\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 85}

\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=85}\\[2pt]
    {\small\ttfamily\bfseries\color{red} TRIG0085}
\end{minipage}%
}
}

\begin{document}

\sloppy

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

 Dans le triangle ABC rectangle en A, on sait que :
\begin{itemize}
\item BC=5,9 cm
\item \(\widehat{ABC}\) = 67\textdegree{}
\end{itemize}

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

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

 Dans le triangle KJN rectangle en K, on sait que :
\begin{itemize}
\item KJ=1,8 cm
\item \(\widehat{JNK}\) =  42\textdegree{}
\end{itemize}

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

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

 Dans le triangle GHF rectangle en G, on sait que :
\begin{itemize}
\item GF=4 cm
\item \(\widehat{HFG}\) =  42\textdegree{}
\end{itemize}

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

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

 Dans le triangle GMA rectangle en G, on sait que :
\begin{itemize}
\item GA=5,9 cm
\item MA=8,7 cm
\end{itemize}

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

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

 Dans le triangle KPB rectangle en K, on sait que :
\begin{itemize}
\item KB=5,2 cm
\item PB=9,4 cm
\end{itemize}

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

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

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,3.040) -- (1.043,2.780) -- (1.057,2.518) -- (1.060,2.259) -- (1.060,1.997) -- (1.063,1.740) -- (1.050,1.477) -- (1.049,1.223) -- (1.050,0.960);
  \draw[black] (1.050,0.960) -- (1.401,0.950) -- (1.752,0.948) -- (2.097,0.972) -- (2.448,0.969) -- (2.802,0.962) -- (3.148,0.948) -- (3.503,0.956) -- (3.849,0.969) -- (4.197,0.961) -- (4.548,0.972) -- (4.898,0.968) -- (5.247,0.956) -- (5.600,0.963) -- (5.950,0.960);
  \draw[black] (5.950,0.960) -- (5.626,1.099) -- (5.290,1.229) -- (4.972,1.378) -- (4.641,1.506) -- (4.315,1.651) -- (3.990,1.800) -- (3.666,1.932) -- (3.337,2.070) -- (3.014,2.220) -- (2.681,2.341) -- (2.357,2.490) -- (2.034,2.635) -- (1.707,2.766) -- (1.378,2.906) -- (1.050,3.040);
  \draw[black] (1.050,1.168) -- (1.050,1.100) -- (1.050,1.028) -- (1.050,0.960);
  \draw[black] (1.050,0.960) -- (1.119,0.955) -- (1.188,0.960) -- (1.258,0.960);
  \draw[black] (1.050,1.168) -- (1.118,1.172) -- (1.188,1.170) -- (1.258,1.168);
  \draw[black] (1.258,1.168) -- (1.253,1.098) -- (1.257,1.031) -- (1.258,0.960);
  \draw[black] (1.420,2.883) -- (1.409,2.844) -- (1.373,2.821) -- (1.352,2.789) -- (1.332,2.755) -- (1.308,2.722) -- (1.266,2.714) -- (1.234,2.692) -- (1.200,2.674) -- (1.166,2.656) -- (1.126,2.656) -- (1.088,2.651) -- (1.050,2.650);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,0.827) {A};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,3.107) {B};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,0.827) {C};
  \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 (3.667,2.000) {5,9 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.418,2.417) {67°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{ABC}\)) = \(\dfrac{AB}{BC}\)
\par\addvspace{.5em}
d'où cos(67\textdegree{}) = \(\dfrac{AB}{5{,}9}\)

