\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 217}

\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=217}\\[2pt]
    {\small\ttfamily\bfseries\color{red} TRIG0217}
\end{minipage}%
}
}

\begin{document}

\sloppy

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

 Dans le triangle RMA rectangle en R, on sait que :
\begin{itemize}
\item RA=1,3 cm
\item \(\widehat{RMA}\) = 50\textdegree{}
\end{itemize}

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

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

 Dans le triangle DFV rectangle en D, on sait que :
\begin{itemize}
\item FV=8,2 cm
\item \(\widehat{DFV}\) = 55\textdegree{}
\end{itemize}

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

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

 Dans le triangle CSK rectangle en C, on sait que :
\begin{itemize}
\item CS=3,2 cm
\item SK=8,4 cm
\end{itemize}

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

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

 Dans le triangle CFL rectangle en C, on sait que :
\begin{itemize}
\item CF=6,3 cm
\item \(\widehat{CFL}\) = 65\textdegree{}
\end{itemize}

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

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

 Dans le triangle PLD rectangle en P, on sait que :
\begin{itemize}
\item PD=4,1 cm
\item LD=7,9 cm
\end{itemize}

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

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

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.832,3.400) -- (1.837,3.048) -- (1.841,2.700) -- (1.830,2.347) -- (1.822,1.999) -- (1.827,1.651) -- (1.830,1.299) -- (1.834,0.948) -- (1.832,0.600);
  \draw[black] (1.832,0.600) -- (2.169,0.613) -- (2.497,0.592) -- (2.831,0.612) -- (3.166,0.601) -- (3.502,0.603) -- (3.836,0.597) -- (4.166,0.595) -- (4.503,0.608) -- (4.837,0.592) -- (5.168,0.600);
  \draw[black] (5.168,0.600) -- (4.915,0.817) -- (4.650,1.022) -- (4.396,1.247) -- (4.143,1.460) -- (3.886,1.676) -- (3.628,1.889) -- (3.377,2.119) -- (3.109,2.314) -- (2.861,2.541) -- (2.601,2.756) -- (2.344,2.970) -- (2.083,3.180) -- (1.832,3.400);
  \draw[black] (1.832,0.880) -- (1.825,0.785) -- (1.829,0.694) -- (1.832,0.600);
  \draw[black] (1.832,0.600) -- (1.925,0.599) -- (2.018,0.606) -- (2.112,0.600);
  \draw[black] (1.832,0.880) -- (1.926,0.878) -- (2.018,0.874) -- (2.112,0.880);
  \draw[black] (2.112,0.880) -- (2.113,0.786) -- (2.113,0.694) -- (2.112,0.600);
  \draw[black] (2.136,3.144) -- (2.116,3.124) -- (2.100,3.099) -- (2.072,3.086) -- (2.052,3.065) -- (2.031,3.044) -- (2.000,3.038) -- (1.973,3.028) -- (1.949,3.009) -- (1.918,3.009) -- (1.889,3.005) -- (1.860,3.007) -- (1.832,3.006);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.565,0.467) {R};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.565,3.467) {M};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.268,0.467) {A};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.267) {1,3 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 (2.113,2.729) {50°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{RMA}\)) =  \(\dfrac{RA}{MA}\)
\par\addvspace{.5em}
 d'où sin(50\textdegree{}) = \(\dfrac{1{,}3}{MA}\)
\par\addvspace{.5em}
 On a donc MA = \(\dfrac{1{,}3}{\sin(50^{\circ})}\) \(\approx\) 1,7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.501,3.400) -- (1.491,3.053) -- (1.506,2.699) -- (1.509,2.348) -- (1.487,2.000) -- (1.489,1.651) -- (1.490,1.299) -- (1.501,0.950) -- (1.501,0.600);
  \draw[black] (1.501,0.600) -- (1.865,0.591) -- (2.226,0.597) -- (2.594,0.613) -- (2.956,0.602) -- (3.316,0.608) -- (3.680,0.607) -- (4.046,0.598) -- (4.406,0.602) -- (4.775,0.612) -- (5.134,0.591) -- (5.499,0.600);
  \draw[black] (5.499,0.600) -- (5.218,0.803) -- (4.927,0.999) -- (4.649,1.209) -- (4.360,1.400) -- (4.074,1.600) -- (3.780,1.796) -- (3.506,2.011) -- (3.210,2.199) -- (2.934,2.408) -- (2.642,2.595) -- (2.350,2.791) -- (2.077,3.003) -- (1.785,3.203) -- (1.501,3.400);
  \draw[black] (1.501,0.880) -- (1.500,0.786) -- (1.500,0.694) -- (1.501,0.600);
  \draw[black] (1.501,0.600) -- (1.593,0.598) -- (1.689,0.600) -- (1.781,0.600);
  \draw[black] (1.501,0.880) -- (1.595,0.878) -- (1.687,0.885) -- (1.781,0.880);
  \draw[black] (1.781,0.880) -- (1.779,0.787) -- (1.777,0.692) -- (1.781,0.600);
  \draw[black] (1.824,3.174) -- (1.807,3.146) -- (1.792,3.117) -- (1.768,3.095) -- (1.741,3.077) -- (1.708,3.069) -- (1.684,3.048) -- (1.656,3.032) -- (1.629,3.011) -- (1.593,3.021) -- (1.564,3.010) -- (1.532,3.010) -- (1.501,2.999);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.234,0.467) {D};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.234,3.467) {F};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.599,0.467) {V};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.001,1.933) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {8,2 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.808,2.742) {55°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{DFV}\)) = \(\dfrac{DF}{FV}\)
\par\addvspace{.5em}
d'où cos(55\textdegree{}) = \(\dfrac{DF}{8{,}2}\)

