\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 185}

\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=185}\\[2pt]
    {\small\ttfamily\bfseries\color{red} TRIG0185}
\end{minipage}%
}
}

\begin{document}

\sloppy

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

 Dans le triangle BKV rectangle en B, on sait que :
\begin{itemize}
\item BK=6,5 cm
\item \(\widehat{BKV}\) = 58\textdegree{}
\end{itemize}

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

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

 Dans le triangle STC rectangle en S, on sait que :
\begin{itemize}
\item ST=9,8 cm
\item \(\widehat{TCS}\) =  13\textdegree{}
\end{itemize}

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

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

 Dans le triangle HZS rectangle en H, on sait que :
\begin{itemize}
\item HZ=1 cm
\item HS=4,9 cm
\end{itemize}

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

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

 Dans le triangle VJB rectangle en V, on sait que :
\begin{itemize}
\item JB=5,4 cm
\item \(\widehat{JBV}\) =  28\textdegree{}
\end{itemize}

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

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

 Dans le triangle SMB rectangle en S, on sait que :
\begin{itemize}
\item SM=2,1 cm
\item MB=7,1 cm
\end{itemize}

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

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

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.260,3.400) -- (1.264,3.049) -- (1.270,2.702) -- (1.260,2.353) -- (1.262,1.998) -- (1.272,1.647) -- (1.256,1.303) -- (1.251,0.947) -- (1.260,0.600);
  \draw[black] (1.260,0.600) -- (1.602,0.603) -- (1.949,0.595) -- (2.292,0.602) -- (2.638,0.595) -- (2.981,0.602) -- (3.331,0.605) -- (3.669,0.611) -- (4.015,0.606) -- (4.363,0.595) -- (4.708,0.606) -- (5.052,0.611) -- (5.393,0.588) -- (5.740,0.600);
  \draw[black] (5.740,0.600) -- (5.443,0.783) -- (5.141,0.976) -- (4.844,1.155) -- (4.544,1.339) -- (4.240,1.527) -- (3.949,1.716) -- (3.645,1.903) -- (3.352,2.101) -- (3.049,2.275) -- (2.762,2.475) -- (2.451,2.654) -- (2.156,2.841) -- (1.854,3.025) -- (1.553,3.202) -- (1.260,3.400);
  \draw[black] (1.260,0.880) -- (1.264,0.787) -- (1.255,0.693) -- (1.260,0.600);
  \draw[black] (1.260,0.600) -- (1.354,0.604) -- (1.446,0.606) -- (1.540,0.600);
  \draw[black] (1.260,0.880) -- (1.353,0.882) -- (1.447,0.874) -- (1.540,0.880);
  \draw[black] (1.540,0.880) -- (1.545,0.786) -- (1.536,0.693) -- (1.540,0.600);
  \draw[black] (1.598,3.189) -- (1.575,3.163) -- (1.559,3.133) -- (1.538,3.106) -- (1.506,3.092) -- (1.485,3.063) -- (1.456,3.045) -- (1.421,3.040) -- (1.394,3.017) -- (1.361,3.007) -- (1.326,3.012) -- (1.294,2.997) -- (1.260,2.991);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.993,0.467) {B};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.993,3.467) {K};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.840,0.467) {V};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.760,1.933) {6,5 cm};
  \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 (1.583,2.750) {58°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{BKV}\)) = \(\dfrac{BV}{BK}\)
\par\addvspace{.5em}
d'où tan(58\textdegree{}) =\(\dfrac{BV}{6{,}5}\)
\par\addvspace{.5em}
On a donc BV = 6,5 \(\times\) tan(58\textdegree{}) \(\approx\) 10,4 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 2}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,2.566) -- (1.045,2.424) -- (1.062,2.283) -- (1.044,2.144) -- (1.047,1.998) -- (1.063,1.856) -- (1.042,1.720) -- (1.052,1.578) -- (1.050,1.434);
  \draw[black] (1.050,1.434) -- (1.397,1.443) -- (1.750,1.433) -- (2.102,1.426) -- (2.449,1.439) -- (2.797,1.426) -- (3.152,1.441) -- (3.499,1.444) -- (3.849,1.428) -- (4.201,1.432) -- (4.547,1.430) -- (4.902,1.427) -- (5.251,1.425) -- (5.598,1.433) -- (5.950,1.434);
  \draw[black] (5.950,1.434) -- (5.621,1.515) -- (5.298,1.589) -- (4.972,1.657) -- (4.642,1.735) -- (4.319,1.809) -- (3.989,1.892) -- (3.668,1.966) -- (3.335,2.039) -- (3.009,2.118) -- (2.686,2.196) -- (2.354,2.257) -- (2.026,2.334) -- (1.705,2.414) -- (1.374,2.483) -- (1.050,2.566);
  \draw[black] (1.050,1.547) -- (1.049,1.511) -- (1.053,1.473) -- (1.050,1.434);
  \draw[black] (1.050,1.434) -- (1.089,1.433) -- (1.124,1.441) -- (1.163,1.434);
  \draw[black] (1.050,1.547) -- (1.089,1.548) -- (1.125,1.541) -- (1.163,1.547);
  \draw[black] (1.163,1.547) -- (1.167,1.510) -- (1.162,1.471) -- (1.163,1.434);
