\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 282}

\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=282}\\[2pt]
    {\small\ttfamily\bfseries\color{red} TRIG0282}
\end{minipage}%
}
}

\begin{document}

\sloppy

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

 Dans le triangle VRS rectangle en V, on sait que :
\begin{itemize}
\item RS=9,6 cm
\item \(\widehat{VRS}\) = 53\textdegree{}
\end{itemize}

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

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

 Dans le triangle KDP rectangle en K, on sait que :
\begin{itemize}
\item KD=1,7 cm
\item DP=7,6 cm
\end{itemize}

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

 Dans le triangle RTC rectangle en R, on sait que :
\begin{itemize}
\item RT=1,7 cm
\item RC=3,9 cm
\end{itemize}

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

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

 Dans le triangle KAR rectangle en K, on sait que :
\begin{itemize}
\item KA=4,1 cm
\item \(\widehat{KAR}\) = 72\textdegree{}
\end{itemize}

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

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

 Dans le triangle NJG rectangle en N, on sait que :
\begin{itemize}
\item JG=2,5 cm
\item \(\widehat{JGN}\) =  42\textdegree{}
\end{itemize}

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

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

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

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.642,3.400) -- (1.645,3.052) -- (1.637,2.697) -- (1.643,2.351) -- (1.650,2.002) -- (1.653,1.649) -- (1.639,1.300) -- (1.633,0.949) -- (1.642,0.600);
  \draw[black] (1.642,0.600) -- (1.980,0.605) -- (2.319,0.604) -- (2.657,0.607) -- (2.994,0.606) -- (3.330,0.588) -- (3.671,0.609) -- (4.009,0.594) -- (4.341,0.592) -- (4.684,0.599) -- (5.018,0.602) -- (5.358,0.600);
  \draw[black] (5.358,0.600) -- (5.068,0.812) -- (4.783,1.021) -- (4.500,1.244) -- (4.210,1.452) -- (3.933,1.687) -- (3.640,1.887) -- (3.365,2.115) -- (3.079,2.334) -- (2.779,2.531) -- (2.502,2.753) -- (2.213,2.966) -- (1.928,3.186) -- (1.642,3.400);
  \draw[black] (1.642,0.880) -- (1.648,0.787) -- (1.643,0.693) -- (1.642,0.600);
  \draw[black] (1.642,0.600) -- (1.735,0.600) -- (1.830,0.598) -- (1.922,0.600);
  \draw[black] (1.642,0.880) -- (1.736,0.874) -- (1.829,0.883) -- (1.922,0.880);
  \draw[black] (1.922,0.880) -- (1.921,0.787) -- (1.927,0.692) -- (1.922,0.600);
  \draw[black] (1.964,3.158) -- (1.946,3.132) -- (1.923,3.111) -- (1.902,3.087) -- (1.871,3.077) -- (1.850,3.053) -- (1.821,3.042) -- (1.794,3.025) -- (1.762,3.023) -- (1.732,3.018) -- (1.705,2.998) -- (1.673,2.999) -- (1.642,2.998);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.375,0.467) {V};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.375,3.467) {R};
  \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) {9,6 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 VRS rectangle en V, on cherche une relation entre l'angle aigu   \(\widehat{VRS}\) son coté opposé et l'hypoténuse du triangle.

\par\addvspace{.5em}
sin(\(\widehat{VRS}\)) =  \(\dfrac{VS}{RS}\)
\par\addvspace{.5em}
 d'où sin(53\textdegree{}) = \(\dfrac{VS}{9{,}6}\)

