\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 244}

\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=244}\\[2pt]
    {\small\ttfamily\bfseries\color{red} TRIG0244}
\end{minipage}%
}
}

\begin{document}

\sloppy

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

 Dans le triangle JKA rectangle en J, on sait que :
\begin{itemize}
\item JA=1,6 cm
\item \(\widehat{KAJ}\) =  24\textdegree{}
\end{itemize}

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

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

 Dans le triangle ARM rectangle en A, on sait que :
\begin{itemize}
\item RM=7,6 cm
\item \(\widehat{ARM}\) = 75\textdegree{}
\end{itemize}

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

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

 Dans le triangle HKB rectangle en H, on sait que :
\begin{itemize}
\item HK=1,9 cm
\item HB=6,4 cm
\end{itemize}

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

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

 Dans le triangle NHJ rectangle en N, on sait que :
\begin{itemize}
\item NJ=4,4 cm
\item HJ=8,8 cm
\end{itemize}

Après avoir fait un schéma, calcule l'arrondi au degré près de la mesure de l'angle \(\widehat{NHJ}\).
\needspace{4\baselineskip}
\section*{
Exercice 5}

 Dans le triangle WVH rectangle en W, on sait que :
\begin{itemize}
\item WV=9,4 cm
\item \(\widehat{VHW}\) =  30\textdegree{}
\end{itemize}

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

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

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

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,3.091) -- (1.051,2.816) -- (1.053,2.542) -- (1.046,2.273) -- (1.063,2.000) -- (1.040,1.729) -- (1.051,1.452) -- (1.041,1.184) -- (1.050,0.909);
  \draw[black] (1.050,0.909) -- (1.402,0.904) -- (1.753,0.914) -- (2.099,0.899) -- (2.447,0.915) -- (2.798,0.918) -- (3.147,0.916) -- (3.498,0.920) -- (3.847,0.899) -- (4.200,0.911) -- (4.550,0.911) -- (4.897,0.903) -- (5.250,0.922) -- (5.601,0.907) -- (5.950,0.909);
  \draw[black] (5.950,0.909) -- (5.647,1.049) -- (5.334,1.179) -- (5.030,1.313) -- (4.731,1.465) -- (4.420,1.586) -- (4.115,1.731) -- (3.803,1.863) -- (3.495,1.988) -- (3.187,2.128) -- (2.884,2.267) -- (2.574,2.400) -- (2.278,2.544) -- (1.971,2.682) -- (1.659,2.806) -- (1.350,2.949) -- (1.050,3.091);
  \draw[black] (1.050,1.127) -- (1.045,1.054) -- (1.044,0.982) -- (1.050,0.909);
  \draw[black] (1.050,0.909) -- (1.124,0.908) -- (1.195,0.903) -- (1.268,0.909);
  \draw[black] (1.050,1.127) -- (1.122,1.131) -- (1.196,1.129) -- (1.268,1.127);
  \draw[black] (1.268,1.127) -- (1.270,1.053) -- (1.265,0.981) -- (1.268,0.909);
  \draw[black] (5.594,1.068) -- (5.586,1.056) -- (5.579,1.044) -- (5.570,1.033) -- (5.566,1.019) -- (5.556,1.007) -- (5.563,0.992) -- (5.556,0.979) -- (5.556,0.965) -- (5.549,0.951) -- (5.548,0.937) -- (5.560,0.923) -- (5.548,0.909);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,0.776) {J};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,3.157) {K};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,0.776) {A};
  \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 (3.500,0.576) {1,6 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.298,0.981) {24°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{JAK}\)) = \(\dfrac{JK}{JA}\)
\par\addvspace{.5em}
d'où  tan(24\textdegree{}) =  \(\dfrac{JK}{1{,}6}\)
\par\addvspace{.5em}
On a donc JK =  1,6 \(\times\) tan(24\textdegree{})  \(\approx\) 0,7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,2.656) -- (1.062,2.491) -- (1.058,2.329) -- (1.042,2.163) -- (1.055,2.001) -- (1.054,1.837) -- (1.043,1.671) -- (1.053,1.508) -- (1.050,1.344);
  \draw[black] (1.050,1.344) -- (1.401,1.356) -- (1.748,1.354) -- (2.101,1.356) -- (2.448,1.342) -- (2.800,1.346) -- (3.149,1.346) -- (3.503,1.332) -- (3.849,1.343) -- (4.200,1.347) -- (4.551,1.338) -- (4.897,1.343) -- (5.247,1.353) -- (5.598,1.332) -- (5.950,1.344);
  \draw[black] (5.950,1.344) -- (5.623,1.429) -- (5.297,1.519) -- (4.974,1.613) -- (4.639,1.684) -- (4.320,1.788) -- (3.991,1.871) -- (3.664,1.959) -- (3.335,2.046) -- (3.008,2.132) -- (2.687,2.225) -- (2.357,2.295) -- (2.027,2.392) -- (1.702,2.490) -- (1.378,2.564) -- (1.050,2.656);
  \draw[black] (1.050,1.475) -- (1.046,1.431) -- (1.049,1.387) -- (1.050,1.344);
  \draw[black] (1.050,1.344) -- (1.095,1.342) -- (1.137,1.348) -- (1.181,1.344);
  \draw[black] (1.050,1.475) -- (1.093,1.475) -- (1.138,1.478) -- (1.181,1.475);
  \draw[black] (1.181,1.475) -- (1.182,1.431) -- (1.182,1.387) -- (1.181,1.344);
  \draw[black] (1.437,2.553) -- (1.424,2.511) -- (1.414,2.467) -- (1.374,2.440) -- (1.357,2.399) -- (1.325,2.369) -- (1.291,2.342) -- (1.255,2.318) -- (1.220,2.292) -- (1.180,2.272) -- (1.137,2.263) -- (1.094,2.258) -- (1.050,2.247);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,1.210) {A};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,2.723) {R};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,1.210) {M};
  \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 (3.667,2.000) {7,6 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.456,2.061) {75°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{ARM}\)) =  \(\dfrac{AM}{RM}\)
\par\addvspace{.5em}
 d'où sin(75\textdegree{}) = \(\dfrac{AM}{7{,}6}\)

