\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 211}

\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=211}\\[2pt]
    {\small\ttfamily\bfseries\color{red} TRIG0211}
\end{minipage}%
}
}

\begin{document}

\sloppy

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

 Dans le triangle PTV rectangle en P, on sait que :
\begin{itemize}
\item PV=5 cm
\item TV=9,5 cm
\end{itemize}

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

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

 Dans le triangle BAH rectangle en B, on sait que :
\begin{itemize}
\item BH=5,9 cm
\item AH=7,6 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{BAH}\).
\needspace{4\baselineskip}
\section*{
Exercice 3}

 Dans le triangle PKD rectangle en P, on sait que :
\begin{itemize}
\item KD=1,4 cm
\item \(\widehat{PKD}\) = 56\textdegree{}
\end{itemize}

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

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

 Dans le triangle GTA rectangle en G, on sait que :
\begin{itemize}
\item GA=9 cm
\item \(\widehat{GTA}\) = 48\textdegree{}
\end{itemize}

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

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

 Dans le triangle FKL rectangle en F, on sait que :
\begin{itemize}
\item FK=4,3 cm
\item \(\widehat{FKL}\) = 64\textdegree{}
\end{itemize}

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

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

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

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (2.633,3.400) -- (2.631,3.047) -- (2.642,2.700) -- (2.621,2.350) -- (2.629,2.001) -- (2.645,1.651) -- (2.636,1.299) -- (2.639,0.948) -- (2.633,0.600);
  \draw[black] (2.633,0.600) -- (2.848,0.606) -- (3.067,0.591) -- (3.282,0.602) -- (3.497,0.599) -- (3.716,0.590) -- (3.932,0.601) -- (4.152,0.609) -- (4.367,0.600);
  \draw[black] (4.367,0.600) -- (4.172,0.911) -- (3.985,1.225) -- (3.788,1.534) -- (3.598,1.847) -- (3.414,2.161) -- (3.209,2.469) -- (3.015,2.775) -- (2.829,3.094) -- (2.633,3.400);
  \draw[black] (2.633,0.773) -- (2.630,0.716) -- (2.633,0.657) -- (2.633,0.600);
  \draw[black] (2.633,0.600) -- (2.690,0.602) -- (2.750,0.601) -- (2.807,0.600);
  \draw[black] (2.633,0.773) -- (2.692,0.768) -- (2.749,0.773) -- (2.807,0.773);
  \draw[black] (2.807,0.773) -- (2.811,0.715) -- (2.810,0.659) -- (2.807,0.600);
  \draw[black] (4.159,0.936) -- (4.133,0.915) -- (4.106,0.894) -- (4.082,0.872) -- (4.059,0.847) -- (4.034,0.824) -- (4.009,0.799) -- (3.997,0.767) -- (3.990,0.733) -- (3.984,0.699) -- (3.975,0.667) -- (3.963,0.634) -- (3.965,0.600);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.367,0.467) {P};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.367,3.467) {T};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (4.467,0.467) {V};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.267) {5 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {9,5 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.784,0.858) {?};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{PVT}\)) = \(\dfrac{PV}{TV}\)
\par\addvspace{.5em}
  d'où    cos(\(\widehat{PVT}\)) =  \(\dfrac{5}{9{,}5}\)
\par\addvspace{.5em}
On a donc \(\widehat{PVT}\) = Arccos (\(\dfrac{5}{9{,}5}\)) \(\approx\) 58\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.776,3.400) -- (1.785,3.052) -- (1.782,2.700) -- (1.764,2.352) -- (1.774,2.002) -- (1.783,1.647) -- (1.786,1.302) -- (1.771,0.947) -- (1.776,0.600);
