\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 0}

\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=0}\\[2pt]
    {\small\ttfamily\bfseries\color{red} TRIG0000}
\end{minipage}%
}
}

\begin{document}

\sloppy

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

 Dans le triangle LNS rectangle en L, on sait que :
\begin{itemize}
\item LS=9,1 cm
\item \(\widehat{LNS}\) = 70\textdegree{}
\end{itemize}

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

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

 Dans le triangle JCK rectangle en J, on sait que :
\begin{itemize}
\item CK=4,9 cm
\item \(\widehat{CKJ}\) =  31\textdegree{}
\end{itemize}

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

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

 Dans le triangle JMV rectangle en J, on sait que :
\begin{itemize}
\item JM=4,7 cm
\item \(\widehat{MVJ}\) =  15\textdegree{}
\end{itemize}

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

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

 Dans le triangle WFH rectangle en W, on sait que :
\begin{itemize}
\item WF=2,2 cm
\item WH=4,9 cm
\end{itemize}

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

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

 Dans le triangle HFR rectangle en H, on sait que :
\begin{itemize}
\item HF=2,1 cm
\item FR=9,7 cm
\end{itemize}

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

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

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,2.892) -- (1.048,2.670) -- (1.055,2.445) -- (1.058,2.223) -- (1.052,2.000) -- (1.062,1.774) -- (1.039,1.557) -- (1.041,1.330) -- (1.050,1.108);
  \draw[black] (1.050,1.108) -- (1.401,1.101) -- (1.752,1.105) -- (2.102,1.113) -- (2.447,1.103) -- (2.798,1.097) -- (3.148,1.100) -- (3.497,1.107) -- (3.853,1.095) -- (4.200,1.095) -- (4.549,1.114) -- (4.899,1.118) -- (5.248,1.097) -- (5.600,1.100) -- (5.950,1.108);
  \draw[black] (5.950,1.108) -- (5.624,1.221) -- (5.301,1.353) -- (4.970,1.458) -- (4.649,1.591) -- (4.323,1.714) -- (3.985,1.811) -- (3.662,1.930) -- (3.335,2.047) -- (3.010,2.184) -- (2.683,2.303) -- (2.357,2.417) -- (2.029,2.528) -- (1.703,2.659) -- (1.377,2.770) -- (1.050,2.892);
  \draw[black] (1.050,1.287) -- (1.050,1.227) -- (1.047,1.166) -- (1.050,1.108);
  \draw[black] (1.050,1.108) -- (1.109,1.110) -- (1.169,1.113) -- (1.228,1.108);
  \draw[black] (1.050,1.287) -- (1.108,1.289) -- (1.168,1.293) -- (1.228,1.287);
  \draw[black] (1.228,1.287) -- (1.226,1.228) -- (1.233,1.168) -- (1.228,1.108);
  \draw[black] (1.418,2.758) -- (1.415,2.715) -- (1.396,2.678) -- (1.363,2.652) -- (1.346,2.613) -- (1.313,2.587) -- (1.284,2.557) -- (1.246,2.541) -- (1.205,2.532) -- (1.168,2.517) -- (1.130,2.506) -- (1.091,2.487) -- (1.050,2.499);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,0.975) {L};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,2.958) {N};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,0.975) {S};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.775) {9,1 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 (1.432,2.279) {70°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{LNS}\)) =  \(\dfrac{LS}{NS}\)
\par\addvspace{.5em}
 d'où sin(70\textdegree{}) = \(\dfrac{9{,}1}{NS}\)
\par\addvspace{.5em}
 On a donc NS = \(\dfrac{9{,}1}{\sin(70^{\circ})}\) \(\approx\) 9,7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.170,3.400) -- (1.165,3.052) -- (1.163,2.698) -- (1.179,2.352) -- (1.169,2.003) -- (1.159,1.649) -- (1.183,1.299) -- (1.159,0.948) -- (1.170,0.600);
  \draw[black] (1.170,0.600) -- (1.529,0.589) -- (1.888,0.603) -- (2.247,0.609) -- (2.601,0.601) -- (2.963,0.594) -- (3.323,0.610) -- (3.680,0.600) -- (4.038,0.595) -- (4.399,0.590) -- (4.753,0.610) -- (5.115,0.593) -- (5.472,0.611) -- (5.830,0.600);
  \draw[black] (5.830,0.600) -- (5.543,0.778) -- (5.244,0.943) -- (4.958,1.130) -- (4.673,1.307) -- (4.378,1.484) -- (4.085,1.657) -- (3.789,1.816) -- (3.503,2.004) -- (3.212,2.185) -- (2.922,2.351) -- (2.621,2.514) -- (2.342,2.706) -- (2.045,2.884) -- (1.751,3.048) -- (1.465,3.237) -- (1.170,3.400);
  \draw[black] (1.170,0.880) -- (1.166,0.788) -- (1.176,0.692) -- (1.170,0.600);
  \draw[black] (1.170,0.600) -- (1.262,0.601) -- (1.358,0.606) -- (1.450,0.600);
  \draw[black] (1.170,0.880) -- (1.264,0.874) -- (1.356,0.879) -- (1.450,0.880);
  \draw[black] (1.450,0.880) -- (1.449,0.788) -- (1.453,0.693) -- (1.450,0.600);
  \draw[black] (5.493,0.803) -- (5.479,0.790) -- (5.462,0.778) -- (5.460,0.759) -- (5.465,0.738) -- (5.442,0.727) -- (5.439,0.708) -- (5.448,0.688) -- (5.441,0.671) -- (5.431,0.654) -- (5.425,0.637) -- (5.428,0.618) -- (5.438,0.600);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.903,0.467) {J};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.903,3.467) {C};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.930,0.467) {K};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.670,1.933) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {4,9 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.188,0.711) {31°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{JKC}\)) = \(\dfrac{JC}{CK}\)
\par\addvspace{.5em}
 D'où sin(31\textdegree{}) = \(\dfrac{JC}{4{,}9}\)