  \draw[black] (5.553,1.526) -- (5.551,1.519) -- (5.552,1.510) -- (5.556,1.502) -- (5.556,1.494) -- (5.557,1.487) -- (5.547,1.480) -- (5.553,1.472) -- (5.541,1.465) -- (5.546,1.457) -- (5.542,1.450) -- (5.557,1.442) -- (5.544,1.434);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,1.301) {S};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,2.632) {T};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,1.301) {C};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {9,8 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 (5.288,1.443) {13°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{SCT}\)) = \(\dfrac{ST}{TC}\)
\par\addvspace{.5em}
 D'où sin(13\textdegree{}) = \(\dfrac{9{,}8}{TC}\)
\par\addvspace{.5em}
On a donc TC =\(\dfrac{9{,}8}{\sin(13^{\circ})}\) \(\approx\) 43,6 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 3}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,2.500) -- (1.040,2.376) -- (1.038,2.250) -- (1.057,2.123) -- (1.057,1.998) -- (1.051,1.873) -- (1.051,1.747) -- (1.039,1.626) -- (1.050,1.500);
  \draw[black] (1.050,1.500) -- (1.402,1.512) -- (1.748,1.508) -- (2.099,1.508) -- (2.453,1.496) -- (2.801,1.500) -- (3.147,1.502) -- (3.502,1.502) -- (3.847,1.491) -- (4.203,1.505) -- (4.549,1.505) -- (4.899,1.502) -- (5.247,1.497) -- (5.603,1.495) -- (5.950,1.500);
  \draw[black] (5.950,1.500) -- (5.619,1.555) -- (5.300,1.638) -- (4.973,1.709) -- (4.646,1.766) -- (4.317,1.825) -- (3.987,1.896) -- (3.662,1.968) -- (3.336,2.042) -- (3.009,2.097) -- (2.685,2.169) -- (2.359,2.228) -- (2.030,2.299) -- (1.703,2.356) -- (1.374,2.429) -- (1.050,2.500);
  \draw[black] (1.050,1.600) -- (1.054,1.568) -- (1.047,1.534) -- (1.050,1.500);
  \draw[black] (1.050,1.500) -- (1.083,1.505) -- (1.116,1.502) -- (1.150,1.500);
  \draw[black] (1.050,1.600) -- (1.084,1.604) -- (1.115,1.600) -- (1.150,1.600);
  \draw[black] (1.150,1.600) -- (1.147,1.568) -- (1.155,1.534) -- (1.150,1.500);
  \draw[black] (5.563,1.579) -- (5.552,1.574) -- (5.548,1.568) -- (5.553,1.560) -- (5.550,1.554) -- (5.556,1.547) -- (5.559,1.540) -- (5.544,1.534) -- (5.551,1.527) -- (5.548,1.520) -- (5.557,1.513) -- (5.551,1.507) -- (5.546,1.500);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,1.367) {H};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,2.567) {Z};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,1.367) {S};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.550,1.933) {1 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (3.500,1.167) {4,9 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.287,1.500) {?};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
tan(\(\widehat{HSZ}\)) = \(\dfrac{HZ}{HS}\)
\par\addvspace{.5em}
d'où  tan(\(\widehat{HSZ}\)) =  \(\dfrac{1}{4{,}9}\)
\par\addvspace{.5em}
 On a donc \(\widehat{HSZ}\) =  ArcTan(\(\dfrac{1}{4{,}9}\)) \(\approx\) 12\textdegree{}
\needspace{4\baselineskip}
\subsection*{
Exercice 4}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,3.303) -- (1.048,2.979) -- (1.053,2.649) -- (1.051,2.324) -- (1.059,1.997) -- (1.059,1.676) -- (1.057,1.350) -- (1.052,1.021) -- (1.050,0.697);
  \draw[black] (1.050,0.697) -- (1.401,0.708) -- (1.751,0.707) -- (2.102,0.694) -- (2.448,0.693) -- (2.803,0.696) -- (3.150,0.687) -- (3.499,0.697) -- (3.848,0.703) -- (4.197,0.693) -- (4.551,0.705) -- (4.897,0.703) -- (5.251,0.709) -- (5.602,0.697) -- (5.950,0.697);
  \draw[black] (5.950,0.697) -- (5.641,0.857) -- (5.333,1.022) -- (5.031,1.181) -- (4.722,1.342) -- (4.423,1.521) -- (4.109,1.665) -- (3.807,1.843) -- (3.506,2.008) -- (3.191,2.159) -- (2.897,2.336) -- (2.579,2.488) -- (2.283,2.661) -- (1.966,2.804) -- (1.662,2.979) -- (1.356,3.140) -- (1.050,3.303);
  \draw[black] (1.050,0.958) -- (1.052,0.870) -- (1.051,0.784) -- (1.050,0.697);
  \draw[black] (1.050,0.697) -- (1.137,0.697) -- (1.223,0.693) -- (1.311,0.697);
  \draw[black] (1.050,0.958) -- (1.138,0.957) -- (1.223,0.959) -- (1.311,0.958);
  \draw[black] (1.311,0.958) -- (1.306,0.873) -- (1.316,0.783) -- (1.311,0.697);
  \draw[black] (5.593,0.887) -- (5.594,0.869) -- (5.581,0.857) -- (5.569,0.843) -- (5.579,0.823) -- (5.559,0.812) -- (5.569,0.792) -- (5.553,0.779) -- (5.550,0.763) -- (5.558,0.745) -- (5.556,0.730) -- (5.547,0.714) -- (5.550,0.697);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,0.564) {V};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,3.369) {J};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,0.564) {B};
  \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 (3.667,2.000) {5,4 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (5.303,0.792) {28°};
\end{tikzpicture}\end{center}


