\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 269}

\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=269}\\[2pt]
    {\small\ttfamily\bfseries\color{red} TRIG0269}
\end{minipage}%
}
}

\begin{document}

\sloppy

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

 Dans le triangle SNT rectangle en S, on sait que :
\begin{itemize}
\item ST=4,7 cm
\item \(\widehat{NTS}\) =  42\textdegree{}
\end{itemize}

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

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

 Dans le triangle DML rectangle en D, on sait que :
\begin{itemize}
\item ML=2 cm
\item \(\widehat{MLD}\) =  22\textdegree{}
\end{itemize}

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

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

 Dans le triangle HKZ rectangle en H, on sait que :
\begin{itemize}
\item HK=7,2 cm
\item \(\widehat{HKZ}\) = 48\textdegree{}
\end{itemize}

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

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

 Dans le triangle BPF rectangle en B, on sait que :
\begin{itemize}
\item BF=3,8 cm
\item PF=8,4 cm
\end{itemize}

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

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

 Dans le triangle ARW rectangle en A, on sait que :
\begin{itemize}
\item AW=4,9 cm
\item RW=9,2 cm
\end{itemize}

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

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

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.945,3.400) -- (1.957,3.048) -- (1.951,2.701) -- (1.936,2.353) -- (1.950,2.002) -- (1.949,1.649) -- (1.940,1.302) -- (1.957,0.950) -- (1.945,0.600);
  \draw[black] (1.945,0.600) -- (2.291,0.606) -- (2.637,0.592) -- (2.980,0.590) -- (3.329,0.604) -- (3.671,0.602) -- (4.020,0.613) -- (4.366,0.597) -- (4.707,0.589) -- (5.055,0.600);
  \draw[black] (5.055,0.600) -- (4.791,0.829) -- (4.537,1.065) -- (4.277,1.303) -- (4.021,1.538) -- (3.754,1.764) -- (3.494,1.994) -- (3.249,2.237) -- (2.980,2.469) -- (2.716,2.692) -- (2.461,2.929) -- (2.206,3.165) -- (1.945,3.400);
  \draw[black] (1.945,0.880) -- (1.942,0.787) -- (1.946,0.695) -- (1.945,0.600);
  \draw[black] (1.945,0.600) -- (2.038,0.600) -- (2.132,0.596) -- (2.225,0.600);
  \draw[black] (1.945,0.880) -- (2.038,0.879) -- (2.133,0.880) -- (2.225,0.880);
  \draw[black] (2.225,0.880) -- (2.227,0.785) -- (2.225,0.693) -- (2.225,0.600);
  \draw[black] (4.755,0.870) -- (4.738,0.852) -- (4.731,0.827) -- (4.722,0.804) -- (4.699,0.789) -- (4.685,0.769) -- (4.682,0.743) -- (4.678,0.719) -- (4.657,0.699) -- (4.669,0.671) -- (4.665,0.648) -- (4.647,0.625) -- (4.646,0.600);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.678,0.467) {S};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.678,3.467) {N};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.155,0.467) {T};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.445,1.933) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.267) {4,7 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (4.432,0.772) {42°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{STN}\)) = \(\dfrac{SN}{ST}\)
\par\addvspace{.5em}
d'où  tan(42\textdegree{}) =  \(\dfrac{SN}{4{,}7}\)
\par\addvspace{.5em}
On a donc SN =  4,7 \(\times\) tan(42\textdegree{})  \(\approx\) 4,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,2.990) -- (1.040,2.741) -- (1.050,2.498) -- (1.047,2.250) -- (1.041,2.000) -- (1.046,1.752) -- (1.059,1.507) -- (1.039,1.259) -- (1.050,1.010);
  \draw[black] (1.050,1.010) -- (1.397,0.999) -- (1.748,1.010) -- (2.100,1.002) -- (2.453,1.013) -- (2.800,1.015) -- (3.149,1.018) -- (3.501,1.007) -- (3.850,1.015) -- (4.201,1.020) -- (4.552,1.017) -- (4.898,1.015) -- (5.249,1.011) -- (5.601,0.999) -- (5.950,1.010);
  \draw[black] (5.950,1.010) -- (5.622,1.146) -- (5.302,1.285) -- (4.972,1.411) -- (4.644,1.547) -- (4.311,1.662) -- (3.992,1.806) -- (3.664,1.932) -- (3.338,2.077) -- (3.017,2.207) -- (2.687,2.336) -- (2.357,2.465) -- (2.028,2.587) -- (1.709,2.737) -- (1.374,2.851) -- (1.050,2.990);
  \draw[black] (1.050,1.208) -- (1.052,1.141) -- (1.049,1.075) -- (1.050,1.010);
  \draw[black] (1.050,1.010) -- (1.117,1.004) -- (1.183,1.008) -- (1.248,1.010);
  \draw[black] (1.050,1.208) -- (1.116,1.212) -- (1.182,1.208) -- (1.248,1.208);
  \draw[black] (1.248,1.208) -- (1.242,1.141) -- (1.252,1.076) -- (1.248,1.010);
  \draw[black] (5.572,1.163) -- (5.582,1.145) -- (5.564,1.138) -- (5.563,1.125) -- (5.555,1.114) -- (5.562,1.099) -- (5.554,1.087) -- (5.545,1.075) -- (5.560,1.060) -- (5.555,1.048) -- (5.541,1.036) -- (5.556,1.023) -- (5.542,1.010);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,0.877) {D};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,3.057) {M};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,0.877) {L};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.677) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {2 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.296,1.071) {22°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{DLM}\)) = \(\dfrac{DL}{ML}\)
\par\addvspace{.5em}
  d'où    cos(22\textdegree{}) =  \(\dfrac{DL}{2}\)

