\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 294}

\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=294}\\[2pt]
    {\small\ttfamily\bfseries\color{red} TRIG0294}
\end{minipage}%
}
}

\begin{document}

\sloppy

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

 Dans le triangle PFA rectangle en P, on sait que :
\begin{itemize}
\item PA=7 cm
\item \(\widehat{FAP}\) =  45\textdegree{}
\end{itemize}

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

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

 Dans le triangle GJN rectangle en G, on sait que :
\begin{itemize}
\item JN=9,4 cm
\item \(\widehat{GJN}\) = 70\textdegree{}
\end{itemize}

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

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

 Dans le triangle FBR rectangle en F, on sait que :
\begin{itemize}
\item FB=2,5 cm
\item BR=6,9 cm
\end{itemize}

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

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

 Dans le triangle KGN rectangle en K, on sait que :
\begin{itemize}
\item KG=3,8 cm
\item \(\widehat{KGN}\) = 54\textdegree{}
\end{itemize}

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

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

 Dans le triangle DGT rectangle en D, on sait que :
\begin{itemize}
\item DG=3,2 cm
\item DT=4,1 cm
\end{itemize}

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

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

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (2.100,3.400) -- (2.106,3.050) -- (2.088,2.702) -- (2.091,2.348) -- (2.103,1.997) -- (2.099,1.649) -- (2.109,1.300) -- (2.107,0.950) -- (2.100,0.600);
  \draw[black] (2.100,0.600) -- (2.450,0.605) -- (2.798,0.593) -- (3.150,0.612) -- (3.500,0.598) -- (3.852,0.609) -- (4.197,0.603) -- (4.552,0.597) -- (4.900,0.600);
  \draw[black] (4.900,0.600) -- (4.654,0.859) -- (4.398,1.113) -- (4.142,1.367) -- (3.889,1.624) -- (3.631,1.872) -- (3.372,2.128) -- (3.126,2.391) -- (2.863,2.640) -- (2.605,2.887) -- (2.360,3.149) -- (2.100,3.400);
  \draw[black] (2.100,0.880) -- (2.103,0.787) -- (2.096,0.694) -- (2.100,0.600);
  \draw[black] (2.100,0.600) -- (2.194,0.599) -- (2.286,0.605) -- (2.380,0.600);
  \draw[black] (2.100,0.880) -- (2.194,0.878) -- (2.287,0.886) -- (2.380,0.880);
  \draw[black] (2.380,0.880) -- (2.385,0.788) -- (2.375,0.693) -- (2.380,0.600);
  \draw[black] (4.619,0.881) -- (4.602,0.862) -- (4.590,0.838) -- (4.561,0.826) -- (4.560,0.796) -- (4.549,0.773) -- (4.526,0.755) -- (4.523,0.728) -- (4.509,0.705) -- (4.510,0.678) -- (4.505,0.652) -- (4.507,0.626) -- (4.499,0.600);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.833,0.467) {P};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.833,3.467) {F};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.000,0.467) {A};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.267) {7 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 (4.284,0.788) {45°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{PAF}\)) = \(\dfrac{PA}{FA}\)
\par\addvspace{.5em}
  d'où    cos(45\textdegree{}) =  \(\dfrac{7}{FA}\)
\par\addvspace{.5em}
 On a donc FA =  \(\dfrac{7}{\cos(45^{\circ})}\) \(\approx\) 9,9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,2.892) -- (1.041,2.669) -- (1.038,2.445) -- (1.063,2.225) -- (1.060,1.999) -- (1.040,1.777) -- (1.037,1.556) -- (1.042,1.333) -- (1.050,1.108);
  \draw[black] (1.050,1.108) -- (1.399,1.122) -- (1.749,1.113) -- (2.097,1.118) -- (2.452,1.109) -- (2.799,1.103) -- (3.148,1.110) -- (3.501,1.104) -- (3.849,1.121) -- (4.197,1.098) -- (4.553,1.098) -- (4.900,1.116) -- (5.253,1.113) -- (5.600,1.116) -- (5.950,1.108);
  \draw[black] (5.950,1.108) -- (5.620,1.216) -- (5.295,1.341) -- (4.967,1.452) -- (4.645,1.597) -- (4.322,1.711) -- (3.997,1.832) -- (3.664,1.939) -- (3.341,2.065) -- (3.010,2.181) -- (2.688,2.304) -- (2.353,2.414) -- (2.030,2.541) -- (1.705,2.654) -- (1.379,2.785) -- (1.050,2.892);
  \draw[black] (1.050,1.287) -- (1.048,1.228) -- (1.055,1.167) -- (1.050,1.108);
  \draw[black] (1.050,1.108) -- (1.110,1.110) -- (1.170,1.114) -- (1.228,1.108);
  \draw[black] (1.050,1.287) -- (1.109,1.287) -- (1.168,1.283) -- (1.228,1.287);
  \draw[black] (1.228,1.287) -- (1.227,1.228) -- (1.225,1.169) -- (1.228,1.108);
  \draw[black] (1.427,2.754) -- (1.404,2.720) -- (1.392,2.681) -- (1.367,2.649) -- (1.337,2.621) -- (1.309,2.592) -- (1.283,2.559) -- (1.241,2.549) -- (1.210,2.520) -- (1.173,2.501) -- (1.133,2.490) -- (1.092,2.485) -- (1.050,2.489);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,0.975) {G};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,2.958) {J};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,0.975) {N};
  \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) {9,4 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.432,2.279) {70°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{GJN}\)) = \(\dfrac{GJ}{JN}\)
\par\addvspace{.5em}
d'où cos(70\textdegree{}) = \(\dfrac{GJ}{9{,}4}\)

