\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 3}

\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=3}\\[2pt]
    {\small\ttfamily\bfseries\color{red} TRIG0003}
\end{minipage}%
}
}

\begin{document}

\sloppy

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

 Dans le triangle KDH rectangle en K, on sait que :
\begin{itemize}
\item KD=1,7 cm
\item DH=10 cm
\end{itemize}

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

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

 Dans le triangle PZD rectangle en P, on sait que :
\begin{itemize}
\item ZD=4,2 cm
\item \(\widehat{ZDP}\) =  23\textdegree{}
\end{itemize}

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

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

 Dans le triangle VZG rectangle en V, on sait que :
\begin{itemize}
\item ZG=7,9 cm
\item \(\widehat{VZG}\) = 56\textdegree{}
\end{itemize}

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

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

 Dans le triangle JBA rectangle en J, on sait que :
\begin{itemize}
\item JA=3,8 cm
\item \(\widehat{JBA}\) = 58\textdegree{}
\end{itemize}

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

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

 Dans le triangle RCV rectangle en R, on sait que :
\begin{itemize}
\item RC=1,3 cm
\item RV=4,4 cm
\end{itemize}

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

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

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,2.423) -- (1.044,2.316) -- (1.062,2.210) -- (1.048,2.107) -- (1.063,1.998) -- (1.051,1.893) -- (1.040,1.787) -- (1.039,1.684) -- (1.050,1.577);
  \draw[black] (1.050,1.577) -- (1.400,1.569) -- (1.753,1.566) -- (2.098,1.588) -- (2.453,1.585) -- (2.801,1.579) -- (3.150,1.583) -- (3.499,1.580) -- (3.851,1.575) -- (4.201,1.567) -- (4.552,1.566) -- (4.901,1.564) -- (5.252,1.586) -- (5.603,1.565) -- (5.950,1.577);
  \draw[black] (5.950,1.577) -- (5.599,1.640) -- (5.247,1.691) -- (4.902,1.760) -- (4.546,1.808) -- (4.205,1.892) -- (3.852,1.947) -- (3.500,1.991) -- (3.149,2.068) -- (2.797,2.109) -- (2.447,2.169) -- (2.097,2.238) -- (1.752,2.315) -- (1.404,2.375) -- (1.050,2.423);
  \draw[black] (1.050,1.662) -- (1.056,1.633) -- (1.056,1.607) -- (1.050,1.577);
  \draw[black] (1.050,1.577) -- (1.079,1.579) -- (1.105,1.582) -- (1.135,1.577);
  \draw[black] (1.050,1.662) -- (1.078,1.657) -- (1.108,1.656) -- (1.135,1.662);
  \draw[black] (1.135,1.662) -- (1.136,1.634) -- (1.134,1.605) -- (1.135,1.577);
  \draw[black] (5.554,1.646) -- (5.554,1.640) -- (5.552,1.634) -- (5.545,1.630) -- (5.552,1.623) -- (5.553,1.617) -- (5.547,1.612) -- (5.556,1.605) -- (5.555,1.600) -- (5.541,1.595) -- (5.550,1.589) -- (5.549,1.583) -- (5.560,1.577);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,1.444) {K};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,2.489) {D};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,1.444) {H};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {1,7 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {10 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.286,1.568) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{KHD}\)) = \(\dfrac{KD}{DH}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{KHD}\)) =\(\dfrac{1{,}7}{10}\)
\par\addvspace{.5em}
 On a donc \(\widehat{KHD}\) = ArcSin(\(\dfrac{1{,}7}{10}\)) \(\approx\) 10\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,3.040) -- (1.043,2.783) -- (1.048,2.518) -- (1.055,2.262) -- (1.042,2.002) -- (1.058,1.740) -- (1.045,1.483) -- (1.055,1.221) -- (1.050,0.960);
  \draw[black] (1.050,0.960) -- (1.400,0.951) -- (1.748,0.951) -- (2.101,0.964) -- (2.450,0.950) -- (2.797,0.967) -- (3.148,0.959) -- (3.502,0.952) -- (3.849,0.968) -- (4.201,0.952) -- (4.549,0.952) -- (4.899,0.970) -- (5.248,0.952) -- (5.599,0.964) -- (5.950,0.960);
  \draw[black] (5.950,0.960) -- (5.623,1.096) -- (5.303,1.247) -- (4.962,1.365) -- (4.644,1.525) -- (4.313,1.653) -- (3.996,1.802) -- (3.664,1.938) -- (3.338,2.073) -- (3.009,2.202) -- (2.683,2.339) -- (2.359,2.487) -- (2.036,2.630) -- (1.706,2.773) -- (1.379,2.913) -- (1.050,3.040);
  \draw[black] (1.050,1.168) -- (1.045,1.098) -- (1.056,1.029) -- (1.050,0.960);
  \draw[black] (1.050,0.960) -- (1.119,0.964) -- (1.189,0.958) -- (1.258,0.960);
  \draw[black] (1.050,1.168) -- (1.121,1.169) -- (1.188,1.167) -- (1.258,1.168);
  \draw[black] (1.258,1.168) -- (1.261,1.097) -- (1.258,1.028) -- (1.258,0.960);
  \draw[black] (5.585,1.115) -- (5.582,1.102) -- (5.572,1.092) -- (5.562,1.081) -- (5.571,1.064) -- (5.555,1.054) -- (5.550,1.041) -- (5.550,1.028) -- (5.547,1.014) -- (5.562,0.999) -- (5.548,0.987) -- (5.543,0.974) -- (5.559,0.960);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,0.827) {P};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,3.107) {Z};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,0.827) {D};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.627) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {4,2 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.297,1.026) {23°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{PDZ}\)) = \(\dfrac{PD}{ZD}\)
\par\addvspace{.5em}
  d'où    cos(23\textdegree{}) =  \(\dfrac{PD}{4{,}2}\)