On a donc DF = 8,2 \(\times\) cos(55\textdegree{}) \(\approx\) 4,7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,3.009) -- (1.053,2.760) -- (1.047,2.503) -- (1.040,2.250) -- (1.042,2.000) -- (1.038,1.748) -- (1.047,1.496) -- (1.042,1.243) -- (1.050,0.991);
  \draw[black] (1.050,0.991) -- (1.397,0.988) -- (1.750,0.999) -- (2.099,0.982) -- (2.447,0.983) -- (2.801,0.991) -- (3.152,0.984) -- (3.502,0.979) -- (3.847,0.979) -- (4.201,1.004) -- (4.548,0.992) -- (4.900,0.989) -- (5.252,0.997) -- (5.601,0.986) -- (5.950,0.991);
  \draw[black] (5.950,0.991) -- (5.620,1.114) -- (5.299,1.267) -- (4.965,1.387) -- (4.638,1.523) -- (4.317,1.659) -- (3.985,1.789) -- (3.659,1.924) -- (3.334,2.064) -- (3.013,2.204) -- (2.689,2.348) -- (2.359,2.468) -- (2.031,2.600) -- (1.704,2.749) -- (1.370,2.866) -- (1.050,3.009);
  \draw[black] (1.050,1.192) -- (1.050,1.124) -- (1.044,1.056) -- (1.050,0.991);
  \draw[black] (1.050,0.991) -- (1.119,0.992) -- (1.186,0.986) -- (1.252,0.991);
  \draw[black] (1.050,1.192) -- (1.117,1.186) -- (1.183,1.199) -- (1.252,1.192);
  \draw[black] (1.252,1.192) -- (1.254,1.126) -- (1.256,1.056) -- (1.252,0.991);
  \draw[black] (5.574,1.145) -- (5.573,1.132) -- (5.570,1.119) -- (5.562,1.108) -- (5.554,1.096) -- (5.551,1.083) -- (5.555,1.069) -- (5.555,1.055) -- (5.552,1.043) -- (5.562,1.029) -- (5.552,1.016) -- (5.559,1.003) -- (5.549,0.991);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,0.857) {C};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,3.076) {S};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,0.857) {K};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {3,2 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 (5.296,1.053) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{CKS}\)) = \(\dfrac{CS}{SK}\)
\par\addvspace{.5em}
d'où sin(\(\widehat{CKS}\)) =\(\dfrac{3{,}2}{8{,}4}\)
\par\addvspace{.5em}
 On a donc \(\widehat{CKS}\) = ArcSin(\(\dfrac{3{,}2}{8{,}4}\)) \(\approx\) 22\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,3.142) -- (1.047,2.859) -- (1.045,2.571) -- (1.045,2.287) -- (1.043,2.000) -- (1.045,1.716) -- (1.037,1.431) -- (1.046,1.143) -- (1.050,0.858);
  \draw[black] (1.050,0.858) -- (1.397,0.845) -- (1.749,0.849) -- (2.097,0.854) -- (2.449,0.848) -- (2.799,0.854) -- (3.151,0.854) -- (3.503,0.864) -- (3.848,0.848) -- (4.203,0.847) -- (4.547,0.858) -- (4.897,0.854) -- (5.250,0.865) -- (5.599,0.867) -- (5.950,0.858);
  \draw[black] (5.950,0.858) -- (5.638,0.993) -- (5.336,1.135) -- (5.031,1.288) -- (4.730,1.439) -- (4.419,1.572) -- (4.113,1.721) -- (3.804,1.850) -- (3.503,2.005) -- (3.187,2.134) -- (2.893,2.297) -- (2.578,2.417) -- (2.275,2.576) -- (1.971,2.724) -- (1.662,2.860) -- (1.359,3.009) -- (1.050,3.142);
  \draw[black] (1.050,1.086) -- (1.045,1.010) -- (1.054,0.935) -- (1.050,0.858);
