,3.467) {H};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.155,0.467) {R};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.267) {4,8 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 (2.216,2.724) {48°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{AHR}\)) =  \(\dfrac{AR}{HR}\)
\par\addvspace{.5em}
 d'où sin(48\textdegree{}) = \(\dfrac{4{,}8}{HR}\)
\par\addvspace{.5em}
 On a donc HR = \(\dfrac{4{,}8}{\sin(48^{\circ})}\) \(\approx\) 6,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,3.142) -- (1.051,2.854) -- (1.049,2.574) -- (1.052,2.288) -- (1.049,1.999) -- (1.045,1.712) -- (1.043,1.431) -- (1.046,1.142) -- (1.050,0.858);
  \draw[black] (1.050,0.858) -- (1.398,0.861) -- (1.749,0.847) -- (2.099,0.846) -- (2.447,0.853) -- (2.799,0.854) -- (3.148,0.848) -- (3.503,0.869) -- (3.851,0.846) -- (4.199,0.855) -- (4.550,0.859) -- (4.897,0.849) -- (5.251,0.860) -- (5.600,0.858) -- (5.950,0.858);
  \draw[black] (5.950,0.858) -- (5.639,0.993) -- (5.335,1.144) -- (5.030,1.278) -- (4.722,1.416) -- (4.412,1.563) -- (4.107,1.707) -- (3.808,1.853) -- (3.498,1.997) -- (3.191,2.140) -- (2.894,2.297) -- (2.583,2.429) -- (2.270,2.562) -- (1.974,2.726) -- (1.666,2.863) -- (1.362,3.003) -- (1.050,3.142);
  \draw[black] (1.050,1.086) -- (1.056,1.011) -- (1.054,0.935) -- (1.050,0.858);
  \draw[black] (1.050,0.858) -- (1.127,0.853) -- (1.202,0.863) -- (1.278,0.858);
  \draw[black] (1.050,1.086) -- (1.128,1.081) -- (1.203,1.084) -- (1.278,1.086);
  \draw[black] (1.278,1.086) -- (1.283,1.010) -- (1.278,0.934) -- (1.278,0.858);
  \draw[black] (5.584,1.028) -- (5.586,1.011) -- (5.575,1.000) -- (5.574,0.985) -- (5.567,0.972) -- (5.565,0.958) -- (5.550,0.946) -- (5.563,0.929) -- (5.552,0.916) -- (5.553,0.901) -- (5.556,0.886) -- (5.554,0.872) -- (5.549,0.858);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,0.724) {D};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,3.209) {J};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,0.724) {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 (3.500,0.524) {5,7 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 DJV rectangle en D, on cherche une relation entre l'angle aigu   \(\widehat{DVJ}\) son coté opposé et son coté adjacent.

\par\addvspace{.5em}
tan(\(\widehat{DVJ}\)) = \(\dfrac{DJ}{DV}\)
\par\addvspace{.5em}
d'où  tan(25\textdegree{}) =  \(\dfrac{DJ}{5{,}7}\)
\par\addvspace{.5em}
On a donc DJ =  5,7 \(\times\) tan(25\textdegree{})  \(\approx\) 2,7 cm\end{document}
