\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 68}

\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=68}\\[2pt]
    {\small\ttfamily\bfseries\color{red} TRIG0068}
\end{minipage}%
}
}

\begin{document}

\sloppy

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

 Dans le triangle WZC rectangle en W, on sait que :
\begin{itemize}
\item WC=8,1 cm
\item \(\widehat{ZCW}\) =  34\textdegree{}
\end{itemize}

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

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

 Dans le triangle DNT rectangle en D, on sait que :
\begin{itemize}
\item DT=5,8 cm
\item NT=9,6 cm
\end{itemize}

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

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

 Dans le triangle WGZ rectangle en W, on sait que :
\begin{itemize}
\item WZ=1,9 cm
\item \(\widehat{WGZ}\) = 68\textdegree{}
\end{itemize}

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

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

 Dans le triangle RHL rectangle en R, on sait que :
\begin{itemize}
\item RL=5,5 cm
\item HL=9,7 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{RHL}\).
\needspace{4\baselineskip}
\section*{
Exercice 5}

 Dans le triangle KPL rectangle en K, on sait que :
\begin{itemize}
\item KP=9 cm
\item \(\widehat{KPL}\) = 54\textdegree{}
\end{itemize}

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

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

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

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.424,3.400) -- (1.417,3.053) -- (1.416,2.700) -- (1.428,2.352) -- (1.420,2.000) -- (1.437,1.647) -- (1.434,1.300) -- (1.434,0.948) -- (1.424,0.600);
  \draw[black] (1.424,0.600) -- (1.773,0.613) -- (2.117,0.603) -- (2.460,0.606) -- (2.808,0.602) -- (3.152,0.589) -- (3.502,0.609) -- (3.849,0.612) -- (4.190,0.603) -- (4.540,0.607) -- (4.883,0.604) -- (5.232,0.587) -- (5.576,0.600);
  \draw[black] (5.576,0.600) -- (5.302,0.796) -- (5.023,0.975) -- (4.743,1.160) -- (4.469,1.347) -- (4.199,1.541) -- (3.907,1.710) -- (3.643,1.913) -- (3.356,2.088) -- (3.084,2.284) -- (2.814,2.475) -- (2.531,2.647) -- (2.249,2.832) -- (1.981,3.027) -- (1.709,3.220) -- (1.424,3.400);
  \draw[black] (1.424,0.880) -- (1.429,0.787) -- (1.419,0.692) -- (1.424,0.600);
  \draw[black] (1.424,0.600) -- (1.517,0.602) -- (1.612,0.597) -- (1.704,0.600);
  \draw[black] (1.424,0.880) -- (1.519,0.882) -- (1.612,0.884) -- (1.704,0.880);
  \draw[black] (1.704,0.880) -- (1.706,0.785) -- (1.708,0.692) -- (1.704,0.600);
  \draw[black] (5.237,0.828) -- (5.230,0.809) -- (5.222,0.790) -- (5.206,0.776) -- (5.208,0.754) -- (5.206,0.733) -- (5.194,0.717) -- (5.178,0.700) -- (5.184,0.678) -- (5.172,0.660) -- (5.185,0.639) -- (5.174,0.620) -- (5.180,0.600);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.158,0.467) {W};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.158,3.467) {Z};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.676,0.467) {C};
  \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) {8,1 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 WZC rectangle en W, on cherche une relation entre l'angle aigu   \(\widehat{WCZ}\) son coté opposé et son coté adjacent.

