\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 111}

\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=111}\\[2pt]
    {\small\ttfamily\bfseries\color{red} TRIG0111}
\end{minipage}%
}
}

\begin{document}

\sloppy

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

 Dans le triangle ANK rectangle en A, on sait que :
\begin{itemize}
\item AN=2,5 cm
\item AK=3,7 cm
\end{itemize}

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

 Dans le triangle VRS rectangle en V, on sait que :
\begin{itemize}
\item RS=0,4 cm
\item \(\widehat{RSV}\) =  32\textdegree{}
\end{itemize}

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

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

 Dans le triangle LKV rectangle en L, on sait que :
\begin{itemize}
\item LK=4 cm
\item \(\widehat{LKV}\) = 48\textdegree{}
\end{itemize}

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

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

 Dans le triangle RSV rectangle en R, on sait que :
\begin{itemize}
\item SV=6,7 cm
\item \(\widehat{RSV}\) = 61\textdegree{}
\end{itemize}

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

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

 Dans le triangle DMZ rectangle en D, on sait que :
\begin{itemize}
\item DM=3,3 cm
\item DZ=6,2 cm
\end{itemize}

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

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

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

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.428,3.400) -- (1.421,3.052) -- (1.426,2.701) -- (1.418,2.348) -- (1.438,2.003) -- (1.429,1.652) -- (1.435,1.297) -- (1.416,0.953) -- (1.428,0.600);
  \draw[black] (1.428,0.600) -- (1.774,0.613) -- (2.121,0.588) -- (2.466,0.595) -- (2.809,0.600) -- (3.152,0.590) -- (3.501,0.595) -- (3.847,0.609) -- (4.194,0.609) -- (4.536,0.593) -- (4.884,0.598) -- (5.225,0.604) -- (5.572,0.600);
  \draw[black] (5.572,0.600) -- (5.296,0.792) -- (5.013,0.967) -- (4.741,1.158) -- (4.464,1.348) -- (4.189,1.537) -- (3.923,1.729) -- (3.643,1.917) -- (3.356,2.086) -- (3.085,2.273) -- (2.803,2.456) -- (2.541,2.660) -- (2.256,2.843) -- (1.977,3.024) -- (1.704,3.216) -- (1.428,3.400);
  \draw[black] (1.428,0.880) -- (1.428,0.785) -- (1.429,0.694) -- (1.428,0.600);
  \draw[black] (1.428,0.600) -- (1.521,0.605) -- (1.616,0.605) -- (1.708,0.600);
  \draw[black] (1.428,0.880) -- (1.522,0.885) -- (1.615,0.876) -- (1.708,0.880);
  \draw[black] (1.708,0.880) -- (1.714,0.788) -- (1.702,0.694) -- (1.708,0.600);
  \draw[black] (1.764,3.173) -- (1.734,3.155) -- (1.712,3.132) -- (1.694,3.105) -- (1.669,3.084) -- (1.641,3.068) -- (1.617,3.044) -- (1.587,3.031) -- (1.556,3.020) -- (1.524,3.015) -- (1.493,3.007) -- (1.460,3.006) -- (1.428,2.999);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.161,0.467) {A};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.161,3.467) {N};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.672,0.467) {K};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.928,1.933) {2,5 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.267) {3,7 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.741,2.745) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{ANK}\)) = \(\dfrac{AK}{AN}\)
\par\addvspace{.5em}
d'où tan(\(\widehat{ANK}\)) = \(\dfrac{3{,}7}{2{,}5}\)
\par\addvspace{.5em}
 On a donc ANK = ArcTan(\(\dfrac{3{,}7}{2{,}5}\)) \(\approx\) 56\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.260,3.400) -- (1.255,3.052) -- (1.259,2.703) -- (1.257,2.353) -- (1.260,2.000) -- (1.264,1.650) -- (1.255,1.297) -- (1.259,0.953) -- (1.260,0.600);
  \draw[black] (1.260,0.600) -- (1.604,0.596) -- (1.950,0.587) -- (2.290,0.606) -- (2.638,0.610) -- (2.982,0.597) -- (3.329,0.608) -- (3.672,0.606) -- (4.018,0.599) -- (4.364,0.589) -- (4.705,0.595) -- (5.051,0.612) -- (5.395,0.593) -- (5.740,0.600);
  \draw[black] (5.740,0.600) -- (5.437,0.778) -- (5.141,0.970) -- (4.848,1.168) -- (4.540,1.340) -- (4.247,1.533) -- (3.948,1.720) -- (3.643,1.901) -- (3.349,2.088) -- (3.054,2.285) -- (2.757,2.476) -- (2.457,2.654) -- (2.153,2.832) -- (1.854,3.019) -- (1.562,3.221) -- (1.260,3.400);
  \draw[black] (1.260,0.880) -- (1.262,0.786) -- (1.256,0.695) -- (1.260,0.600);
  \draw[black] (1.260,0.600) -- (1.352,0.597) -- (1.448,0.596) -- (1.540,0.600);
  \draw[black] (1.260,0.880) -- (1.354,0.886) -- (1.447,0.885) -- (1.540,0.880);
  \draw[black] (1.540,0.880) -- (1.543,0.786) -- (1.540,0.693) -- (1.540,0.600);
  \draw[black] (5.399,0.813) -- (5.392,0.796) -- (5.383,0.780) -- (5.373,0.764) -- (5.369,0.745) -- (5.370,0.725) -- (5.348,0.712) -- (5.344,0.694) -- (5.342,0.675) -- (5.348,0.655) -- (5.337,0.638) -- (5.343,0.618) -- (5.340,0.600);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.993,0.467) {V};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.993,3.467) {R};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.840,0.467) {S};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.760,1.933) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {0,4 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.100,0.717) {32°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{VSR}\)) = \(\dfrac{VR}{RS}\)
\par\addvspace{.5em}
 D'où sin(32\textdegree{}) = \(\dfrac{VR}{0{,}4}\)