  \draw[black] (1.776,0.600) -- (2.119,0.600) -- (2.464,0.605) -- (2.811,0.601) -- (3.158,0.595) -- (3.499,0.613) -- (3.843,0.595) -- (4.188,0.603) -- (4.532,0.594) -- (4.881,0.596) -- (5.224,0.600);
  \draw[black] (5.224,0.600) -- (4.952,0.811) -- (4.703,1.038) -- (4.430,1.253) -- (4.162,1.458) -- (3.895,1.669) -- (3.642,1.902) -- (3.365,2.103) -- (3.094,2.316) -- (2.841,2.545) -- (2.570,2.750) -- (2.300,2.963) -- (2.048,3.194) -- (1.776,3.400);
  \draw[black] (1.776,0.880) -- (1.779,0.787) -- (1.777,0.693) -- (1.776,0.600);
  \draw[black] (1.776,0.600) -- (1.869,0.603) -- (1.963,0.596) -- (2.056,0.600);
  \draw[black] (1.776,0.880) -- (1.870,0.882) -- (1.963,0.886) -- (2.056,0.880);
  \draw[black] (2.056,0.880) -- (2.061,0.788) -- (2.055,0.695) -- (2.056,0.600);
  \draw[black] (2.083,3.151) -- (2.071,3.121) -- (2.049,3.101) -- (2.019,3.090) -- (1.995,3.075) -- (1.978,3.046) -- (1.945,3.045) -- (1.919,3.030) -- (1.895,3.008) -- (1.865,3.003) -- (1.835,3.001) -- (1.805,3.009) -- (1.776,3.000);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.509,0.467) {B};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.509,3.467) {A};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.324,0.467) {H};
  \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) {7,6 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.062,2.731) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{BAH}\)) =  \(\dfrac{BH}{AH}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{BAH}\)) = \(\dfrac{5{,}9}{7{,}6}\)
\par\addvspace{.5em}
On a donc \(\widehat{BAH}\) =  ArcSin(\(\dfrac{5{,}9}{7{,}6}\)) \(\approx\) 51\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.424,3.400) -- (1.435,3.049) -- (1.420,2.703) -- (1.420,2.353) -- (1.436,2.002) -- (1.411,1.650) -- (1.428,1.298) -- (1.435,0.952) -- (1.424,0.600);
  \draw[black] (1.424,0.600) -- (1.768,0.591) -- (2.115,0.613) -- (2.461,0.593) -- (2.808,0.600) -- (3.153,0.599) -- (3.503,0.597) -- (3.848,0.603) -- (4.190,0.611) -- (4.535,0.598) -- (4.885,0.609) -- (5.228,0.608) -- (5.576,0.600);
  \draw[black] (5.576,0.600) -- (5.292,0.780) -- (5.026,0.984) -- (4.744,1.160) -- (4.478,1.355) -- (4.188,1.527) -- (3.921,1.726) -- (3.639,1.902) -- (3.367,2.102) -- (3.080,2.277) -- (2.813,2.477) -- (2.529,2.646) -- (2.258,2.839) -- (1.986,3.035) -- (1.695,3.210) -- (1.424,3.400);
  \draw[black] (1.424,0.880) -- (1.420,0.788) -- (1.428,0.694) -- (1.424,0.600);
  \draw[black] (1.424,0.600) -- (1.519,0.603) -- (1.610,0.604) -- (1.704,0.600);
  \draw[black] (1.424,0.880) -- (1.519,0.884) -- (1.612,0.876) -- (1.704,0.880);
  \draw[black] (1.704,0.880) -- (1.711,0.788) -- (1.704,0.693) -- (1.704,0.600);
  \draw[black] (1.754,3.178) -- (1.741,3.147) -- (1.720,3.121) -- (1.691,3.104) -- (1.667,3.082) -- (1.638,3.066) -- (1.610,3.051) -- (1.581,3.038) -- (1.552,3.023) -- (1.521,3.013) -- (1.491,2.997) -- (1.457,2.998) -- (1.424,2.990);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.158,0.467) {P};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.158,3.467) {K};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.676,0.467) {D};
  \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) {1,4 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 PKD rectangle en P, on cherche une relation entre l'angle aigu   \(\widehat{PKD}\) son coté adjacent et l'hypoténuse du triangle.

cos(\(\widehat{PKD}\)) = \(\dfrac{PK}{KD}\)
\par\addvspace{.5em}
d'où cos(56\textdegree{}) = \(\dfrac{PK}{1{,}4}\)