On a donc DL = 2 \(\times\) cos(22\textdegree{}) \(\approx\) 1{,}9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.945,3.400) -- (1.943,3.048) -- (1.937,2.701) -- (1.938,2.350) -- (1.958,2.003) -- (1.950,1.652) -- (1.933,1.301) -- (1.949,0.953) -- (1.945,0.600);
  \draw[black] (1.945,0.600) -- (2.291,0.606) -- (2.639,0.606) -- (2.983,0.594) -- (3.330,0.596) -- (3.670,0.589) -- (4.016,0.596) -- (4.362,0.592) -- (4.707,0.603) -- (5.055,0.600);
  \draw[black] (5.055,0.600) -- (4.799,0.836) -- (4.528,1.058) -- (4.274,1.292) -- (4.023,1.534) -- (3.754,1.765) -- (3.499,1.995) -- (3.246,2.239) -- (2.976,2.457) -- (2.723,2.698) -- (2.459,2.933) -- (2.208,3.172) -- (1.945,3.400);
  \draw[black] (1.945,0.880) -- (1.946,0.788) -- (1.951,0.693) -- (1.945,0.600);
  \draw[black] (1.945,0.600) -- (2.037,0.603) -- (2.131,0.603) -- (2.225,0.600);
  \draw[black] (1.945,0.880) -- (2.039,0.880) -- (2.130,0.878) -- (2.225,0.880);
  \draw[black] (2.225,0.880) -- (2.224,0.786) -- (2.230,0.694) -- (2.225,0.600);
  \draw[black] (2.247,3.129) -- (2.225,3.110) -- (2.202,3.094) -- (2.182,3.074) -- (2.159,3.058) -- (2.137,3.039) -- (2.107,3.036) -- (2.083,3.021) -- (2.058,3.007) -- (2.028,3.009) -- (2.000,3.010) -- (1.974,2.993) -- (1.945,3.009);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.678,0.467) {H};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.678,3.467) {K};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.155,0.467) {Z};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.445,1.933) {7,2 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 HKZ rectangle en H, on cherche une relation entre l'angle aigu   \(\widehat{HKZ}\) son coté adjacent et l'hypoténuse du triangle.