On a donc AM = 7,6 \(\times\) sin(75\textdegree{}) \(\approx\) 7,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,2.727) -- (1.053,2.544) -- (1.048,2.365) -- (1.049,2.185) -- (1.055,1.998) -- (1.047,1.821) -- (1.057,1.639) -- (1.053,1.456) -- (1.050,1.273);
  \draw[black] (1.050,1.273) -- (1.397,1.273) -- (1.752,1.264) -- (2.099,1.275) -- (2.448,1.285) -- (2.798,1.271) -- (3.152,1.282) -- (3.497,1.264) -- (3.851,1.268) -- (4.202,1.264) -- (4.551,1.278) -- (4.900,1.278) -- (5.250,1.266) -- (5.603,1.269) -- (5.950,1.273);
  \draw[black] (5.950,1.273) -- (5.617,1.359) -- (5.296,1.469) -- (4.964,1.553) -- (4.642,1.650) -- (4.314,1.751) -- (3.992,1.862) -- (3.658,1.941) -- (3.339,2.058) -- (3.014,2.156) -- (2.683,2.236) -- (2.352,2.334) -- (2.028,2.425) -- (1.706,2.540) -- (1.379,2.632) -- (1.050,2.727);
  \draw[black] (1.050,1.418) -- (1.050,1.368) -- (1.052,1.321) -- (1.050,1.273);
  \draw[black] (1.050,1.273) -- (1.100,1.267) -- (1.146,1.278) -- (1.195,1.273);
  \draw[black] (1.050,1.418) -- (1.100,1.422) -- (1.147,1.415) -- (1.195,1.418);
  \draw[black] (1.195,1.418) -- (1.199,1.370) -- (1.189,1.320) -- (1.195,1.273);
  \draw[black] (5.575,1.384) -- (5.563,1.378) -- (5.552,1.370) -- (5.562,1.358) -- (5.565,1.348) -- (5.559,1.339) -- (5.546,1.331) -- (5.554,1.320) -- (5.552,1.311) -- (5.559,1.301) -- (5.544,1.292) -- (5.549,1.282) -- (5.552,1.273);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,1.139) {H};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,2.794) {K};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,1.139) {B};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {1,9 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.939) {6,4 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.290,1.302) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{HBK}\)) = \(\dfrac{HK}{HB}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{HBK}\)) =  \(\dfrac{1{,}9}{6{,}4}\)
\par\addvspace{.5em}
 On a donc \(\widehat{HBK}\) =  ArcTan(\(\dfrac{1{,}9}{6{,}4}\)) \(\approx\) 17\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (2.692,3.400) -- (2.700,3.050) -- (2.699,2.703) -- (2.681,2.348) -- (2.702,2.002) -- (2.697,1.647) -- (2.692,1.300) -- (2.699,0.948) -- (2.692,0.600);
  \draw[black] (2.692,0.600) -- (2.895,0.601) -- (3.097,0.604) -- (3.296,0.608) -- (3.500,0.600) -- (3.705,0.590) -- (3.906,0.600) -- (4.109,0.601) -- (4.308,0.600);
  \draw[black] (4.308,0.600) -- (4.121,0.904) -- (3.951,1.225) -- (3.766,1.529) -- (3.585,1.845) -- (3.419,2.160) -- (3.220,2.465) -- (3.053,2.777) -- (2.871,3.087) -- (2.692,3.400);
  \draw[black] (2.692,0.762) -- (2.688,0.708) -- (2.688,0.655) -- (2.692,0.600);
  \draw[black] (2.692,0.600) -- (2.747,0.601) -- (2.801,0.602) -- (2.853,0.600);
  \draw[black] (2.692,0.762) -- (2.746,0.761) -- (2.798,0.761) -- (2.853,0.762);
  \draw[black] (2.853,0.762) -- (2.847,0.709) -- (2.854,0.653) -- (2.853,0.600);
