\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 371}

\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=371}\\[2pt]
    {\small\ttfamily\bfseries\color{red} TRIG0371}
\end{minipage}%
}
}

\begin{document}

\sloppy

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

 Dans le triangle KJC rectangle en K, on sait que :
\begin{itemize}
\item KJ=1,7 cm
\item KC=3,9 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{KJC}\).
\needspace{4\baselineskip}
\section*{
Exercice 2}

 Dans le triangle HPV rectangle en H, on sait que :
\begin{itemize}
\item HP=3,3 cm
\item HV=5,4 cm
\end{itemize}

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

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

 Dans le triangle DWR rectangle en D, on sait que :
\begin{itemize}
\item DR=1 cm
\item \(\widehat{DWR}\) = 58\textdegree{}
\end{itemize}

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

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

 Dans le triangle FKS rectangle en F, on sait que :
\begin{itemize}
\item KS=10 cm
\item \(\widehat{FKS}\) = 53\textdegree{}
\end{itemize}

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

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

 Dans le triangle HAP rectangle en H, on sait que :
\begin{itemize}
\item HP=2 cm
\item \(\widehat{HAP}\) = 76\textdegree{}
\end{itemize}

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

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

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

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,3.068) -- (1.045,2.801) -- (1.047,2.537) -- (1.045,2.264) -- (1.045,2.002) -- (1.060,1.731) -- (1.041,1.466) -- (1.038,1.198) -- (1.050,0.932);
  \draw[black] (1.050,0.932) -- (1.402,0.944) -- (1.752,0.939) -- (2.098,0.925) -- (2.448,0.923) -- (2.801,0.924) -- (3.151,0.945) -- (3.499,0.943) -- (3.852,0.943) -- (4.203,0.943) -- (4.550,0.943) -- (4.897,0.932) -- (5.251,0.938) -- (5.602,0.929) -- (5.950,0.932);
  \draw[black] (5.950,0.932) -- (5.645,1.062) -- (5.337,1.194) -- (5.033,1.330) -- (4.727,1.467) -- (4.420,1.603) -- (4.108,1.727) -- (3.813,1.877) -- (3.506,2.005) -- (3.194,2.133) -- (2.885,2.263) -- (2.587,2.412) -- (2.274,2.538) -- (1.970,2.661) -- (1.658,2.790) -- (1.357,2.935) -- (1.050,3.068);
  \draw[black] (1.050,1.146) -- (1.054,1.075) -- (1.057,1.002) -- (1.050,0.932);
  \draw[black] (1.050,0.932) -- (1.121,0.938) -- (1.193,0.927) -- (1.264,0.932);
  \draw[black] (1.050,1.146) -- (1.122,1.143) -- (1.193,1.139) -- (1.264,1.146);
  \draw[black] (1.264,1.146) -- (1.259,1.074) -- (1.260,1.004) -- (1.264,0.932);
  \draw[black] (1.411,2.911) -- (1.405,2.871) -- (1.386,2.836) -- (1.363,2.804) -- (1.329,2.782) -- (1.301,2.756) -- (1.271,2.731) -- (1.233,2.719) -- (1.201,2.697) -- (1.166,2.680) -- (1.127,2.675) -- (1.089,2.663) -- (1.050,2.674);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,0.799) {K};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,3.135) {J};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,0.799) {C};
  \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 (3.500,0.599) {3,9 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.415,2.444) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{KJC}\)) = \(\dfrac{KC}{KJ}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{KJC}\)) = \(\dfrac{3{,}9}{1{,}7}\)
\par\addvspace{.5em}
 On a donc KJC = ArcTan(\(\dfrac{3{,}9}{1{,}7}\)) \(\approx\) 66\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.209,3.400) -- (1.207,3.047) -- (1.203,2.701) -- (1.213,2.349) -- (1.222,1.997) -- (1.201,1.647) -- (1.210,1.298) -- (1.208,0.950) -- (1.209,0.600);