On a donc GJ = 9,4 \(\times\) cos(70\textdegree{}) \(\approx\) 3,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,2.952) -- (1.063,2.712) -- (1.057,2.477) -- (1.060,2.239) -- (1.043,2.003) -- (1.061,1.762) -- (1.056,1.523) -- (1.057,1.289) -- (1.050,1.048);
  \draw[black] (1.050,1.048) -- (1.400,1.041) -- (1.749,1.039) -- (2.102,1.056) -- (2.451,1.041) -- (2.803,1.038) -- (3.150,1.042) -- (3.498,1.037) -- (3.849,1.045) -- (4.202,1.036) -- (4.548,1.047) -- (4.901,1.056) -- (5.247,1.052) -- (5.602,1.046) -- (5.950,1.048);
  \draw[black] (5.950,1.048) -- (5.623,1.173) -- (5.293,1.298) -- (4.972,1.435) -- (4.645,1.556) -- (4.311,1.677) -- (3.992,1.820) -- (3.666,1.939) -- (3.334,2.057) -- (3.012,2.196) -- (2.680,2.316) -- (2.359,2.445) -- (2.034,2.581) -- (1.701,2.700) -- (1.372,2.817) -- (1.050,2.952);
  \draw[black] (1.050,1.238) -- (1.046,1.174) -- (1.052,1.112) -- (1.050,1.048);
  \draw[black] (1.050,1.048) -- (1.114,1.047) -- (1.176,1.047) -- (1.240,1.048);
  \draw[black] (1.050,1.238) -- (1.112,1.238) -- (1.175,1.237) -- (1.240,1.238);
  \draw[black] (1.240,1.238) -- (1.247,1.174) -- (1.238,1.112) -- (1.240,1.048);
  \draw[black] (5.570,1.195) -- (5.581,1.178) -- (5.574,1.168) -- (5.564,1.158) -- (5.565,1.145) -- (5.555,1.134) -- (5.552,1.122) -- (5.554,1.109) -- (5.552,1.097) -- (5.547,1.085) -- (5.558,1.072) -- (5.541,1.060) -- (5.558,1.048);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,0.914) {F};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,3.019) {B};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,0.914) {R};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {2,5 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {6,9 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.295,1.104) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{FRB}\)) = \(\dfrac{FB}{BR}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{FRB}\)) =\(\dfrac{2{,}5}{6{,}9}\)
\par\addvspace{.5em}
 On a donc \(\widehat{FRB}\) = ArcSin(\(\dfrac{2{,}5}{6{,}9}\)) \(\approx\) 21\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.573,3.400) -- (1.582,3.052) -- (1.569,2.703) -- (1.575,2.350) -- (1.581,2.003) -- (1.563,1.652) -- (1.561,1.297) -- (1.562,0.949) -- (1.573,0.600);
  \draw[black] (1.573,0.600) -- (1.923,0.605) -- (2.272,0.587) -- (2.624,0.604) -- (2.978,0.605) -- (3.323,0.588) -- (3.672,0.601) -- (4.023,0.602) -- (4.377,0.589) -- (4.725,0.609) -- (5.079,0.587) -- (5.427,0.600);
  \draw[black] (5.427,0.600) -- (5.145,0.792) -- (4.881,1.004) -- (4.599,1.193) -- (4.319,1.390) -- (4.054,1.602) -- (3.775,1.795) -- (3.509,2.009) -- (3.226,2.206) -- (2.954,2.401) -- (2.682,2.608) -- (2.394,2.791) -- (2.118,2.994) -- (1.848,3.202) -- (1.573,3.400);
  \draw[black] (1.573,0.880) -- (1.579,0.788) -- (1.573,0.693) -- (1.573,0.600);
  \draw[black] (1.573,0.600) -- (1.666,0.602) -- (1.761,0.594) -- (1.853,0.600);