cos(\(\widehat{HKZ}\)) = \(\dfrac{HK}{KZ}\)
\par\addvspace{.5em}
 d'où cos(48\textdegree{}) = \(\dfrac{7{,}2}{KZ}\)
\par\addvspace{.5em}
On a donc KZ = \(\dfrac{7{,}2}{\cos(48^{\circ})}\)\(\approx\) 10,8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (2.790,3.400) -- (2.793,3.050) -- (2.790,2.703) -- (2.789,2.348) -- (2.788,1.998) -- (2.802,1.651) -- (2.795,1.303) -- (2.782,0.952) -- (2.790,0.600);
  \draw[black] (2.790,0.600) -- (2.966,0.598) -- (3.147,0.589) -- (3.325,0.588) -- (3.499,0.608) -- (3.675,0.594) -- (3.853,0.598) -- (4.031,0.589) -- (4.210,0.600);
  \draw[black] (4.210,0.600) -- (4.060,0.915) -- (3.893,1.220) -- (3.748,1.537) -- (3.580,1.842) -- (3.428,2.158) -- (3.263,2.467) -- (3.111,2.781) -- (2.937,3.085) -- (2.790,3.400);
  \draw[black] (2.790,0.742) -- (2.785,0.695) -- (2.793,0.647) -- (2.790,0.600);
  \draw[black] (2.790,0.600) -- (2.839,0.603) -- (2.886,0.604) -- (2.932,0.600);
  \draw[black] (2.790,0.742) -- (2.836,0.746) -- (2.884,0.740) -- (2.932,0.742);
  \draw[black] (2.932,0.742) -- (2.929,0.694) -- (2.926,0.648) -- (2.932,0.600);
  \draw[black] (4.029,0.956) -- (3.996,0.941) -- (3.970,0.913) -- (3.935,0.899) -- (3.920,0.862) -- (3.888,0.841) -- (3.861,0.814) -- (3.857,0.775) -- (3.839,0.743) -- (3.827,0.708) -- (3.820,0.672) -- (3.822,0.636) -- (3.806,0.600);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.523,0.467) {B};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.523,3.467) {P};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (4.310,0.467) {F};
  \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 (3.667,2.000) {8,4 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.642,0.882) {?};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{BFP}\)) = \(\dfrac{BF}{PF}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{BFP}\)) =  \(\dfrac{3{,}8}{8{,}4}\)
\par\addvspace{.5em}
On a donc \(\widehat{BFP}\) = Arccos (\(\dfrac{3{,}8}{8{,}4}\)) \(\approx\) 63\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (2.619,3.400) -- (2.621,3.053) -- (2.622,2.698) -- (2.610,2.347) -- (2.613,1.997) -- (2.609,1.651) -- (2.610,1.298) -- (2.628,0.950) -- (2.619,0.600);
  \draw[black] (2.619,0.600) -- (2.841,0.608) -- (3.058,0.593) -- (3.283,0.606) -- (3.501,0.610) -- (3.718,0.603) -- (3.937,0.609) -- (4.163,0.608) -- (4.381,0.600);
  \draw[black] (4.381,0.600) -- (4.186,0.912) -- (4.002,1.226) -- (3.802,1.538) -- (3.588,1.839) -- (3.403,2.159) -- (3.200,2.463) -- (3.023,2.783) -- (2.808,3.087) -- (2.619,3.400);
  \draw[black] (2.619,0.776) -- (2.617,0.719) -- (2.619,0.660) -- (2.619,0.600);
  \draw[black] (2.619,0.600) -- (2.678,0.601) -- (2.738,0.598) -- (2.795,0.600);
  \draw[black] (2.619,0.776) -- (2.677,0.780) -- (2.738,0.778) -- (2.795,0.776);
  \draw[black] (2.795,0.776) -- (2.794,0.716) -- (2.792,0.660) -- (2.795,0.600);
  \draw[black] (2.836,3.055) -- (2.812,3.059) -- (2.796,3.049) -- (2.781,3.038) -- (2.766,3.025) -- (2.750,3.016) -- (2.732,3.007) -- (2.713,3.005) -- (2.694,3.005) -- (2.674,3.013) -- (2.656,3.009) -- (2.638,2.998) -- (2.619,3.002);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.352,0.467) {A};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.352,3.467) {R};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (4.481,0.467) {W};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.267) {4,9 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {9,2 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.804,2.693) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{ARW}\)) =  \(\dfrac{AW}{RW}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{ARW}\)) = \(\dfrac{4{,}9}{9{,}2}\)
\par\addvspace{.5em}
On a donc \(\widehat{ARW}\) =  ArcSin(\(\dfrac{4{,}9}{9{,}2}\)) \(\approx\) 32\textdegree{}\end{document}