On a donc VS = 9,6 \(\times\) sin(53\textdegree{}) \(\approx\) 7,7 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,2.562) -- (1.041,2.422) -- (1.044,2.280) -- (1.044,2.143) -- (1.059,1.999) -- (1.050,1.857) -- (1.044,1.718) -- (1.038,1.578) -- (1.050,1.438);
  \draw[black] (1.050,1.438) -- (1.402,1.433) -- (1.752,1.445) -- (2.103,1.445) -- (2.452,1.427) -- (2.803,1.446) -- (3.149,1.438) -- (3.499,1.428) -- (3.848,1.427) -- (4.198,1.431) -- (4.552,1.425) -- (4.900,1.444) -- (5.253,1.437) -- (5.601,1.448) -- (5.950,1.438);
  \draw[black] (5.950,1.438) -- (5.621,1.504) -- (5.295,1.588) -- (4.970,1.666) -- (4.645,1.732) -- (4.320,1.817) -- (3.985,1.876) -- (3.665,1.965) -- (3.335,2.027) -- (3.011,2.112) -- (2.686,2.196) -- (2.358,2.257) -- (2.033,2.344) -- (1.699,2.407) -- (1.381,2.497) -- (1.050,2.562);
  \draw[black] (1.050,1.550) -- (1.049,1.511) -- (1.044,1.476) -- (1.050,1.438);
  \draw[black] (1.050,1.438) -- (1.087,1.433) -- (1.124,1.432) -- (1.162,1.438);
  \draw[black] (1.050,1.550) -- (1.088,1.550) -- (1.124,1.547) -- (1.162,1.550);
  \draw[black] (1.162,1.550) -- (1.161,1.514) -- (1.162,1.474) -- (1.162,1.438);
  \draw[black] (1.440,2.473) -- (1.426,2.430) -- (1.410,2.388) -- (1.393,2.346) -- (1.362,2.313) -- (1.337,2.275) -- (1.293,2.257) -- (1.258,2.231) -- (1.224,2.201) -- (1.180,2.190) -- (1.140,2.168) -- (1.095,2.164) -- (1.050,2.154);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,1.304) {K};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,2.629) {D};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,1.304) {P};
  \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 (3.667,2.000) {7,6 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.465,1.974) {?};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{KDP}\)) = \(\dfrac{KD}{DP}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{KDP}\)) = \(\dfrac{1{,}7}{7{,}6}\)
\par\addvspace{.5em}
On a donc \(\widehat{KDP}\) =ArcCos(\(\dfrac{1{,}7}{7{,}6}\)) \(\approx\)77\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,3.068) -- (1.051,2.799) -- (1.060,2.536) -- (1.044,2.268) -- (1.055,1.997) -- (1.049,1.731) -- (1.049,1.467) -- (1.062,1.198) -- (1.050,0.932);
  \draw[black] (1.050,0.932) -- (1.400,0.928) -- (1.752,0.922) -- (2.102,0.941) -- (2.449,0.927) -- (2.802,0.938) -- (3.150,0.919) -- (3.497,0.922) -- (3.852,0.939) -- (4.202,0.929) -- (4.550,0.922) -- (4.901,0.945) -- (5.249,0.921) -- (5.599,0.923) -- (5.950,0.932);
  \draw[black] (5.950,0.932) -- (5.640,1.057) -- (5.345,1.207) -- (5.029,1.333) -- (4.725,1.472) -- (4.424,1.608) -- (4.114,1.734) -- (3.809,1.869) -- (3.497,1.989) -- (3.195,2.135) -- (2.888,2.270) -- (2.579,2.395) -- (2.270,2.526) -- (1.969,2.669) -- (1.670,2.809) -- (1.350,2.926) -- (1.050,3.068);
  \draw[black] (1.050,1.146) -- (1.049,1.076) -- (1.052,1.002) -- (1.050,0.932);
  \draw[black] (1.050,0.932) -- (1.121,0.931) -- (1.193,0.935) -- (1.264,0.932);
  \draw[black] (1.050,1.146) -- (1.121,1.151) -- (1.192,1.145) -- (1.264,1.146);
  \draw[black] (1.264,1.146) -- (1.267,1.073) -- (1.257,1.003) -- (1.264,0.932);
  \draw[black] (5.583,1.092) -- (5.585,1.077) -- (5.577,1.065) -- (5.570,1.053) -- (5.566,1.040) -- (5.566,1.026) -- (5.567,1.012) -- (5.556,1.000) -- (5.545,0.988) -- (5.543,0.974) -- (5.560,0.959) -- (5.550,0.946) -- (5.544,0.932);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,0.799) {R};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,3.135) {T};
  \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 (5.297,1.001) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{RCT}\)) = \(\dfrac{RT}{RC}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{RCT}\)) =  \(\dfrac{1{,}7}{3{,}9}\)
\par\addvspace{.5em}