On a donc AB = 5,9 \(\times\) cos(67\textdegree{}) \(\approx\) 2,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.945,3.400) -- (1.938,3.048) -- (1.946,2.702) -- (1.957,2.350) -- (1.953,2.003) -- (1.957,1.649) -- (1.945,1.301) -- (1.941,0.950) -- (1.945,0.600);
  \draw[black] (1.945,0.600) -- (2.293,0.600) -- (2.637,0.599) -- (2.981,0.595) -- (3.326,0.587) -- (3.671,0.605) -- (4.020,0.604) -- (4.362,0.603) -- (4.709,0.597) -- (5.055,0.600);
  \draw[black] (5.055,0.600) -- (4.806,0.841) -- (4.539,1.067) -- (4.284,1.310) -- (4.020,1.534) -- (3.754,1.762) -- (3.509,2.007) -- (3.235,2.223) -- (2.988,2.474) -- (2.718,2.693) -- (2.456,2.927) -- (2.202,3.159) -- (1.945,3.400);
  \draw[black] (1.945,0.880) -- (1.948,0.785) -- (1.941,0.694) -- (1.945,0.600);
  \draw[black] (1.945,0.600) -- (2.040,0.598) -- (2.132,0.606) -- (2.225,0.600);
  \draw[black] (1.945,0.880) -- (2.037,0.879) -- (2.131,0.883) -- (2.225,0.880);
  \draw[black] (2.225,0.880) -- (2.226,0.786) -- (2.226,0.694) -- (2.225,0.600);
  \draw[black] (4.758,0.868) -- (4.744,0.848) -- (4.723,0.832) -- (4.710,0.811) -- (4.707,0.785) -- (4.683,0.770) -- (4.685,0.742) -- (4.671,0.721) -- (4.660,0.698) -- (4.665,0.672) -- (4.650,0.650) -- (4.652,0.625) -- (4.660,0.600);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.678,0.467) {K};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.678,3.467) {J};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.155,0.467) {N};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.445,1.933) {1,8 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 (4.432,0.772) {42°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{KNJ}\)) = \(\dfrac{KJ}{KN}\)
\par\addvspace{.5em}
d'où  tan(42\textdegree{}) =  \(\dfrac{1{,}8}{KN}\)
\par\addvspace{.5em}
On a donc KN = \(\dfrac{1{,}8}{\tan(42^{\circ})}\) \(\approx\) 2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.945,3.400) -- (1.948,3.047) -- (1.934,2.699) -- (1.934,2.351) -- (1.951,2.001) -- (1.947,1.650) -- (1.939,1.300) -- (1.958,0.951) -- (1.945,0.600);
  \draw[black] (1.945,0.600) -- (2.289,0.587) -- (2.639,0.588) -- (2.981,0.594) -- (3.328,0.598) -- (3.672,0.608) -- (4.016,0.608) -- (4.365,0.588) -- (4.709,0.598) -- (5.055,0.600);
  \draw[black] (5.055,0.600) -- (4.794,0.834) -- (4.532,1.059) -- (4.283,1.305) -- (4.017,1.532) -- (3.759,1.770) -- (3.498,2.001) -- (3.240,2.235) -- (2.973,2.458) -- (2.728,2.704) -- (2.461,2.928) -- (2.201,3.163) -- (1.945,3.400);
  \draw[black] (1.945,0.880) -- (1.951,0.787) -- (1.944,0.692) -- (1.945,0.600);
  \draw[black] (1.945,0.600) -- (2.039,0.599) -- (2.132,0.602) -- (2.225,0.600);
  \draw[black] (1.945,0.880) -- (2.039,0.875) -- (2.133,0.886) -- (2.225,0.880);
  \draw[black] (2.225,0.880) -- (2.227,0.788) -- (2.224,0.692) -- (2.225,0.600);
  \draw[black] (4.762,0.864) -- (4.744,0.847) -- (4.733,0.826) -- (4.713,0.809) -- (4.699,0.789) -- (4.692,0.766) -- (4.673,0.747) -- (4.676,0.719) -- (4.661,0.698) -- (4.658,0.674) -- (4.659,0.649) -- (4.660,0.624) -- (4.648,0.600);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.678,0.467) {G};
  \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) {F};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.267) {4 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 (4.432,0.772) {42°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{GFH}\)) = \(\dfrac{GF}{HF}\)
\par\addvspace{.5em}
  d'où    cos(42\textdegree{}) =  \(\dfrac{4}{HF}\)
\par\addvspace{.5em}