On a donc PD = 4,2 \(\times\) cos(23\textdegree{}) \(\approx\) 3{,}9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.424,3.400) -- (1.431,3.047) -- (1.426,2.702) -- (1.438,2.349) -- (1.417,2.000) -- (1.437,1.650) -- (1.420,1.300) -- (1.421,0.953) -- (1.424,0.600);
  \draw[black] (1.424,0.600) -- (1.767,0.590) -- (2.118,0.597) -- (2.465,0.605) -- (2.805,0.602) -- (3.153,0.608) -- (3.502,0.598) -- (3.843,0.607) -- (4.191,0.613) -- (4.538,0.587) -- (4.881,0.590) -- (5.233,0.595) -- (5.576,0.600);
  \draw[black] (5.576,0.600) -- (5.300,0.793) -- (5.023,0.974) -- (4.741,1.151) -- (4.467,1.349) -- (4.195,1.538) -- (3.912,1.714) -- (3.638,1.908) -- (3.365,2.096) -- (3.091,2.286) -- (2.816,2.473) -- (2.526,2.645) -- (2.259,2.844) -- (1.977,3.021) -- (1.708,3.225) -- (1.424,3.400);
  \draw[black] (1.424,0.880) -- (1.424,0.788) -- (1.424,0.695) -- (1.424,0.600);
  \draw[black] (1.424,0.600) -- (1.518,0.599) -- (1.610,0.595) -- (1.704,0.600);
  \draw[black] (1.424,0.880) -- (1.519,0.886) -- (1.611,0.878) -- (1.704,0.880);
  \draw[black] (1.704,0.880) -- (1.702,0.785) -- (1.708,0.693) -- (1.704,0.600);
  \draw[black] (1.763,3.171) -- (1.738,3.149) -- (1.722,3.119) -- (1.690,3.105) -- (1.664,3.086) -- (1.645,3.056) -- (1.615,3.041) -- (1.587,3.024) -- (1.552,3.023) -- (1.522,3.011) -- (1.490,3.004) -- (1.458,2.993) -- (1.424,2.996);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.158,0.467) {V};
  \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) {G};
  \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 (3.667,2.000) {7,9 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.737,2.745) {56°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{VZG}\)) = \(\dfrac{VZ}{ZG}\)
\par\addvspace{.5em}
d'où cos(56\textdegree{}) = \(\dfrac{VZ}{7{,}9}\)