On a donc JC = 4,9 \(\times\) sin(31\textdegree{}) \(\approx\) 2,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,2.656) -- (1.056,2.495) -- (1.048,2.329) -- (1.037,2.167) -- (1.060,1.998) -- (1.038,1.837) -- (1.040,1.670) -- (1.055,1.505) -- (1.050,1.344);
  \draw[black] (1.050,1.344) -- (1.397,1.349) -- (1.750,1.342) -- (2.102,1.348) -- (2.447,1.335) -- (2.800,1.354) -- (3.149,1.356) -- (3.498,1.339) -- (3.850,1.352) -- (4.203,1.335) -- (4.552,1.355) -- (4.898,1.352) -- (5.247,1.347) -- (5.599,1.342) -- (5.950,1.344);
  \draw[black] (5.950,1.344) -- (5.624,1.442) -- (5.300,1.524) -- (4.969,1.595) -- (4.646,1.707) -- (4.316,1.772) -- (3.985,1.859) -- (3.669,1.965) -- (3.335,2.045) -- (3.008,2.124) -- (2.681,2.213) -- (2.358,2.320) -- (2.033,2.396) -- (1.704,2.480) -- (1.381,2.577) -- (1.050,2.656);
  \draw[black] (1.050,1.475) -- (1.049,1.431) -- (1.049,1.387) -- (1.050,1.344);
  \draw[black] (1.050,1.344) -- (1.094,1.345) -- (1.139,1.341) -- (1.181,1.344);
  \draw[black] (1.050,1.475) -- (1.095,1.470) -- (1.139,1.473) -- (1.181,1.475);
  \draw[black] (1.181,1.475) -- (1.181,1.430) -- (1.188,1.387) -- (1.181,1.344);
  \draw[black] (5.559,1.448) -- (5.565,1.438) -- (5.563,1.429) -- (5.557,1.422) -- (5.549,1.414) -- (5.553,1.405) -- (5.560,1.395) -- (5.555,1.387) -- (5.556,1.378) -- (5.547,1.370) -- (5.551,1.361) -- (5.544,1.352) -- (5.550,1.344);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,1.210) {J};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,2.723) {M};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,1.210) {V};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {4,7 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,1.010) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.289,1.364) {15°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{JVM}\)) = \(\dfrac{JM}{JV}\)
\par\addvspace{.5em}
d'où  tan(15\textdegree{}) =  \(\dfrac{4{,}7}{JV}\)
\par\addvspace{.5em}
On a donc JV = \(\dfrac{4{,}7}{\tan(15^{\circ})}\) \(\approx\) 17,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,3.100) -- (1.040,2.823) -- (1.055,2.550) -- (1.054,2.275) -- (1.056,1.998) -- (1.043,1.726) -- (1.053,1.452) -- (1.038,1.177) -- (1.050,0.900);
  \draw[black] (1.050,0.900) -- (1.399,0.894) -- (1.750,0.888) -- (2.103,0.910) -- (2.447,0.897) -- (2.802,0.895) -- (3.147,0.904) -- (3.500,0.901) -- (3.852,0.911) -- (4.201,0.913) -- (4.549,0.910) -- (4.901,0.895) -- (5.252,0.908) -- (5.597,0.893) -- (5.950,0.900);
  \draw[black] (5.950,0.900) -- (5.638,1.027) -- (5.336,1.179) -- (5.030,1.317) -- (4.729,1.455) -- (4.420,1.597) -- (4.113,1.722) -- (3.806,1.870) -- (3.508,2.011) -- (3.195,2.148) -- (2.883,2.263) -- (2.582,2.409) -- (2.273,2.547) -- (1.971,2.700) -- (1.659,2.813) -- (1.351,2.951) -- (1.050,3.100);
  \draw[black] (1.050,1.120) -- (1.049,1.045) -- (1.054,0.972) -- (1.050,0.900);
  \draw[black] (1.050,0.900) -- (1.124,0.895) -- (1.197,0.900) -- (1.270,0.900);
  \draw[black] (1.050,1.120) -- (1.125,1.120) -- (1.198,1.125) -- (1.270,1.120);