On a donc VR = 0,4 \(\times\) sin(32\textdegree{}) \(\approx\) 0,2 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.945,3.400) -- (1.936,3.049) -- (1.945,2.702) -- (1.953,2.353) -- (1.942,2.002) -- (1.958,1.649) -- (1.941,1.303) -- (1.952,0.947) -- (1.945,0.600);
  \draw[black] (1.945,0.600) -- (2.293,0.595) -- (2.633,0.607) -- (2.982,0.596) -- (3.324,0.598) -- (3.675,0.597) -- (4.018,0.600) -- (4.363,0.609) -- (4.706,0.608) -- (5.055,0.600);
  \draw[black] (5.055,0.600) -- (4.805,0.842) -- (4.529,1.060) -- (4.270,1.295) -- (4.014,1.527) -- (3.758,1.761) -- (3.505,2.007) -- (3.241,2.238) -- (2.973,2.460) -- (2.717,2.698) -- (2.472,2.942) -- (2.211,3.173) -- (1.945,3.400);
  \draw[black] (1.945,0.880) -- (1.950,0.787) -- (1.943,0.693) -- (1.945,0.600);
  \draw[black] (1.945,0.600) -- (2.037,0.600) -- (2.133,0.598) -- (2.225,0.600);
  \draw[black] (1.945,0.880) -- (2.038,0.878) -- (2.133,0.877) -- (2.225,0.880);
  \draw[black] (2.225,0.880) -- (2.227,0.787) -- (2.226,0.693) -- (2.225,0.600);
  \draw[black] (2.249,3.126) -- (2.227,3.108) -- (2.196,3.101) -- (2.178,3.080) -- (2.158,3.059) -- (2.137,3.039) -- (2.109,3.032) -- (2.084,3.018) -- (2.054,3.022) -- (2.027,3.013) -- (2.000,3.008) -- (1.972,3.009) -- (1.945,2.996);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.678,0.467) {L};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.678,3.467) {K};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.155,0.467) {V};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.445,1.933) {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 (2.216,2.724) {48°};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{LKV}\)) = \(\dfrac{LK}{KV}\)
\par\addvspace{.5em}
 d'où cos(48\textdegree{}) = \(\dfrac{4}{KV}\)
\par\addvspace{.5em}
On a donc KV = \(\dfrac{4}{\cos(48^{\circ})}\)\(\approx\) 6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,3.358) -- (1.037,3.018) -- (1.054,2.678) -- (1.058,2.341) -- (1.051,2.003) -- (1.052,1.663) -- (1.043,1.320) -- (1.041,0.984) -- (1.050,0.642);
  \draw[black] (1.050,0.642) -- (1.401,0.652) -- (1.749,0.639) -- (2.100,0.654) -- (2.452,0.629) -- (2.801,0.653) -- (3.149,0.635) -- (3.502,0.653) -- (3.850,0.655) -- (4.202,0.654) -- (4.550,0.647) -- (4.901,0.638) -- (5.248,0.637) -- (5.601,0.653) -- (5.950,0.642);
  \draw[black] (5.950,0.642) -- (5.640,0.800) -- (5.336,0.983) -- (5.027,1.141) -- (4.717,1.311) -- (4.415,1.483) -- (4.113,1.655) -- (3.814,1.839) -- (3.493,1.989) -- (3.189,2.158) -- (2.887,2.336) -- (2.584,2.520) -- (2.276,2.682) -- (1.968,2.853) -- (1.661,3.015) -- (1.360,3.193) -- (1.050,3.358);
  \draw[black] (1.050,0.914) -- (1.052,0.822) -- (1.049,0.734) -- (1.050,0.642);
  \draw[black] (1.050,0.642) -- (1.139,0.644) -- (1.232,0.642) -- (1.322,0.642);
  \draw[black] (1.050,0.914) -- (1.141,0.912) -- (1.231,0.919) -- (1.322,0.914);
  \draw[black] (1.322,0.914) -- (1.316,0.823) -- (1.317,0.733) -- (1.322,0.642);
  \draw[black] (1.392,3.169) -- (1.374,3.139) -- (1.357,3.108) -- (1.335,3.081) -- (1.310,3.056) -- (1.281,3.036) -- (1.256,3.008) -- (1.223,2.994) -- (1.190,2.980) -- (1.153,2.981) -- (1.122,2.955) -- (1.085,2.960) -- (1.050,2.961);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,0.509) {R};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,3.425) {S};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,0.509) {V};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.309) {?};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.667,2.000) {6,7 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.388,2.717) {61°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{RSV}\)) =  \(\dfrac{RV}{SV}\)
\par\addvspace{.5em}
 d'où sin(61\textdegree{}) = \(\dfrac{RV}{6{,}7}\)