  \draw[black] (1.209,0.600) -- (1.564,0.604) -- (1.916,0.591) -- (2.269,0.603) -- (2.618,0.600) -- (2.974,0.607) -- (3.323,0.609) -- (3.676,0.607) -- (4.028,0.603) -- (4.383,0.587) -- (4.735,0.603) -- (5.089,0.607) -- (5.441,0.604) -- (5.791,0.600);
  \draw[black] (5.791,0.600) -- (5.507,0.774) -- (5.216,0.941) -- (4.936,1.135) -- (4.641,1.290) -- (4.360,1.475) -- (4.075,1.654) -- (3.783,1.816) -- (3.507,2.007) -- (3.213,2.172) -- (2.933,2.358) -- (2.638,2.517) -- (2.359,2.711) -- (2.065,2.871) -- (1.785,3.059) -- (1.491,3.216) -- (1.209,3.400);
  \draw[black] (1.209,0.880) -- (1.208,0.786) -- (1.205,0.694) -- (1.209,0.600);
  \draw[black] (1.209,0.600) -- (1.303,0.598) -- (1.397,0.595) -- (1.489,0.600);
  \draw[black] (1.209,0.880) -- (1.302,0.878) -- (1.397,0.884) -- (1.489,0.880);
  \draw[black] (1.489,0.880) -- (1.489,0.787) -- (1.486,0.692) -- (1.489,0.600);
  \draw[black] (5.450,0.809) -- (5.436,0.795) -- (5.440,0.773) -- (5.416,0.764) -- (5.423,0.741) -- (5.411,0.726) -- (5.400,0.710) -- (5.405,0.690) -- (5.402,0.672) -- (5.396,0.654) -- (5.400,0.636) -- (5.384,0.619) -- (5.381,0.600);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.942,0.467) {H};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.942,3.467) {P};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.891,0.467) {V};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.709,1.933) {3,3 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.267) {5,4 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.149,0.714) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{HVP}\)) = \(\dfrac{HP}{HV}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{HVP}\)) =  \(\dfrac{3{,}3}{5{,}4}\)
\par\addvspace{.5em}
 On a donc \(\widehat{HVP}\) =  ArcTan(\(\dfrac{3{,}3}{5{,}4}\)) \(\approx\) 31\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.260,3.400) -- (1.253,3.049) -- (1.246,2.698) -- (1.258,2.347) -- (1.267,2.002) -- (1.270,1.647) -- (1.259,1.299) -- (1.256,0.953) -- (1.260,0.600);
  \draw[black] (1.260,0.600) -- (1.603,0.597) -- (1.946,0.595) -- (2.292,0.597) -- (2.635,0.613) -- (2.985,0.595) -- (3.327,0.591) -- (3.675,0.587) -- (4.019,0.593) -- (4.359,0.602) -- (4.707,0.608) -- (5.048,0.599) -- (5.396,0.607) -- (5.740,0.600);
  \draw[black] (5.740,0.600) -- (5.434,0.778) -- (5.140,0.965) -- (4.844,1.165) -- (4.550,1.355) -- (4.254,1.542) -- (3.945,1.717) -- (3.653,1.916) -- (3.346,2.086) -- (3.061,2.289) -- (2.753,2.469) -- (2.453,2.649) -- (2.157,2.836) -- (1.859,3.032) -- (1.551,3.207) -- (1.260,3.400);
  \draw[black] (1.260,0.880) -- (1.263,0.786) -- (1.262,0.695) -- (1.260,0.600);
  \draw[black] (1.260,0.600) -- (1.352,0.600) -- (1.446,0.602) -- (1.540,0.600);
  \draw[black] (1.260,0.880) -- (1.352,0.886) -- (1.448,0.887) -- (1.540,0.880);
  \draw[black] (1.540,0.880) -- (1.544,0.787) -- (1.537,0.694) -- (1.540,0.600);
  \draw[black] (1.606,3.183) -- (1.577,3.162) -- (1.559,3.134) -- (1.529,3.116) -- (1.513,3.083) -- (1.479,3.073) -- (1.450,3.057) -- (1.420,3.043) -- (1.389,3.032) -- (1.362,3.004) -- (1.326,3.011) -- (1.293,2.999) -- (1.260,3.000);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.993,0.467) {D};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.993,3.467) {W};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.840,0.467) {R};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.267) {1 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 DWR rectangle en D, on cherche une relation entre l'angle aigu   \(\widehat{DWR}\) son coté opposé et l'hypoténuse du triangle.

