\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 256}

\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=256}\\[2pt]
    {\small\ttfamily\bfseries\color{red} TRIG0256}
\end{minipage}%
}
}

\begin{document}

\sloppy

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

 Dans le triangle WVB rectangle en W, on sait que :
\begin{itemize}
\item WV=3,2 cm
\item WB=5,2 cm
\end{itemize}

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

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

 Dans le triangle HTR rectangle en H, on sait que :
\begin{itemize}
\item HT=0,9 cm
\item \(\widehat{HTR}\) = 80\textdegree{}
\end{itemize}

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

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

 Dans le triangle HSJ rectangle en H, on sait que :
\begin{itemize}
\item HJ=6,3 cm
\item SJ=8,8 cm
\end{itemize}

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

 Dans le triangle HVN rectangle en H, on sait que :
\begin{itemize}
\item VN=9,2 cm
\item \(\widehat{HVN}\) = 53\textdegree{}
\end{itemize}

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

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

 Dans le triangle JGP rectangle en J, on sait que :
\begin{itemize}
\item JG=9,2 cm
\item \(\widehat{JGP}\) = 63\textdegree{}
\end{itemize}

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

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

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

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.225,3.400) -- (1.226,3.053) -- (1.216,2.697) -- (1.233,2.350) -- (1.218,1.998) -- (1.221,1.653) -- (1.235,1.301) -- (1.212,0.950) -- (1.225,0.600);
  \draw[black] (1.225,0.600) -- (1.575,0.591) -- (1.922,0.588) -- (2.273,0.599) -- (2.626,0.603) -- (2.978,0.607) -- (3.326,0.591) -- (3.672,0.593) -- (4.027,0.612) -- (4.372,0.606) -- (4.723,0.589) -- (5.073,0.587) -- (5.427,0.603) -- (5.775,0.600);
  \draw[black] (5.775,0.600) -- (5.497,0.782) -- (5.212,0.962) -- (4.920,1.124) -- (4.641,1.305) -- (4.350,1.476) -- (4.067,1.653) -- (3.790,1.835) -- (3.492,1.989) -- (3.216,2.174) -- (2.936,2.356) -- (2.648,2.521) -- (2.366,2.702) -- (2.079,2.879) -- (1.802,3.059) -- (1.506,3.217) -- (1.225,3.400);
  \draw[black] (1.225,0.880) -- (1.228,0.786) -- (1.229,0.694) -- (1.225,0.600);
  \draw[black] (1.225,0.600) -- (1.317,0.601) -- (1.413,0.603) -- (1.505,0.600);
  \draw[black] (1.225,0.880) -- (1.318,0.884) -- (1.412,0.881) -- (1.505,0.880);
  \draw[black] (1.505,0.880) -- (1.507,0.787) -- (1.505,0.694) -- (1.505,0.600);
  \draw[black] (5.434,0.810) -- (5.424,0.794) -- (5.416,0.778) -- (5.417,0.757) -- (5.409,0.741) -- (5.399,0.725) -- (5.383,0.711) -- (5.379,0.693) -- (5.380,0.673) -- (5.382,0.655) -- (5.386,0.636) -- (5.366,0.619) -- (5.367,0.600);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.958,0.467) {W};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.958,3.467) {V};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.875,0.467) {B};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.725,1.933) {3,2 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.267) {5,2 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.134,0.715) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{WBV}\)) = \(\dfrac{WV}{WB}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{WBV}\)) =  \(\dfrac{3{,}2}{5{,}2}\)
\par\addvspace{.5em}
 On a donc \(\widehat{WBV}\) =  ArcTan(\(\dfrac{3{,}2}{5{,}2}\)) \(\approx\) 32\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,2.432) -- (1.059,2.324) -- (1.056,2.214) -- (1.062,2.110) -- (1.052,1.998) -- (1.048,1.891) -- (1.051,1.787) -- (1.042,1.674) -- (1.050,1.568);
  \draw[black] (1.050,1.568) -- (1.402,1.562) -- (1.753,1.575) -- (2.099,1.555) -- (2.450,1.573) -- (2.800,1.562) -- (3.147,1.574) -- (3.497,1.569) -- (3.851,1.563) -- (4.201,1.571) -- (4.548,1.563) -- (4.903,1.562) -- (5.249,1.564) -- (5.600,1.571) -- (5.950,1.568);
  \draw[black] (5.950,1.568) -- (5.605,1.639) -- (5.250,1.688) -- (4.899,1.751) -- (4.548,1.803) -- (4.200,1.880) -- (3.851,1.928) -- (3.503,2.007) -- (3.155,2.073) -- (2.797,2.113) -- (2.451,2.179) -- (2.098,2.249) -- (1.751,2.319) -- (1.400,2.375) -- (1.050,2.432);
  \draw[black] (1.050,1.654) -- (1.048,1.626) -- (1.045,1.596) -- (1.050,1.568);
  \draw[black] (1.050,1.568) -- (1.078,1.564) -- (1.109,1.565) -- (1.136,1.568);
  \draw[black] (1.050,1.654) -- (1.078,1.654) -- (1.106,1.660) -- (1.136,1.654);