  \draw[black] (1.573,0.880) -- (1.668,0.875) -- (1.759,0.876) -- (1.853,0.880);
  \draw[black] (1.853,0.880) -- (1.848,0.788) -- (1.850,0.694) -- (1.853,0.600);
  \draw[black] (1.892,3.168) -- (1.877,3.140) -- (1.853,3.120) -- (1.828,3.102) -- (1.806,3.080) -- (1.782,3.058) -- (1.754,3.046) -- (1.723,3.038) -- (1.698,3.014) -- (1.669,3.001) -- (1.636,3.003) -- (1.604,3.004) -- (1.573,2.992);
  \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) {G};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.527,0.467) {N};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.073,1.933) {3,8 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 (1.876,2.739) {54°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{KGN}\)) = \(\dfrac{KN}{KG}\)
\par\addvspace{.5em}
d'où tan(54\textdegree{}) =\(\dfrac{KN}{3{,}8}\)
\par\addvspace{.5em}
On a donc KN = 3,8 \(\times\) tan(54\textdegree{}) \(\approx\) 5,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.706,3.400) -- (1.711,3.050) -- (1.694,2.697) -- (1.706,2.347) -- (1.694,1.999) -- (1.699,1.649) -- (1.705,1.299) -- (1.718,0.951) -- (1.706,0.600);
  \draw[black] (1.706,0.600) -- (2.064,0.590) -- (2.423,0.610) -- (2.779,0.611) -- (3.142,0.599) -- (3.503,0.601) -- (3.855,0.600) -- (4.216,0.609) -- (4.579,0.600) -- (4.936,0.590) -- (5.294,0.600);
  \draw[black] (5.294,0.600) -- (5.018,0.812) -- (4.747,1.037) -- (4.472,1.254) -- (4.189,1.461) -- (3.911,1.675) -- (3.634,1.887) -- (3.358,2.104) -- (3.081,2.316) -- (2.816,2.549) -- (2.525,2.745) -- (2.265,2.975) -- (1.982,3.182) -- (1.706,3.400);
  \draw[black] (1.706,0.880) -- (1.701,0.787) -- (1.705,0.694) -- (1.706,0.600);
  \draw[black] (1.706,0.600) -- (1.800,0.600) -- (1.894,0.604) -- (1.986,0.600);
  \draw[black] (1.706,0.880) -- (1.800,0.882) -- (1.894,0.885) -- (1.986,0.880);
  \draw[black] (1.986,0.880) -- (1.990,0.786) -- (1.989,0.694) -- (1.986,0.600);
  \draw[black] (2.016,3.158) -- (2.009,3.124) -- (1.982,3.107) -- (1.955,3.093) -- (1.930,3.076) -- (1.904,3.063) -- (1.879,3.046) -- (1.851,3.035) -- (1.828,3.009) -- (1.795,3.015) -- (1.768,2.996) -- (1.736,3.008) -- (1.706,3.008);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.440,0.467) {D};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.440,3.467) {G};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.394,0.467) {T};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.206,1.933) {3,2 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.267) {4,1 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.999,2.734) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{DGT}\)) = \(\dfrac{DT}{DG}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{DGT}\)) = \(\dfrac{4{,}1}{3{,}2}\)
\par\addvspace{.5em}
 On a donc DGT = ArcTan(\(\dfrac{4{,}1}{3{,}2}\)) \(\approx\) 52\textdegree{}\end{document}