  \draw[black] (1.050,0.858) -- (1.128,0.854) -- (1.202,0.857) -- (1.278,0.858);
  \draw[black] (1.050,1.086) -- (1.127,1.088) -- (1.204,1.083) -- (1.278,1.086);
  \draw[black] (1.278,1.086) -- (1.278,1.009) -- (1.273,0.933) -- (1.278,0.858);
  \draw[black] (1.404,2.977) -- (1.387,2.944) -- (1.378,2.906) -- (1.344,2.884) -- (1.322,2.855) -- (1.292,2.831) -- (1.263,2.808) -- (1.230,2.791) -- (1.195,2.779) -- (1.160,2.764) -- (1.125,2.753) -- (1.089,2.735) -- (1.050,2.743);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,0.724) {C};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,3.209) {F};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,0.724) {L};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {6,3 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.524) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.408,2.514) {65°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{CFL}\)) = \(\dfrac{CL}{CF}\)
\par\addvspace{.5em}
d'où tan(65\textdegree{}) =\(\dfrac{CL}{6{,}3}\)
\par\addvspace{.5em}
On a donc CL = 6,3 \(\times\) tan(65\textdegree{}) \(\approx\) 13,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (2.650,3.400) -- (2.645,3.049) -- (2.652,2.700) -- (2.642,2.347) -- (2.638,2.002) -- (2.658,1.650) -- (2.641,1.298) -- (2.649,0.953) -- (2.650,0.600);
  \draw[black] (2.650,0.600) -- (2.864,0.609) -- (3.073,0.592) -- (3.290,0.606) -- (3.497,0.609) -- (3.710,0.602) -- (3.928,0.600) -- (4.140,0.588) -- (4.350,0.600);
  \draw[black] (4.350,0.600) -- (4.171,0.921) -- (3.970,1.221) -- (3.775,1.525) -- (3.590,1.840) -- (3.408,2.155) -- (3.220,2.469) -- (3.025,2.779) -- (2.838,3.085) -- (2.650,3.400);
  \draw[black] (2.650,0.770) -- (2.653,0.715) -- (2.644,0.657) -- (2.650,0.600);
  \draw[black] (2.650,0.600) -- (2.706,0.599) -- (2.765,0.597) -- (2.820,0.600);
  \draw[black] (2.650,0.770) -- (2.708,0.775) -- (2.762,0.773) -- (2.820,0.770);
  \draw[black] (2.820,0.770) -- (2.817,0.714) -- (2.818,0.655) -- (2.820,0.600);
  \draw[black] (2.857,3.060) -- (2.844,3.045) -- (2.823,3.047) -- (2.812,3.027) -- (2.792,3.027) -- (2.776,3.017) -- (2.755,3.024) -- (2.741,3.008) -- (2.722,3.007) -- (2.705,2.997) -- (2.687,2.999) -- (2.668,3.001) -- (2.650,2.995);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.383,0.467) {P};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.383,3.467) {L};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (4.450,0.467) {D};
  \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 (3.667,2.000) {7,9 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.830,2.691) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{PLD}\)) =  \(\dfrac{PD}{LD}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{PLD}\)) = \(\dfrac{4{,}1}{7{,}9}\)
\par\addvspace{.5em}
On a donc \(\widehat{PLD}\) =  ArcSin(\(\dfrac{4{,}1}{7{,}9}\)) \(\approx\) 31\textdegree{}\end{document}