  \draw[black] (1.136,1.654) -- (1.134,1.627) -- (1.141,1.596) -- (1.136,1.568);
  \draw[black] (1.453,2.361) -- (1.428,2.319) -- (1.417,2.274) -- (1.390,2.236) -- (1.369,2.195) -- (1.347,2.152) -- (1.301,2.133) -- (1.275,2.090) -- (1.231,2.072) -- (1.186,2.059) -- (1.141,2.049) -- (1.097,2.033) -- (1.050,2.036);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,1.435) {H};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,2.499) {T};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,1.435) {R};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {0,9 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,1.235) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.479,1.855) {80°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{HTR}\)) = \(\dfrac{HR}{HT}\)
\par\addvspace{.5em}
d'où tan(80\textdegree{}) =\(\dfrac{HR}{0{,}9}\)
\par\addvspace{.5em}
On a donc HR = 0,9 \(\times\) tan(80\textdegree{}) \(\approx\) 5,1 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (2.064,3.400) -- (2.058,3.053) -- (2.066,2.703) -- (2.063,2.353) -- (2.058,2.000) -- (2.070,1.651) -- (2.072,1.298) -- (2.053,0.949) -- (2.064,0.600);
  \draw[black] (2.064,0.600) -- (2.420,0.611) -- (2.783,0.612) -- (3.144,0.610) -- (3.503,0.606) -- (3.859,0.595) -- (4.220,0.613) -- (4.575,0.603) -- (4.936,0.600);
  \draw[black] (4.936,0.600) -- (4.698,0.835) -- (4.456,1.064) -- (4.209,1.295) -- (3.971,1.529) -- (3.733,1.759) -- (3.494,1.993) -- (3.267,2.237) -- (3.024,2.465) -- (2.787,2.707) -- (2.538,2.928) -- (2.305,3.163) -- (2.064,3.400);
  \draw[black] (2.064,0.880) -- (2.065,0.788) -- (2.061,0.692) -- (2.064,0.600);
  \draw[black] (2.064,0.600) -- (2.157,0.593) -- (2.250,0.594) -- (2.344,0.600);
  \draw[black] (2.064,0.880) -- (2.159,0.882) -- (2.253,0.886) -- (2.344,0.880);
  \draw[black] (2.344,0.880) -- (2.347,0.787) -- (2.339,0.694) -- (2.344,0.600);
  \draw[black] (2.354,3.117) -- (2.328,3.107) -- (2.307,3.091) -- (2.292,3.066) -- (2.265,3.059) -- (2.242,3.047) -- (2.219,3.033) -- (2.193,3.029) -- (2.171,3.011) -- (2.144,3.008) -- (2.116,3.012) -- (2.091,3.003) -- (2.064,2.998);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.798,0.467) {H};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.798,3.467) {S};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.036,0.467) {J};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.267) {6,3 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 (2.323,2.719) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{HSJ}\)) =  \(\dfrac{HJ}{SJ}\)
\par\addvspace{.5em}
 d'où sin(\(\widehat{HSJ}\)) = \(\dfrac{6{,}3}{8{,}8}\)
\par\addvspace{.5em}
On a donc \(\widehat{HSJ}\) =  ArcSin(\(\dfrac{6{,}3}{8{,}8}\)) \(\approx\) 46\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.642,3.400) -- (1.653,3.053) -- (1.647,2.698) -- (1.637,2.353) -- (1.646,2.002) -- (1.640,1.650) -- (1.639,1.299) -- (1.633,0.952) -- (1.642,0.600);
  \draw[black] (1.642,0.600) -- (1.979,0.603) -- (2.321,0.599) -- (2.658,0.613) -- (2.997,0.598) -- (3.334,0.604) -- (3.667,0.593) -- (4.005,0.613) -- (4.347,0.603) -- (4.685,0.604) -- (5.019,0.610) -- (5.358,0.600);
  \draw[black] (5.358,0.600) -- (5.076,0.821) -- (4.789,1.040) -- (4.506,1.251) -- (4.215,1.466) -- (3.928,1.678) -- (3.646,1.893) -- (3.355,2.108) -- (3.068,2.320) -- (2.787,2.542) -- (2.494,2.747) -- (2.211,2.961) -- (1.925,3.182) -- (1.642,3.400);
  \draw[black] (1.642,0.880) -- (1.636,0.787) -- (1.641,0.695) -- (1.642,0.600);
  \draw[black] (1.642,0.600) -- (1.735,0.604) -- (1.828,0.597) -- (1.922,0.600);
  \draw[black] (1.642,0.880) -- (1.734,0.885) -- (1.829,0.885) -- (1.922,0.880);
  \draw[black] (1.922,0.880) -- (1.926,0.786) -- (1.924,0.694) -- (1.922,0.600);
  \draw[black] (1.961,3.160) -- (1.948,3.130) -- (1.928,3.106) -- (1.903,3.087) -- (1.873,3.074) -- (1.852,3.050) -- (1.819,3.046) -- (1.791,3.033) -- (1.765,3.016) -- (1.735,3.005) -- (1.703,3.010) -- (1.672,3.008) -- (1.642,3.010);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.375,0.467) {H};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.375,3.467) {V};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.458,0.467) {N};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.142,1.933) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {9,2 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 HVN rectangle en H, on cherche une relation entre l'angle aigu   \(\widehat{HVN}\) son coté adjacent et l'hypoténuse du triangle.