\par\addvspace{.5em}
tan(\(\widehat{WCZ}\)) = \(\dfrac{WZ}{WC}\)
\par\addvspace{.5em}
d'où  tan(34\textdegree{}) =  \(\dfrac{WZ}{8{,}1}\)
\par\addvspace{.5em}
On a donc WZ =  8,1 \(\times\) tan(34\textdegree{})  \(\approx\) 5,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (2.439,3.400) -- (2.443,3.050) -- (2.427,2.702) -- (2.448,2.347) -- (2.433,1.999) -- (2.444,1.651) -- (2.426,1.299) -- (2.428,0.951) -- (2.439,0.600);
  \draw[black] (2.439,0.600) -- (2.707,0.591) -- (2.966,0.604) -- (3.233,0.599) -- (3.500,0.607) -- (3.762,0.608) -- (4.032,0.591) -- (4.294,0.613) -- (4.561,0.600);
  \draw[black] (4.561,0.600) -- (4.352,0.882) -- (4.126,1.154) -- (3.916,1.435) -- (3.708,1.720) -- (3.501,2.000) -- (3.294,2.285) -- (3.079,2.563) -- (2.863,2.844) -- (2.654,3.121) -- (2.439,3.400);
  \draw[black] (2.439,0.812) -- (2.434,0.742) -- (2.441,0.670) -- (2.439,0.600);
  \draw[black] (2.439,0.600) -- (2.509,0.595) -- (2.581,0.595) -- (2.651,0.600);
  \draw[black] (2.439,0.812) -- (2.509,0.814) -- (2.579,0.816) -- (2.651,0.812);
  \draw[black] (2.651,0.812) -- (2.647,0.743) -- (2.645,0.669) -- (2.651,0.600);
  \draw[black] (4.324,0.913) -- (4.292,0.904) -- (4.269,0.883) -- (4.250,0.858) -- (4.239,0.827) -- (4.224,0.801) -- (4.198,0.780) -- (4.190,0.750) -- (4.185,0.720) -- (4.179,0.690) -- (4.167,0.661) -- (4.171,0.630) -- (4.158,0.600);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.172,0.467) {D};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.172,3.467) {N};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (4.661,0.467) {T};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.267) {5,8 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {9,6 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.964,0.830) {?};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{DTN}\)) = \(\dfrac{DT}{NT}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{DTN}\)) =  \(\dfrac{5{,}8}{9{,}6}\)
\par\addvspace{.5em}
On a donc \(\widehat{DTN}\) = Arccos (\(\dfrac{5{,}8}{9{,}6}\)) \(\approx\) 53\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,2.990) -- (1.049,2.739) -- (1.060,2.492) -- (1.063,2.246) -- (1.044,1.999) -- (1.058,1.754) -- (1.059,1.502) -- (1.051,1.260) -- (1.050,1.010);
  \draw[black] (1.050,1.010) -- (1.397,0.998) -- (1.750,1.018) -- (2.102,1.009) -- (2.450,1.000) -- (2.803,1.023) -- (3.150,1.007) -- (3.502,1.002) -- (3.850,1.021) -- (4.201,1.003) -- (4.549,1.003) -- (4.898,1.015) -- (5.250,1.023) -- (5.598,1.010) -- (5.950,1.010);
  \draw[black] (5.950,1.010) -- (5.622,1.135) -- (5.304,1.283) -- (4.972,1.419) -- (4.641,1.526) -- (4.314,1.661) -- (3.988,1.795) -- (3.668,1.946) -- (3.334,2.058) -- (3.009,2.205) -- (2.685,2.328) -- (2.354,2.450) -- (2.027,2.591) -- (1.707,2.734) -- (1.373,2.856) -- (1.050,2.990);
  \draw[black] (1.050,1.208) -- (1.055,1.141) -- (1.050,1.077) -- (1.050,1.010);
  \draw[black] (1.050,1.010) -- (1.117,1.009) -- (1.181,1.010) -- (1.248,1.010);
  \draw[black] (1.050,1.208) -- (1.117,1.212) -- (1.183,1.208) -- (1.248,1.208);
  \draw[black] (1.248,1.208) -- (1.248,1.141) -- (1.250,1.078) -- (1.248,1.010);
  \draw[black] (1.419,2.841) -- (1.400,2.806) -- (1.380,2.773) -- (1.364,2.735) -- (1.341,2.702) -- (1.302,2.686) -- (1.273,2.659) -- (1.240,2.638) -- (1.202,2.625) -- (1.164,2.616) -- (1.127,2.607) -- (1.090,2.590) -- (1.050,2.596);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,0.877) {W};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,3.057) {G};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,0.877) {Z};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.677) {1,9 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 (1.423,2.371) {68°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{WGZ}\)) =  \(\dfrac{WZ}{GZ}\)
\par\addvspace{.5em}
 d'où sin(68\textdegree{}) = \(\dfrac{1{,}9}{GZ}\)