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

\par\addvspace{.5em}
sin(\(\widehat{VBJ}\)) = \(\dfrac{VJ}{JB}\)
\par\addvspace{.5em}
 D'où sin(28\textdegree{}) = \(\dfrac{VJ}{5{,}4}\)

On a donc VJ = 5,4 \(\times\) sin(28\textdegree{}) \(\approx\) 2,5 cm
\needspace{4\baselineskip}
\subsection*{
Exercice 5}

\begin{center}\begin{tikzpicture}[line width=0.6pt]
  \draw[black] (1.050,2.759) -- (1.062,2.571) -- (1.041,2.379) -- (1.037,2.187) -- (1.050,1.997) -- (1.047,1.809) -- (1.047,1.620) -- (1.057,1.430) -- (1.050,1.241);
  \draw[black] (1.050,1.241) -- (1.398,1.240) -- (1.753,1.235) -- (2.103,1.237) -- (2.451,1.238) -- (2.801,1.254) -- (3.152,1.230) -- (3.500,1.231) -- (3.850,1.243) -- (4.198,1.236) -- (4.552,1.240) -- (4.903,1.236) -- (5.249,1.234) -- (5.602,1.237) -- (5.950,1.241);
  \draw[black] (5.950,1.241) -- (5.622,1.341) -- (5.298,1.449) -- (4.974,1.558) -- (4.646,1.648) -- (4.315,1.736) -- (3.988,1.836) -- (3.663,1.956) -- (3.338,2.063) -- (3.007,2.152) -- (2.686,2.258) -- (2.360,2.359) -- (2.035,2.464) -- (1.704,2.556) -- (1.378,2.657) -- (1.050,2.759);
  \draw[black] (1.050,1.393) -- (1.054,1.341) -- (1.047,1.292) -- (1.050,1.241);
  \draw[black] (1.050,1.241) -- (1.100,1.247) -- (1.152,1.237) -- (1.202,1.241);
  \draw[black] (1.050,1.393) -- (1.100,1.391) -- (1.152,1.390) -- (1.202,1.393);
  \draw[black] (1.202,1.393) -- (1.205,1.342) -- (1.195,1.293) -- (1.202,1.241);
  \draw[black] (1.430,2.641) -- (1.420,2.600) -- (1.391,2.567) -- (1.371,2.531) -- (1.344,2.499) -- (1.316,2.468) -- (1.285,2.440) -- (1.255,2.408) -- (1.217,2.388) -- (1.177,2.373) -- (1.136,2.361) -- (1.091,2.369) -- (1.050,2.353);
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,1.108) {S};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (0.783,2.825) {M};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (6.050,1.108) {B};
  \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) {7,1 cm};
  \node[black, anchor=base, font=\fontsize{22.8}{27.3}\selectfont] at (1.446,2.155) {?};
\end{tikzpicture}\end{center}


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

cos(\(\widehat{SMB}\)) = \(\dfrac{SM}{MB}\)
\par\addvspace{.5em}
 d'où cos(\(\widehat{SMB}\)) = \(\dfrac{2{,}1}{7{,}1}\)
\par\addvspace{.5em}
On a donc \(\widehat{SMB}\) =ArcCos(\(\dfrac{2{,}1}{7{,}1}\)) \(\approx\)73\textdegree{}\end{document}