cos(\(\widehat{HVN}\)) = \(\dfrac{HV}{VN}\)
\par\addvspace{.5em}
d'où cos(53\textdegree{}) = \(\dfrac{HV}{9{,}2}\)

On a donc HV = 9,2 \(\times\) cos(53\textdegree{}) \(\approx\) 5,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,3.248) -- (1.045,2.935) -- (1.057,2.622) -- (1.055,2.313) -- (1.054,1.998) -- (1.047,1.691) -- (1.055,1.374) -- (1.051,1.067) -- (1.050,0.752);
  \draw[black] (1.050,0.752) -- (1.399,0.739) -- (1.749,0.741) -- (2.103,0.757) -- (2.448,0.739) -- (2.797,0.759) -- (3.148,0.739) -- (3.501,0.755) -- (3.850,0.763) -- (4.201,0.755) -- (4.553,0.754) -- (4.897,0.749) -- (5.251,0.755) -- (5.600,0.740) -- (5.950,0.752);
  \draw[black] (5.950,0.752) -- (5.645,0.917) -- (5.344,1.075) -- (5.034,1.229) -- (4.718,1.364) -- (4.414,1.527) -- (4.113,1.694) -- (3.810,1.850) -- (3.497,1.996) -- (3.196,2.160) -- (2.881,2.305) -- (2.585,2.470) -- (2.271,2.615) -- (1.964,2.769) -- (1.668,2.944) -- (1.350,3.083) -- (1.050,3.248);
  \draw[black] (1.050,1.001) -- (1.052,0.918) -- (1.055,0.835) -- (1.050,0.752);
  \draw[black] (1.050,0.752) -- (1.132,0.749) -- (1.217,0.756) -- (1.300,0.752);
  \draw[black] (1.050,1.001) -- (1.132,0.996) -- (1.216,0.997) -- (1.300,1.001);
  \draw[black] (1.300,1.001) -- (1.301,0.919) -- (1.297,0.835) -- (1.300,0.752);
  \draw[black] (1.413,3.063) -- (1.394,3.031) -- (1.366,3.006) -- (1.339,2.981) -- (1.323,2.945) -- (1.291,2.926) -- (1.257,2.911) -- (1.227,2.889) -- (1.190,2.884) -- (1.160,2.859) -- (1.122,2.861) -- (1.087,2.848) -- (1.050,2.842);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,0.618) {J};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,3.315) {G};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,0.618) {P};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {9,2 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 (1.398,2.613) {63°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{JGP}\)) = \(\dfrac{JG}{GP}\)
\par\addvspace{.5em}
 d'où cos(63\textdegree{}) = \(\dfrac{9{,}2}{GP}\)
\par\addvspace{.5em}
On a donc GP = \(\dfrac{9{,}2}{\cos(63^{\circ})}\)\(\approx\) 20,3 cm\end{document}