 On a donc HF =  \(\dfrac{4}{\cos(42^{\circ})}\) \(\approx\) 5,4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (2.208,3.400) -- (2.220,3.051) -- (2.202,2.700) -- (2.216,2.353) -- (2.211,2.001) -- (2.219,1.648) -- (2.198,1.299) -- (2.211,0.949) -- (2.208,0.600);
  \draw[black] (2.208,0.600) -- (2.531,0.602) -- (2.853,0.589) -- (3.178,0.603) -- (3.502,0.608) -- (3.820,0.591) -- (4.147,0.597) -- (4.472,0.604) -- (4.792,0.600);
  \draw[black] (4.792,0.600) -- (4.553,0.852) -- (4.321,1.107) -- (4.091,1.366) -- (3.844,1.613) -- (3.627,1.881) -- (3.390,2.130) -- (3.145,2.380) -- (2.908,2.629) -- (2.674,2.890) -- (2.440,3.140) -- (2.208,3.400);
  \draw[black] (2.208,0.858) -- (2.215,0.771) -- (2.213,0.687) -- (2.208,0.600);
  \draw[black] (2.208,0.600) -- (2.296,0.598) -- (2.379,0.607) -- (2.466,0.600);
  \draw[black] (2.208,0.858) -- (2.295,0.860) -- (2.381,0.864) -- (2.466,0.858);
  \draw[black] (2.466,0.858) -- (2.470,0.774) -- (2.463,0.687) -- (2.466,0.600);
  \draw[black] (4.526,0.889) -- (4.501,0.875) -- (4.486,0.852) -- (4.474,0.827) -- (4.457,0.806) -- (4.436,0.786) -- (4.424,0.761) -- (4.420,0.733) -- (4.413,0.707) -- (4.406,0.681) -- (4.394,0.655) -- (4.397,0.627) -- (4.383,0.600);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.941,0.467) {G};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.941,3.467) {M};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (4.892,0.467) {A};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.267) {5,9 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {8,7 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (4.181,0.801) {?};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{GAM}\)) = \(\dfrac{GA}{MA}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{GAM}\)) =  \(\dfrac{5{,}9}{8{,}7}\)
\par\addvspace{.5em}
On a donc \(\widehat{GAM}\) = Arccos (\(\dfrac{5{,}9}{8{,}7}\)) \(\approx\) 47\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (2.570,3.400) -- (2.569,3.050) -- (2.569,2.702) -- (2.580,2.348) -- (2.557,1.999) -- (2.571,1.653) -- (2.557,1.302) -- (2.574,0.950) -- (2.570,0.600);
  \draw[black] (2.570,0.600) -- (2.801,0.596) -- (3.036,0.593) -- (3.265,0.601) -- (3.497,0.599) -- (3.735,0.601) -- (3.965,0.589) -- (4.196,0.600) -- (4.430,0.600);
  \draw[black] (4.430,0.600) -- (4.251,0.881) -- (4.051,1.157) -- (3.880,1.448) -- (3.688,1.723) -- (3.493,1.994) -- (3.316,2.280) -- (3.127,2.557) -- (2.937,2.835) -- (2.755,3.121) -- (2.570,3.400);
  \draw[black] (2.570,0.786) -- (2.571,0.723) -- (2.564,0.662) -- (2.570,0.600);
  \draw[black] (2.570,0.600) -- (2.632,0.598) -- (2.695,0.595) -- (2.756,0.600);
  \draw[black] (2.570,0.786) -- (2.632,0.780) -- (2.694,0.785) -- (2.756,0.786);
  \draw[black] (2.756,0.786) -- (2.756,0.725) -- (2.755,0.664) -- (2.756,0.600);
  \draw[black] (2.788,3.072) -- (2.778,3.052) -- (2.760,3.044) -- (2.738,3.043) -- (2.725,3.026) -- (2.707,3.017) -- (2.683,3.026) -- (2.667,3.010) -- (2.650,2.999) -- (2.628,3.008) -- (2.610,2.996) -- (2.590,2.991) -- (2.570,2.991);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.304,0.467) {K};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.304,3.467) {P};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (4.530,0.467) {B};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.267) {5,2 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {9,4 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.763,2.695) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{KPB}\)) =  \(\dfrac{KB}{PB}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{KPB}\)) = \(\dfrac{5{,}2}{9{,}4}\)
\par\addvspace{.5em}
On a donc \(\widehat{KPB}\) =  ArcSin(\(\dfrac{5{,}2}{9{,}4}\)) \(\approx\) 34\textdegree{}\end{document}