\par\addvspace{.5em}
 On a donc GZ = \(\dfrac{1{,}9}{\sin(68^{\circ})}\) \(\approx\) 2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (2.536,3.400) -- (2.548,3.050) -- (2.533,2.697) -- (2.524,2.347) -- (2.524,2.003) -- (2.537,1.648) -- (2.547,1.303) -- (2.541,0.950) -- (2.536,0.600);
  \draw[black] (2.536,0.600) -- (2.778,0.610) -- (3.019,0.605) -- (3.260,0.610) -- (3.501,0.612) -- (3.738,0.610) -- (3.984,0.591) -- (4.225,0.611) -- (4.464,0.600);
  \draw[black] (4.464,0.600) -- (4.282,0.884) -- (4.071,1.156) -- (3.895,1.448) -- (3.695,1.721) -- (3.493,1.995) -- (3.315,2.284) -- (3.106,2.550) -- (2.934,2.844) -- (2.740,3.124) -- (2.536,3.400);
  \draw[black] (2.536,0.793) -- (2.534,0.730) -- (2.535,0.664) -- (2.536,0.600);
  \draw[black] (2.536,0.600) -- (2.600,0.604) -- (2.666,0.602) -- (2.729,0.600);
  \draw[black] (2.536,0.793) -- (2.599,0.799) -- (2.663,0.794) -- (2.729,0.793);
  \draw[black] (2.729,0.793) -- (2.735,0.727) -- (2.732,0.664) -- (2.729,0.600);
  \draw[black] (2.766,3.067) -- (2.748,3.056) -- (2.732,3.043) -- (2.710,3.042) -- (2.690,3.038) -- (2.677,3.017) -- (2.653,3.024) -- (2.636,3.012) -- (2.616,3.007) -- (2.597,2.998) -- (2.577,2.998) -- (2.557,2.996) -- (2.536,2.991);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.270,0.467) {R};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.270,3.467) {H};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (4.564,0.467) {L};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.267) {5,5 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {9,7 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.734,2.697) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{RHL}\)) =  \(\dfrac{RL}{HL}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{RHL}\)) = \(\dfrac{5{,}5}{9{,}7}\)
\par\addvspace{.5em}
On a donc \(\widehat{RHL}\) =  ArcSin(\(\dfrac{5{,}5}{9{,}7}\)) \(\approx\) 35\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.573,3.400) -- (1.568,3.051) -- (1.566,2.698) -- (1.569,2.349) -- (1.580,2.001) -- (1.572,1.651) -- (1.565,1.301) -- (1.580,0.949) -- (1.573,0.600);
  \draw[black] (1.573,0.600) -- (1.925,0.602) -- (2.275,0.603) -- (2.623,0.589) -- (2.976,0.596) -- (3.325,0.598) -- (3.677,0.593) -- (4.026,0.607) -- (4.378,0.594) -- (4.727,0.613) -- (5.074,0.604) -- (5.427,0.600);
  \draw[black] (5.427,0.600) -- (5.158,0.811) -- (4.869,0.991) -- (4.596,1.192) -- (4.320,1.394) -- (4.044,1.589) -- (3.784,1.808) -- (3.494,1.992) -- (3.233,2.210) -- (2.944,2.389) -- (2.672,2.595) -- (2.395,2.792) -- (2.117,2.991) -- (1.855,3.206) -- (1.573,3.400);
  \draw[black] (1.573,0.880) -- (1.571,0.786) -- (1.570,0.694) -- (1.573,0.600);
  \draw[black] (1.573,0.600) -- (1.666,0.598) -- (1.758,0.599) -- (1.853,0.600);
  \draw[black] (1.573,0.880) -- (1.667,0.887) -- (1.761,0.874) -- (1.853,0.880);
  \draw[black] (1.853,0.880) -- (1.857,0.788) -- (1.857,0.692) -- (1.853,0.600);
  \draw[black] (1.892,3.168) -- (1.884,3.135) -- (1.862,3.111) -- (1.836,3.092) -- (1.812,3.071) -- (1.780,3.062) -- (1.759,3.035) -- (1.726,3.031) -- (1.694,3.029) -- (1.667,3.010) -- (1.635,3.007) -- (1.604,3.001) -- (1.573,2.995);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.306,0.467) {K};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.306,3.467) {P};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.527,0.467) {L};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.073,1.933) {9 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.267) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.876,2.739) {54°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{KPL}\)) = \(\dfrac{KL}{KP}\)
\par\addvspace{.5em}
d'où tan(54\textdegree{}) =\(\dfrac{KL}{9}\)
\par\addvspace{.5em}
On a donc KL = 9 \(\times\) tan(54\textdegree{}) \(\approx\) 12,4 cm\end{document}