\par\addvspace{.5em}
sin(\(\widehat{DWR}\)) =  \(\dfrac{DR}{WR}\)
\par\addvspace{.5em}
 d'où sin(58\textdegree{}) = \(\dfrac{1}{WR}\)
\par\addvspace{.5em}
 On a donc WR = \(\dfrac{1}{\sin(58^{\circ})}\) \(\approx\) 1,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.642,3.400) -- (1.632,3.050) -- (1.638,2.701) -- (1.633,2.349) -- (1.640,2.001) -- (1.642,1.647) -- (1.641,1.298) -- (1.643,0.952) -- (1.642,0.600);
  \draw[black] (1.642,0.600) -- (1.979,0.600) -- (2.317,0.592) -- (2.657,0.596) -- (2.994,0.588) -- (3.328,0.609) -- (3.670,0.597) -- (4.007,0.595) -- (4.342,0.604) -- (4.685,0.588) -- (5.020,0.600) -- (5.358,0.600);
  \draw[black] (5.358,0.600) -- (5.065,0.804) -- (4.789,1.034) -- (4.507,1.255) -- (4.218,1.466) -- (3.922,1.672) -- (3.648,1.898) -- (3.363,2.112) -- (3.076,2.335) -- (2.784,2.532) -- (2.502,2.758) -- (2.213,2.969) -- (1.928,3.185) -- (1.642,3.400);
  \draw[black] (1.642,0.880) -- (1.647,0.787) -- (1.644,0.694) -- (1.642,0.600);
  \draw[black] (1.642,0.600) -- (1.735,0.604) -- (1.829,0.594) -- (1.922,0.600);
  \draw[black] (1.642,0.880) -- (1.736,0.883) -- (1.828,0.879) -- (1.922,0.880);
  \draw[black] (1.922,0.880) -- (1.921,0.788) -- (1.925,0.694) -- (1.922,0.600);
  \draw[black] (1.968,3.155) -- (1.936,3.141) -- (1.917,3.117) -- (1.896,3.095) -- (1.874,3.073) -- (1.848,3.057) -- (1.817,3.049) -- (1.793,3.029) -- (1.764,3.019) -- (1.734,3.011) -- (1.702,3.013) -- (1.673,3.005) -- (1.642,2.990);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.375,0.467) {F};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.375,3.467) {K};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.458,0.467) {S};
  \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 (3.667,2.000) {10 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.940,2.737) {53°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{FKS}\)) =  \(\dfrac{FS}{KS}\)
\par\addvspace{.5em}
 d'où sin(53\textdegree{}) = \(\dfrac{FS}{10}\)

On a donc FS = 10 \(\times\) sin(53\textdegree{}) \(\approx\) 8 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,2.611) -- (1.039,2.461) -- (1.058,2.306) -- (1.047,2.153) -- (1.055,1.997) -- (1.040,1.845) -- (1.045,1.696) -- (1.043,1.542) -- (1.050,1.389);
  \draw[black] (1.050,1.389) -- (1.400,1.384) -- (1.751,1.392) -- (2.102,1.393) -- (2.452,1.400) -- (2.803,1.397) -- (3.152,1.398) -- (3.499,1.382) -- (3.849,1.395) -- (4.198,1.393) -- (4.552,1.388) -- (4.903,1.402) -- (5.250,1.394) -- (5.599,1.378) -- (5.950,1.389);
  \draw[black] (5.950,1.389) -- (5.624,1.484) -- (5.297,1.544) -- (4.965,1.622) -- (4.643,1.712) -- (4.316,1.800) -- (3.990,1.869) -- (3.661,1.949) -- (3.340,2.049) -- (3.011,2.126) -- (2.684,2.201) -- (2.355,2.278) -- (2.030,2.375) -- (1.703,2.455) -- (1.381,2.540) -- (1.050,2.611);
  \draw[black] (1.050,1.511) -- (1.054,1.471) -- (1.050,1.431) -- (1.050,1.389);
  \draw[black] (1.050,1.389) -- (1.091,1.395) -- (1.133,1.387) -- (1.172,1.389);
  \draw[black] (1.050,1.511) -- (1.092,1.506) -- (1.130,1.506) -- (1.172,1.511);
  \draw[black] (1.172,1.511) -- (1.173,1.469) -- (1.168,1.430) -- (1.172,1.389);
  \draw[black] (1.444,2.513) -- (1.421,2.473) -- (1.403,2.433) -- (1.377,2.398) -- (1.354,2.362) -- (1.332,2.322) -- (1.299,2.292) -- (1.263,2.266) -- (1.219,2.255) -- (1.180,2.234) -- (1.139,2.217) -- (1.094,2.213) -- (1.050,2.203);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,1.256) {H};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,2.678) {A};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,1.256) {P};
  \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 (3.500,1.056) {2 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.460,2.019) {76°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{HAP}\)) = \(\dfrac{HP}{HA}\)
\par\addvspace{.5em}
tan(76\textdegree{}) = \(\dfrac{2}{HA}\)
\par\addvspace{.5em}
 On a donc HA = \(\dfrac{2}{\tan(76^{\circ})}\) \(\approx\) 0,5 cm\end{document}