On a donc RV = 6,7 \(\times\) sin(61\textdegree{}) \(\approx\) 5,9 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,3.304) -- (1.041,2.976) -- (1.054,2.652) -- (1.038,2.326) -- (1.038,1.997) -- (1.057,1.677) -- (1.052,1.348) -- (1.052,1.021) -- (1.050,0.696);
  \draw[black] (1.050,0.696) -- (1.402,0.694) -- (1.751,0.708) -- (2.102,0.707) -- (2.448,0.691) -- (2.802,0.689) -- (3.149,0.696) -- (3.498,0.693) -- (3.851,0.700) -- (4.202,0.696) -- (4.553,0.683) -- (4.899,0.690) -- (5.249,0.709) -- (5.599,0.699) -- (5.950,0.696);
  \draw[black] (5.950,0.696) -- (5.648,0.864) -- (5.339,1.029) -- (5.026,1.182) -- (4.722,1.338) -- (4.416,1.505) -- (4.120,1.683) -- (3.808,1.843) -- (3.496,1.996) -- (3.196,2.166) -- (2.888,2.326) -- (2.582,2.494) -- (2.279,2.659) -- (1.963,2.804) -- (1.658,2.976) -- (1.361,3.148) -- (1.050,3.304);
  \draw[black] (1.050,0.957) -- (1.051,0.871) -- (1.052,0.784) -- (1.050,0.696);
  \draw[black] (1.050,0.696) -- (1.136,0.702) -- (1.224,0.696) -- (1.311,0.696);
  \draw[black] (1.050,0.957) -- (1.136,0.954) -- (1.224,0.958) -- (1.311,0.957);
  \draw[black] (1.311,0.957) -- (1.307,0.871) -- (1.304,0.784) -- (1.311,0.696);
  \draw[black] (5.590,0.887) -- (5.591,0.869) -- (5.574,0.858) -- (5.575,0.840) -- (5.568,0.825) -- (5.565,0.809) -- (5.569,0.791) -- (5.560,0.777) -- (5.555,0.761) -- (5.557,0.744) -- (5.548,0.729) -- (5.553,0.712) -- (5.548,0.696);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,0.563) {D};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,3.371) {M};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,0.563) {Z};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {3,3 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,0.363) {6,2 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.303,0.791) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{DZM}\)) = \(\dfrac{DM}{DZ}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{DZM}\)) =  \(\dfrac{3{,}3}{6{,}2}\)
\par\addvspace{.5em}
 On a donc \(\widehat{DZM}\) =  ArcTan(\(\dfrac{3{,}3}{6{,}2}\)) \(\approx\) 28\textdegree{}\end{document}