 On a donc \(\widehat{RCT}\) =  ArcTan(\(\dfrac{1{,}7}{3{,}9}\)) \(\approx\) 24\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,2.796) -- (1.060,2.594) -- (1.044,2.401) -- (1.060,2.199) -- (1.045,1.998) -- (1.050,1.798) -- (1.056,1.604) -- (1.048,1.405) -- (1.050,1.204);
  \draw[black] (1.050,1.204) -- (1.397,1.207) -- (1.749,1.207) -- (2.099,1.199) -- (2.448,1.208) -- (2.802,1.211) -- (3.151,1.202) -- (3.501,1.197) -- (3.849,1.213) -- (4.200,1.197) -- (4.551,1.193) -- (4.898,1.215) -- (5.252,1.192) -- (5.597,1.199) -- (5.950,1.204);
  \draw[black] (5.950,1.204) -- (5.628,1.322) -- (5.298,1.428) -- (4.968,1.522) -- (4.643,1.629) -- (4.319,1.743) -- (3.988,1.830) -- (3.665,1.956) -- (3.335,2.054) -- (3.007,2.152) -- (2.686,2.268) -- (2.351,2.364) -- (2.028,2.472) -- (1.697,2.574) -- (1.373,2.677) -- (1.050,2.796);
  \draw[black] (1.050,1.363) -- (1.052,1.311) -- (1.053,1.255) -- (1.050,1.204);
  \draw[black] (1.050,1.204) -- (1.102,1.200) -- (1.156,1.197) -- (1.209,1.204);
  \draw[black] (1.050,1.363) -- (1.104,1.369) -- (1.156,1.366) -- (1.209,1.363);
  \draw[black] (1.209,1.363) -- (1.212,1.309) -- (1.209,1.256) -- (1.209,1.204);
  \draw[black] (1.428,2.673) -- (1.412,2.635) -- (1.388,2.601) -- (1.381,2.556) -- (1.348,2.528) -- (1.311,2.506) -- (1.283,2.475) -- (1.248,2.453) -- (1.209,2.440) -- (1.177,2.406) -- (1.132,2.412) -- (1.092,2.400) -- (1.050,2.395);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,1.071) {K};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,2.863) {A};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,1.071) {R};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {4,1 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 (1.442,2.190) {72°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{KAR}\)) = \(\dfrac{KA}{AR}\)
\par\addvspace{.5em}
 d'où cos(72\textdegree{}) = \(\dfrac{4{,}1}{AR}\)
\par\addvspace{.5em}
On a donc AR = \(\dfrac{4{,}1}{\cos(72^{\circ})}\)\(\approx\) 13,3 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.945,3.400) -- (1.952,3.051) -- (1.938,2.701) -- (1.950,2.348) -- (1.949,1.999) -- (1.948,1.649) -- (1.945,1.297) -- (1.935,0.950) -- (1.945,0.600);
  \draw[black] (1.945,0.600) -- (2.292,0.599) -- (2.633,0.612) -- (2.982,0.598) -- (3.325,0.597) -- (3.670,0.592) -- (4.020,0.597) -- (4.361,0.601) -- (4.711,0.599) -- (5.055,0.600);
  \draw[black] (5.055,0.600) -- (4.787,0.823) -- (4.527,1.060) -- (4.285,1.303) -- (4.017,1.527) -- (3.754,1.764) -- (3.493,1.991) -- (3.243,2.236) -- (2.976,2.464) -- (2.719,2.701) -- (2.460,2.926) -- (2.203,3.170) -- (1.945,3.400);
  \draw[black] (1.945,0.880) -- (1.950,0.786) -- (1.945,0.695) -- (1.945,0.600);
  \draw[black] (1.945,0.600) -- (2.038,0.598) -- (2.133,0.606) -- (2.225,0.600);
  \draw[black] (1.945,0.880) -- (2.040,0.875) -- (2.131,0.876) -- (2.225,0.880);
  \draw[black] (2.225,0.880) -- (2.223,0.788) -- (2.226,0.694) -- (2.225,0.600);
  \draw[black] (4.758,0.867) -- (4.743,0.848) -- (4.721,0.834) -- (4.720,0.805) -- (4.710,0.784) -- (4.695,0.764) -- (4.685,0.742) -- (4.679,0.719) -- (4.674,0.695) -- (4.665,0.672) -- (4.653,0.649) -- (4.663,0.624) -- (4.658,0.600);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.678,0.467) {N};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.678,3.467) {J};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.155,0.467) {G};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.445,1.933) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {2,5 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (4.432,0.772) {42°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{NGJ}\)) = \(\dfrac{NJ}{JG}\)
\par\addvspace{.5em}
 D'où sin(42\textdegree{}) = \(\dfrac{NJ}{2{,}5}\)

On a donc NJ = 2,5 \(\times\) sin(42\textdegree{}) \(\approx\) 1,7 cm\end{document}