  \draw[black] (2.893,3.052) -- (2.876,3.046) -- (2.858,3.044) -- (2.843,3.035) -- (2.828,3.025) -- (2.810,3.026) -- (2.795,3.016) -- (2.778,3.013) -- (2.760,3.010) -- (2.745,2.997) -- (2.727,2.994) -- (2.709,2.993) -- (2.692,3.005);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.425,0.467) {N};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.425,3.467) {H};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (4.408,0.467) {J};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.267) {4,4 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {8,8 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (2.864,2.689) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{NHJ}\)) =  \(\dfrac{NJ}{HJ}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{NHJ}\)) = \(\dfrac{4{,}4}{8{,}8}\)
\par\addvspace{.5em}
On a donc \(\widehat{NHJ}\) =  ArcSin(\(\dfrac{4{,}4}{8{,}8}\)) \(\approx\) 30\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.075,3.400) -- (1.072,3.047) -- (1.086,2.697) -- (1.066,2.349) -- (1.078,2.001) -- (1.082,1.652) -- (1.085,1.301) -- (1.087,0.949) -- (1.075,0.600);
  \draw[black] (1.075,0.600) -- (1.419,0.588) -- (1.766,0.609) -- (2.117,0.593) -- (2.462,0.589) -- (2.805,0.604) -- (3.153,0.608) -- (3.502,0.597) -- (3.845,0.593) -- (4.191,0.612) -- (4.541,0.606) -- (4.886,0.593) -- (5.234,0.595) -- (5.580,0.601) -- (5.925,0.600);
  \draw[black] (5.925,0.600) -- (5.631,0.784) -- (5.319,0.952) -- (5.023,1.132) -- (4.716,1.302) -- (4.411,1.479) -- (4.104,1.649) -- (3.807,1.833) -- (3.498,2.001) -- (3.192,2.171) -- (2.892,2.350) -- (2.599,2.536) -- (2.289,2.705) -- (1.989,2.887) -- (1.675,3.039) -- (1.370,3.215) -- (1.075,3.400);
  \draw[black] (1.075,0.880) -- (1.073,0.788) -- (1.070,0.695) -- (1.075,0.600);
  \draw[black] (1.075,0.600) -- (1.169,0.601) -- (1.263,0.598) -- (1.355,0.600);
  \draw[black] (1.075,0.880) -- (1.168,0.883) -- (1.261,0.884) -- (1.355,0.880);
  \draw[black] (1.355,0.880) -- (1.359,0.787) -- (1.353,0.692) -- (1.355,0.600);
  \draw[black] (5.577,0.801) -- (5.568,0.786) -- (5.560,0.770) -- (5.558,0.752) -- (5.541,0.740) -- (5.534,0.723) -- (5.534,0.705) -- (5.533,0.687) -- (5.531,0.670) -- (5.523,0.653) -- (5.518,0.636) -- (5.518,0.618) -- (5.531,0.600);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.808,0.467) {W};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.808,3.467) {V};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.025,0.467) {H};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.575,1.933) {9,4 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 (5.281,0.706) {30°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{WHV}\)) = \(\dfrac{WV}{VH}\)
\par\addvspace{.5em}
 D'où sin(30\textdegree{}) = \(\dfrac{9{,}4}{VH}\)
\par\addvspace{.5em}
On a donc VH =\(\dfrac{9{,}4}{\sin(30^{\circ})}\) \(\approx\) 18,8 cm\end{document}