  \draw[black] (1.270,1.120) -- (1.271,1.048) -- (1.275,0.975) -- (1.270,0.900);
  \draw[black] (5.587,1.063) -- (5.586,1.048) -- (5.566,1.041) -- (5.564,1.026) -- (5.567,1.011) -- (5.564,0.997) -- (5.559,0.984) -- (5.549,0.971) -- (5.557,0.956) -- (5.557,0.942) -- (5.554,0.928) -- (5.558,0.914) -- (5.546,0.900);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,0.767) {W};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,3.167) {F};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,0.767) {H};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {2,2 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.567) {4,9 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.298,0.973) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{WHF}\)) = \(\dfrac{WF}{WH}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{WHF}\)) =  \(\dfrac{2{,}2}{4{,}9}\)
\par\addvspace{.5em}
 On a donc \(\widehat{WHF}\) =  ArcTan(\(\dfrac{2{,}2}{4{,}9}\)) \(\approx\) 24\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,2.543) -- (1.055,2.405) -- (1.049,2.273) -- (1.048,2.139) -- (1.049,2.001) -- (1.053,1.867) -- (1.060,1.730) -- (1.050,1.594) -- (1.050,1.457);
  \draw[black] (1.050,1.457) -- (1.399,1.465) -- (1.752,1.452) -- (2.099,1.454) -- (2.448,1.451) -- (2.799,1.462) -- (3.151,1.445) -- (3.499,1.462) -- (3.851,1.463) -- (4.200,1.453) -- (4.547,1.460) -- (4.903,1.444) -- (5.249,1.458) -- (5.597,1.445) -- (5.950,1.457);
  \draw[black] (5.950,1.457) -- (5.624,1.520) -- (5.294,1.592) -- (4.967,1.665) -- (4.648,1.757) -- (4.318,1.816) -- (3.994,1.903) -- (3.664,1.964) -- (3.340,2.039) -- (3.013,2.115) -- (2.686,2.190) -- (2.358,2.252) -- (2.025,2.317) -- (1.701,2.393) -- (1.375,2.469) -- (1.050,2.543);
  \draw[black] (1.050,1.565) -- (1.055,1.529) -- (1.044,1.493) -- (1.050,1.457);
  \draw[black] (1.050,1.457) -- (1.087,1.458) -- (1.122,1.463) -- (1.159,1.457);
  \draw[black] (1.050,1.565) -- (1.085,1.563) -- (1.124,1.567) -- (1.159,1.565);
  \draw[black] (1.159,1.565) -- (1.156,1.529) -- (1.155,1.493) -- (1.159,1.457);
  \draw[black] (1.448,2.455) -- (1.432,2.412) -- (1.406,2.374) -- (1.383,2.336) -- (1.357,2.301) -- (1.330,2.265) -- (1.301,2.231) -- (1.260,2.212) -- (1.227,2.177) -- (1.185,2.159) -- (1.141,2.148) -- (1.094,2.154) -- (1.050,2.136);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,1.323) {H};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,2.610) {F};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,1.323) {R};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {2,1 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {9,7 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.467,1.957) {?};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{HFR}\)) = \(\dfrac{HF}{FR}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{HFR}\)) = \(\dfrac{2{,}1}{9{,}7}\)
\par\addvspace{.5em}
On a donc \(\widehat{HFR}\) =ArcCos(\(\dfrac{2{,}1}{9{,}7}\)) \(\approx\)77\textdegree{}\end{document}