On a donc VZ = 7,9 \(\times\) cos(56\textdegree{}) \(\approx\) 4,4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.260,3.400) -- (1.268,3.050) -- (1.266,2.700) -- (1.250,2.351) -- (1.254,1.999) -- (1.267,1.649) -- (1.249,1.299) -- (1.272,0.947) -- (1.260,0.600);
  \draw[black] (1.260,0.600) -- (1.607,0.611) -- (1.951,0.592) -- (2.296,0.612) -- (2.639,0.590) -- (2.980,0.593) -- (3.326,0.608) -- (3.674,0.590) -- (4.018,0.610) -- (4.359,0.595) -- (4.704,0.588) -- (5.049,0.612) -- (5.398,0.611) -- (5.740,0.600);
  \draw[black] (5.740,0.600) -- (5.438,0.777) -- (5.151,0.981) -- (4.842,1.160) -- (4.546,1.341) -- (4.245,1.529) -- (3.941,1.711) -- (3.653,1.912) -- (3.346,2.084) -- (3.060,2.287) -- (2.753,2.465) -- (2.456,2.653) -- (2.155,2.841) -- (1.860,3.030) -- (1.557,3.213) -- (1.260,3.400);
  \draw[black] (1.260,0.880) -- (1.257,0.785) -- (1.255,0.695) -- (1.260,0.600);
  \draw[black] (1.260,0.600) -- (1.352,0.594) -- (1.446,0.599) -- (1.540,0.600);
  \draw[black] (1.260,0.880) -- (1.353,0.878) -- (1.446,0.886) -- (1.540,0.880);
  \draw[black] (1.540,0.880) -- (1.536,0.786) -- (1.538,0.693) -- (1.540,0.600);
  \draw[black] (1.593,3.192) -- (1.578,3.162) -- (1.565,3.128) -- (1.534,3.111) -- (1.505,3.094) -- (1.483,3.067) -- (1.458,3.043) -- (1.424,3.032) -- (1.393,3.020) -- (1.359,3.014) -- (1.326,3.010) -- (1.293,3.008) -- (1.260,3.000);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.993,0.467) {J};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.993,3.467) {B};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.840,0.467) {A};
  \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) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.583,2.750) {58°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{JBA}\)) =  \(\dfrac{JA}{BA}\)
\par\addvspace{.5em}
 d'où sin(58\textdegree{}) = \(\dfrac{3{,}8}{BA}\)
\par\addvspace{.5em}
 On a donc BA = \(\dfrac{3{,}8}{\sin(58^{\circ})}\) \(\approx\) 4,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,2.724) -- (1.061,2.545) -- (1.043,2.360) -- (1.058,2.183) -- (1.052,1.998) -- (1.042,1.817) -- (1.048,1.640) -- (1.053,1.459) -- (1.050,1.276);
  \draw[black] (1.050,1.276) -- (1.400,1.275) -- (1.748,1.284) -- (2.097,1.269) -- (2.453,1.268) -- (2.799,1.264) -- (3.151,1.275) -- (3.503,1.273) -- (3.849,1.271) -- (4.200,1.267) -- (4.552,1.273) -- (4.902,1.289) -- (5.250,1.272) -- (5.600,1.276) -- (5.950,1.276);
  \draw[black] (5.950,1.276) -- (5.621,1.365) -- (5.293,1.457) -- (4.971,1.569) -- (4.641,1.665) -- (4.322,1.768) -- (3.991,1.863) -- (3.661,1.940) -- (3.337,2.041) -- (3.008,2.147) -- (2.680,2.237) -- (2.360,2.342) -- (2.032,2.444) -- (1.705,2.542) -- (1.379,2.633) -- (1.050,2.724);
  \draw[black] (1.050,1.421) -- (1.046,1.371) -- (1.045,1.325) -- (1.050,1.276);
  \draw[black] (1.050,1.276) -- (1.098,1.273) -- (1.145,1.281) -- (1.195,1.276);
  \draw[black] (1.050,1.421) -- (1.097,1.427) -- (1.147,1.425) -- (1.195,1.421);
  \draw[black] (1.195,1.421) -- (1.192,1.373) -- (1.196,1.324) -- (1.195,1.276);
  \draw[black] (1.430,2.612) -- (1.419,2.570) -- (1.405,2.529) -- (1.377,2.496) -- (1.347,2.466) -- (1.328,2.425) -- (1.292,2.400) -- (1.253,2.382) -- (1.215,2.362) -- (1.176,2.344) -- (1.135,2.331) -- (1.092,2.332) -- (1.050,2.334);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,1.143) {R};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,2.791) {C};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,1.143) {V};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {1,3 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.943) {4,4 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.449,2.123) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{RCV}\)) = \(\dfrac{RV}{RC}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{RCV}\)) = \(\dfrac{4{,}4}{1{,}3}\)
\par\addvspace{.5em}
 On a donc RCV = ArcTan(\(\dfrac{4{,}4}{1{,}3}\)) \(\approx\) 74\textdegree{}\end{document}