On a donc PK = 1,4 \(\times\) cos(56\textdegree{}) \(\approx\) 0,8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.945,3.400) -- (1.948,3.050) -- (1.940,2.699) -- (1.936,2.350) -- (1.940,1.999) -- (1.950,1.650) -- (1.946,1.297) -- (1.940,0.950) -- (1.945,0.600);
  \draw[black] (1.945,0.600) -- (2.293,0.609) -- (2.637,0.605) -- (2.983,0.587) -- (3.326,0.600) -- (3.674,0.607) -- (4.018,0.593) -- (4.364,0.597) -- (4.710,0.609) -- (5.055,0.600);
  \draw[black] (5.055,0.600) -- (4.801,0.836) -- (4.541,1.071) -- (4.277,1.295) -- (4.015,1.529) -- (3.761,1.767) -- (3.501,1.998) -- (3.241,2.235) -- (2.987,2.477) -- (2.723,2.697) -- (2.470,2.942) -- (2.210,3.177) -- (1.945,3.400);
  \draw[black] (1.945,0.880) -- (1.951,0.785) -- (1.941,0.692) -- (1.945,0.600);
  \draw[black] (1.945,0.600) -- (2.037,0.606) -- (2.133,0.603) -- (2.225,0.600);
  \draw[black] (1.945,0.880) -- (2.039,0.882) -- (2.131,0.882) -- (2.225,0.880);
  \draw[black] (2.225,0.880) -- (2.229,0.786) -- (2.230,0.692) -- (2.225,0.600);
  \draw[black] (2.241,3.134) -- (2.217,3.119) -- (2.200,3.096) -- (2.179,3.078) -- (2.156,3.063) -- (2.137,3.039) -- (2.110,3.030) -- (2.084,3.019) -- (2.056,3.012) -- (2.027,3.016) -- (2.001,3.000) -- (1.974,2.994) -- (1.945,3.008);
  \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) {T};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.155,0.467) {A};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.267) {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 (2.216,2.724) {48°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{GTA}\)) =  \(\dfrac{GA}{TA}\)
\par\addvspace{.5em}
 d'où sin(48\textdegree{}) = \(\dfrac{9}{TA}\)
\par\addvspace{.5em}
 On a donc TA = \(\dfrac{9}{\sin(48^{\circ})}\) \(\approx\) 12,1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,3.195) -- (1.063,2.897) -- (1.053,2.601) -- (1.062,2.299) -- (1.051,1.997) -- (1.049,1.702) -- (1.037,1.403) -- (1.043,1.101) -- (1.050,0.805);
  \draw[black] (1.050,0.805) -- (1.402,0.814) -- (1.752,0.792) -- (2.097,0.818) -- (2.452,0.808) -- (2.797,0.815) -- (3.149,0.802) -- (3.500,0.817) -- (3.851,0.815) -- (4.201,0.795) -- (4.552,0.795) -- (4.902,0.800) -- (5.249,0.795) -- (5.602,0.811) -- (5.950,0.805);
  \draw[black] (5.950,0.805) -- (5.638,0.948) -- (5.331,1.094) -- (5.033,1.255) -- (4.723,1.400) -- (4.418,1.548) -- (4.106,1.691) -- (3.812,1.855) -- (3.498,1.990) -- (3.197,2.161) -- (2.882,2.295) -- (2.577,2.438) -- (2.277,2.609) -- (1.970,2.745) -- (1.659,2.893) -- (1.357,3.052) -- (1.050,3.195);
  \draw[black] (1.050,1.044) -- (1.050,0.965) -- (1.055,0.884) -- (1.050,0.805);
  \draw[black] (1.050,0.805) -- (1.130,0.802) -- (1.209,0.809) -- (1.289,0.805);
  \draw[black] (1.050,1.044) -- (1.130,1.038) -- (1.208,1.047) -- (1.289,1.044);
  \draw[black] (1.289,1.044) -- (1.287,0.964) -- (1.289,0.884) -- (1.289,0.805);
  \draw[black] (1.414,3.017) -- (1.393,2.986) -- (1.365,2.960) -- (1.349,2.926) -- (1.320,2.902) -- (1.287,2.885) -- (1.266,2.850) -- (1.233,2.831) -- (1.196,2.821) -- (1.159,2.814) -- (1.125,2.794) -- (1.088,2.790) -- (1.050,2.805);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,0.672) {F};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,3.262) {K};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,0.672) {L};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {4,3 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.472) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.403,2.563) {64°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{FKL}\)) = \(\dfrac{FL}{FK}\)
\par\addvspace{.5em}
d'où tan(64\textdegree{}) =\(\dfrac{FL}{4{,}3}\)
\par\addvspace{.5em}
On a donc FL = 4,3 \(\times\) tan(64\textdegree{}) \(\approx\) 8,8 cm\end{document}
